कम से कम लागत फ़्लो सॉल्वर का इस्तेमाल करके, असाइनमेंट में होने वाली समस्या के बारे में ज़्यादा जानें.
दरअसल, कम से कम लागत फ़्लो अक्सर MIP या CP-SAT सॉल्वर. हालांकि, MIP और CP-SAT बड़े स्तर की समस्याओं को हल कर सकते हैं कम से कम लागत फ़्लो के लिए है, इसलिए ज़्यादातर मामलों में MIP या CP-SAT सबसे अच्छे विकल्प होते हैं.
नीचे दिए गए सेक्शन में ऐसे Python प्रोग्राम के बारे में बताया गया है जो इन समस्याओं को हल करते हैं कम से कम कॉस्ट फ़्लो सॉल्वर का इस्तेमाल करके, असाइनमेंट से जुड़े सवालों के जवाब दें:
- कम से कम लीनियर असाइनमेंट का उदाहरण.
- कर्मचारियों की टीम के साथ असाइनमेंट से जुड़ी समस्या.
लीनियर असाइनमेंट का उदाहरण
इस सेक्शन में, इस सेक्शन में बताए गए उदाहरण को हल करने का तरीका बताया गया है लीनियर असाइनमेंट सॉल्वर, एक मिनट के तौर पर लागत के फ़्लो से जुड़ी समस्या.
लाइब्रेरी इंपोर्ट करें
यहां दिया गया कोड ज़रूरी लाइब्रेरी इंपोर्ट करता है.
Python
from ortools.graph.python import min_cost_flow
C++
#include <cstdint> #include <vector> #include "ortools/graph/min_cost_flow.h"
Java
import com.google.ortools.Loader; import com.google.ortools.graph.MinCostFlow; import com.google.ortools.graph.MinCostFlowBase;
C#
using System; using Google.OrTools.Graph;
सॉल्वर के बारे में बताओ
यहां दिया गया कोड, कम से कम कॉस्ट फ़्लो सॉल्वर बनाता है.
Python
# Instantiate a SimpleMinCostFlow solver. smcf = min_cost_flow.SimpleMinCostFlow()
C++
// Instantiate a SimpleMinCostFlow solver. SimpleMinCostFlow min_cost_flow;
Java
// Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow();
C#
// Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow();
डेटा बनाना
सवाल के फ़्लो डायग्राम में, लागत के लिए दो हिस्सों वाला ग्राफ़ होता है मैट्रिक्स ( असाइनमेंट के बारे में खास जानकारी अलग उदाहरण), एक स्रोत और सिंक के साथ जोड़ा गया है.
डेटा में स्टार्ट नोड से जुड़े ये चार अरै होते हैं, समाप्ति नोड, क्षमता, और समस्या की लागत. हर कलेक्शन की लंबाई है आप ग्राफ़ में चाप की संख्या देख सकते हैं.
Python
# Define the directed graph for the flow. start_nodes = ( [0, 0, 0, 0] + [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4] + [5, 6, 7, 8] ) end_nodes = ( [1, 2, 3, 4] + [5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8] + [9, 9, 9, 9] ) capacities = ( [1, 1, 1, 1] + [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] + [1, 1, 1, 1] ) costs = ( [0, 0, 0, 0] + [90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115] + [0, 0, 0, 0] ) source = 0 sink = 9 tasks = 4 supplies = [tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks]
C++
// Define four parallel arrays: sources, destinations, capacities, // and unit costs between each pair. For instance, the arc from node 0 // to node 1 has a capacity of 15. const std::vector<int64_t> start_nodes = {0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8}; const std::vector<int64_t> end_nodes = {1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9}; const std::vector<int64_t> capacities = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}; const std::vector<int64_t> unit_costs = {0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 0, 0, 0, 0}; const int64_t source = 0; const int64_t sink = 9; const int64_t tasks = 4; // Define an array of supplies at each node. const std::vector<int64_t> supplies = {tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};
Java
// Define four parallel arrays: sources, destinations, capacities, and unit costs // between each pair. int[] startNodes = new int[] {0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8}; int[] endNodes = new int[] {1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9}; int[] capacities = new int[] {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}; int[] unitCosts = new int[] { 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 0, 0, 0, 0}; int source = 0; int sink = 9; int tasks = 4; // Define an array of supplies at each node. int[] supplies = new int[] {tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};
C#
// Define four parallel arrays: sources, destinations, capacities, and unit costs // between each pair. int[] startNodes = { 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8 }; int[] endNodes = { 1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9 }; int[] capacities = { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 }; int[] unitCosts = { 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 0, 0, 0, 0 }; int source = 0; int sink = 9; int tasks = 4; // Define an array of supplies at each node. int[] supplies = { tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks };
यह साफ़ तौर पर बताने के लिए कि डेटा को कैसे सेट अप किया गया है, हर अरे को तीन हिस्सों में बांटा गया है सब-कलेक्शन:
- पहला अरे, सोर्स से बाहर ले जाने वाले चापों से मेल खाता है.
- दूसरा अरे, वर्कर और टास्क के बीच के चापों से मेल खाता है.
costsके लिए, यह सिर्फ़ लागत का मैट्रिक्स (लीनियर असाइनमेंट सॉल्वर में इस्तेमाल किया जाता है), वेक्टर में फ़्लैट किया जाता है. - तीसरा अरे, सिंक की ओर ले जाने वाले चापों से मेल खाता है.
इस डेटा में वेक्टर supplies भी शामिल है, जो हर
नोड के लिए अलग-अलग हैं.
कम से कम लागत फ़्लो का सवाल, असाइनमेंट से जुड़े सवाल को कैसे दिखाता है
ऊपर दी गई कम से कम लागत फ़्लो की समस्या, असाइनमेंट से जुड़ी किसी समस्या को कैसे दिखाती है? सबसे पहले, हर चाप की धारिता 1 होती है. इसलिए, स्रोत पर 4 की क्षमता का बल होता है चार चापों में से 1 के फ़्लो के लिए मज़दूर को ले जाया जाता है.
इसके बाद, फ़्लो-इन-इसके बराबर-फ़्लो-आउट शर्त हर वर्कर को बाहर ले जाने के लिए मजबूर करती है होना है. अगर मुमकिन हो, तो सॉल्वर उस फ़्लो को कम से कम लागत में ले जाएगा जो हर कर्मचारी के बाहर रखा जा रहा है. हालांकि, सॉल्वर इन फ़्लो को डायरेक्ट नहीं कर सकता एक ही टास्क पर दो अलग-अलग लोग काम कर रहे हों. अगर ऐसा होता, तो कुल उस टास्क पर 2 का फ़्लो, जिसे एक आर्क में नहीं भेजा जा सकता सिंक तक की क्षमता 1. इसका मतलब है कि सॉल्वर सिर्फ़ एक वर्कर को टास्क असाइन कर सकता है, जैसे कि जो असाइनमेंट में पूछे गए सवाल के हिसाब से ज़रूरी होते हैं.
आखिर में, फ़्लो-इन-इसके बराबर-फ़्लो-आउट शर्त हर टास्क को 1 का आउटफ़्लो है, ताकि हर टास्क कुछ वर्कर ने पूरा किया हो.
ग्राफ़ और शर्तें बनाना
यह कोड, ग्राफ़ और कंस्ट्रेंट बनाता है.
Python
# Add each arc. for i in range(len(start_nodes)): smcf.add_arc_with_capacity_and_unit_cost( start_nodes[i], end_nodes[i], capacities[i], costs[i] ) # Add node supplies. for i in range(len(supplies)): smcf.set_node_supply(i, supplies[i])
C++
// Add each arc. for (int i = 0; i < start_nodes.size(); ++i) { int arc = min_cost_flow.AddArcWithCapacityAndUnitCost( start_nodes[i], end_nodes[i], capacities[i], unit_costs[i]); if (arc != i) LOG(FATAL) << "Internal error"; } // Add node supplies. for (int i = 0; i < supplies.size(); ++i) { min_cost_flow.SetNodeSupply(i, supplies[i]); }
Java
// Add each arc. for (int i = 0; i < startNodes.length; ++i) { int arc = minCostFlow.addArcWithCapacityAndUnitCost( startNodes[i], endNodes[i], capacities[i], unitCosts[i]); if (arc != i) { throw new Exception("Internal error"); } } // Add node supplies. for (int i = 0; i < supplies.length; ++i) { minCostFlow.setNodeSupply(i, supplies[i]); }
C#
// Add each arc. for (int i = 0; i < startNodes.Length; ++i) { int arc = minCostFlow.AddArcWithCapacityAndUnitCost(startNodes[i], endNodes[i], capacities[i], unitCosts[i]); if (arc != i) throw new Exception("Internal error"); } // Add node supplies. for (int i = 0; i < supplies.Length; ++i) { minCostFlow.SetNodeSupply(i, supplies[i]); }
सॉल्वर आज़माएं
यहां दिया गया कोड, सॉल्वर को शुरू करता है और समाधान दिखाता है.
Python
# Find the minimum cost flow between node 0 and node 10. status = smcf.solve()
C++
// Find the min cost flow. int status = min_cost_flow.Solve();
Java
// Find the min cost flow. MinCostFlowBase.Status status = minCostFlow.solve();
C#
// Find the min cost flow. MinCostFlow.Status status = minCostFlow.Solve();
समाधान में कर्मचारियों और असाइन किए गए टास्क के बीच चाप शामिल होते हैं सॉल्वर में से 1 का फ़्लो. (स्रोत या सिंक से जुड़े आर्क इसका हिस्सा नहीं हैं समाधान.)
प्रोग्राम हर आर्क की जांच करके देखता है कि उसका फ़्लो 1 है या नहीं. अगर हां, तो
आर्क का Tail (स्टार्ट नोड) और Head (आखिरी नोड) है, जो ए
असाइनमेंट में कर रहे लोग और टास्क जोड़ें.
प्रोग्राम का आउटपुट
Python
if status == smcf.OPTIMAL: print("Total cost = ", smcf.optimal_cost()) print() for arc in range(smcf.num_arcs()): # Can ignore arcs leading out of source or into sink. if smcf.tail(arc) != source and smcf.head(arc) != sink: # Arcs in the solution have a flow value of 1. Their start and end nodes # give an assignment of worker to task. if smcf.flow(arc) > 0: print( "Worker %d assigned to task %d. Cost = %d" % (smcf.tail(arc), smcf.head(arc), smcf.unit_cost(arc)) ) else: print("There was an issue with the min cost flow input.") print(f"Status: {status}")
C++
if (status == MinCostFlow::OPTIMAL) { LOG(INFO) << "Total cost: " << min_cost_flow.OptimalCost(); LOG(INFO) << ""; for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) { // Can ignore arcs leading out of source or into sink. if (min_cost_flow.Tail(i) != source && min_cost_flow.Head(i) != sink) { // Arcs in the solution have a flow value of 1. Their start and end // nodes give an assignment of worker to task. if (min_cost_flow.Flow(i) > 0) { LOG(INFO) << "Worker " << min_cost_flow.Tail(i) << " assigned to task " << min_cost_flow.Head(i) << " Cost: " << min_cost_flow.UnitCost(i); } } } } else { LOG(INFO) << "Solving the min cost flow problem failed."; LOG(INFO) << "Solver status: " << status; }
Java
if (status == MinCostFlow.Status.OPTIMAL) { System.out.println("Total cost: " + minCostFlow.getOptimalCost()); System.out.println(); for (int i = 0; i < minCostFlow.getNumArcs(); ++i) { // Can ignore arcs leading out of source or into sink. if (minCostFlow.getTail(i) != source && minCostFlow.getHead(i) != sink) { // Arcs in the solution have a flow value of 1. Their start and end nodes // give an assignment of worker to task. if (minCostFlow.getFlow(i) > 0) { System.out.println("Worker " + minCostFlow.getTail(i) + " assigned to task " + minCostFlow.getHead(i) + " Cost: " + minCostFlow.getUnitCost(i)); } } } } else { System.out.println("Solving the min cost flow problem failed."); System.out.println("Solver status: " + status); }
C#
if (status == MinCostFlow.Status.OPTIMAL) { Console.WriteLine("Total cost: " + minCostFlow.OptimalCost()); Console.WriteLine(""); for (int i = 0; i < minCostFlow.NumArcs(); ++i) { // Can ignore arcs leading out of source or into sink. if (minCostFlow.Tail(i) != source && minCostFlow.Head(i) != sink) { // Arcs in the solution have a flow value of 1. Their start and end nodes // give an assignment of worker to task. if (minCostFlow.Flow(i) > 0) { Console.WriteLine("Worker " + minCostFlow.Tail(i) + " assigned to task " + minCostFlow.Head(i) + " Cost: " + minCostFlow.UnitCost(i)); } } } } else { Console.WriteLine("Solving the min cost flow problem failed."); Console.WriteLine("Solver status: " + status); }
ये रहे प्रोग्राम का आउटपुट.
Total cost = 265 Worker 1 assigned to task 8. Cost = 70 Worker 2 assigned to task 7. Cost = 55 Worker 3 assigned to task 6. Cost = 95 Worker 4 assigned to task 5. Cost = 45 Time = 0.000245 seconds
नतीजा वैसा ही होता है, जैसा कि लीनियर असाइनमेंट सॉल्वर ( कर्मचारियों की संख्या और लागत). लीनियर असाइनमेंट सॉल्वर थोड़ा तेज़ काम करता है कम से कम लागत फ़्लो से ज़्यादा — 0.000147 सेकंड बनाम 0.000458 सेकंड.
पूरा प्रोग्राम
पूरा प्रोग्राम नीचे दिखाया गया है.
Python
"""Linear assignment example.""" from ortools.graph.python import min_cost_flow def main(): """Solving an Assignment Problem with MinCostFlow.""" # Instantiate a SimpleMinCostFlow solver. smcf = min_cost_flow.SimpleMinCostFlow() # Define the directed graph for the flow. start_nodes = ( [0, 0, 0, 0] + [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4] + [5, 6, 7, 8] ) end_nodes = ( [1, 2, 3, 4] + [5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8] + [9, 9, 9, 9] ) capacities = ( [1, 1, 1, 1] + [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] + [1, 1, 1, 1] ) costs = ( [0, 0, 0, 0] + [90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115] + [0, 0, 0, 0] ) source = 0 sink = 9 tasks = 4 supplies = [tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks] # Add each arc. for i in range(len(start_nodes)): smcf.add_arc_with_capacity_and_unit_cost( start_nodes[i], end_nodes[i], capacities[i], costs[i] ) # Add node supplies. for i in range(len(supplies)): smcf.set_node_supply(i, supplies[i]) # Find the minimum cost flow between node 0 and node 10. status = smcf.solve() if status == smcf.OPTIMAL: print("Total cost = ", smcf.optimal_cost()) print() for arc in range(smcf.num_arcs()): # Can ignore arcs leading out of source or into sink. if smcf.tail(arc) != source and smcf.head(arc) != sink: # Arcs in the solution have a flow value of 1. Their start and end nodes # give an assignment of worker to task. if smcf.flow(arc) > 0: print( "Worker %d assigned to task %d. Cost = %d" % (smcf.tail(arc), smcf.head(arc), smcf.unit_cost(arc)) ) else: print("There was an issue with the min cost flow input.") print(f"Status: {status}") if __name__ == "__main__": main()
C++
#include <cstdint> #include <vector> #include "ortools/graph/min_cost_flow.h" namespace operations_research { // MinCostFlow simple interface example. void AssignmentMinFlow() { // Instantiate a SimpleMinCostFlow solver. SimpleMinCostFlow min_cost_flow; // Define four parallel arrays: sources, destinations, capacities, // and unit costs between each pair. For instance, the arc from node 0 // to node 1 has a capacity of 15. const std::vector<int64_t> start_nodes = {0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8}; const std::vector<int64_t> end_nodes = {1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9}; const std::vector<int64_t> capacities = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}; const std::vector<int64_t> unit_costs = {0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 0, 0, 0, 0}; const int64_t source = 0; const int64_t sink = 9; const int64_t tasks = 4; // Define an array of supplies at each node. const std::vector<int64_t> supplies = {tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks}; // Add each arc. for (int i = 0; i < start_nodes.size(); ++i) { int arc = min_cost_flow.AddArcWithCapacityAndUnitCost( start_nodes[i], end_nodes[i], capacities[i], unit_costs[i]); if (arc != i) LOG(FATAL) << "Internal error"; } // Add node supplies. for (int i = 0; i < supplies.size(); ++i) { min_cost_flow.SetNodeSupply(i, supplies[i]); } // Find the min cost flow. int status = min_cost_flow.Solve(); if (status == MinCostFlow::OPTIMAL) { LOG(INFO) << "Total cost: " << min_cost_flow.OptimalCost(); LOG(INFO) << ""; for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) { // Can ignore arcs leading out of source or into sink. if (min_cost_flow.Tail(i) != source && min_cost_flow.Head(i) != sink) { // Arcs in the solution have a flow value of 1. Their start and end // nodes give an assignment of worker to task. if (min_cost_flow.Flow(i) > 0) { LOG(INFO) << "Worker " << min_cost_flow.Tail(i) << " assigned to task " << min_cost_flow.Head(i) << " Cost: " << min_cost_flow.UnitCost(i); } } } } else { LOG(INFO) << "Solving the min cost flow problem failed."; LOG(INFO) << "Solver status: " << status; } } } // namespace operations_research int main() { operations_research::AssignmentMinFlow(); return EXIT_SUCCESS; }
Java
package com.google.ortools.graph.samples; import com.google.ortools.Loader; import com.google.ortools.graph.MinCostFlow; import com.google.ortools.graph.MinCostFlowBase; /** Minimal Assignment Min Flow. */ public class AssignmentMinFlow { public static void main(String[] args) throws Exception { Loader.loadNativeLibraries(); // Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow(); // Define four parallel arrays: sources, destinations, capacities, and unit costs // between each pair. int[] startNodes = new int[] {0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8}; int[] endNodes = new int[] {1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9}; int[] capacities = new int[] {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}; int[] unitCosts = new int[] { 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 0, 0, 0, 0}; int source = 0; int sink = 9; int tasks = 4; // Define an array of supplies at each node. int[] supplies = new int[] {tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks}; // Add each arc. for (int i = 0; i < startNodes.length; ++i) { int arc = minCostFlow.addArcWithCapacityAndUnitCost( startNodes[i], endNodes[i], capacities[i], unitCosts[i]); if (arc != i) { throw new Exception("Internal error"); } } // Add node supplies. for (int i = 0; i < supplies.length; ++i) { minCostFlow.setNodeSupply(i, supplies[i]); } // Find the min cost flow. MinCostFlowBase.Status status = minCostFlow.solve(); if (status == MinCostFlow.Status.OPTIMAL) { System.out.println("Total cost: " + minCostFlow.getOptimalCost()); System.out.println(); for (int i = 0; i < minCostFlow.getNumArcs(); ++i) { // Can ignore arcs leading out of source or into sink. if (minCostFlow.getTail(i) != source && minCostFlow.getHead(i) != sink) { // Arcs in the solution have a flow value of 1. Their start and end nodes // give an assignment of worker to task. if (minCostFlow.getFlow(i) > 0) { System.out.println("Worker " + minCostFlow.getTail(i) + " assigned to task " + minCostFlow.getHead(i) + " Cost: " + minCostFlow.getUnitCost(i)); } } } } else { System.out.println("Solving the min cost flow problem failed."); System.out.println("Solver status: " + status); } } private AssignmentMinFlow() {} }
C#
using System; using Google.OrTools.Graph; public class AssignmentMinFlow { static void Main() { // Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow(); // Define four parallel arrays: sources, destinations, capacities, and unit costs // between each pair. int[] startNodes = { 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8 }; int[] endNodes = { 1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9 }; int[] capacities = { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 }; int[] unitCosts = { 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 0, 0, 0, 0 }; int source = 0; int sink = 9; int tasks = 4; // Define an array of supplies at each node. int[] supplies = { tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks }; // Add each arc. for (int i = 0; i < startNodes.Length; ++i) { int arc = minCostFlow.AddArcWithCapacityAndUnitCost(startNodes[i], endNodes[i], capacities[i], unitCosts[i]); if (arc != i) throw new Exception("Internal error"); } // Add node supplies. for (int i = 0; i < supplies.Length; ++i) { minCostFlow.SetNodeSupply(i, supplies[i]); } // Find the min cost flow. MinCostFlow.Status status = minCostFlow.Solve(); if (status == MinCostFlow.Status.OPTIMAL) { Console.WriteLine("Total cost: " + minCostFlow.OptimalCost()); Console.WriteLine(""); for (int i = 0; i < minCostFlow.NumArcs(); ++i) { // Can ignore arcs leading out of source or into sink. if (minCostFlow.Tail(i) != source && minCostFlow.Head(i) != sink) { // Arcs in the solution have a flow value of 1. Their start and end nodes // give an assignment of worker to task. if (minCostFlow.Flow(i) > 0) { Console.WriteLine("Worker " + minCostFlow.Tail(i) + " assigned to task " + minCostFlow.Head(i) + " Cost: " + minCostFlow.UnitCost(i)); } } } } else { Console.WriteLine("Solving the min cost flow problem failed."); Console.WriteLine("Solver status: " + status); } } }
कर्मचारियों की टीम के साथ असाइनमेंट
इस सेक्शन में, असाइनमेंट से जुड़ी एक और सामान्य समस्या के बारे में बताया गया है. इस सवाल के लिए, छह कर्मचारियों को दो टीमों में बांटा जाता है. समस्या यह है कि इसलिए, ताकि सभी टीमों के बीच वर्कलोड समान रूप से संतुलित हो जाए — इसलिए, हर टीम दो काम करती है.
इस सवाल का MIP सॉल्वर समाधान देखने के लिए देखें टीम के सदस्यों के साथ असाइनमेंट.
नीचे दिए गए सेक्शन एक ऐसे प्रोग्राम के बारे में बताते हैं जो कम से कम वैल्यू का इस्तेमाल करके सवाल को हल करता है कॉस्ट फ़्लो सॉल्वर.
लाइब्रेरी इंपोर्ट करें
यहां दिया गया कोड ज़रूरी लाइब्रेरी इंपोर्ट करता है.
Python
from ortools.graph.python import min_cost_flow
C++
#include <cstdint> #include <vector> #include "ortools/graph/min_cost_flow.h"
Java
import com.google.ortools.Loader; import com.google.ortools.graph.MinCostFlow; import com.google.ortools.graph.MinCostFlowBase;
C#
using System; using Google.OrTools.Graph;
सॉल्वर के बारे में बताओ
यहां दिया गया कोड, कम से कम कॉस्ट फ़्लो सॉल्वर बनाता है.
Python
smcf = min_cost_flow.SimpleMinCostFlow()
C++
// Instantiate a SimpleMinCostFlow solver. SimpleMinCostFlow min_cost_flow;
Java
// Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow();
C#
// Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow();
डेटा बनाना
यहां दिया गया कोड, प्रोग्राम के लिए डेटा बनाता है.
Python
# Define the directed graph for the flow. team_a = [1, 3, 5] team_b = [2, 4, 6] start_nodes = ( # fmt: off [0, 0] + [11, 11, 11] + [12, 12, 12] + [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6] + [7, 8, 9, 10] # fmt: on ) end_nodes = ( # fmt: off [11, 12] + team_a + team_b + [7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10] + [13, 13, 13, 13] # fmt: on ) capacities = ( # fmt: off [2, 2] + [1, 1, 1] + [1, 1, 1] + [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] + [1, 1, 1, 1] # fmt: on ) costs = ( # fmt: off [0, 0] + [0, 0, 0] + [0, 0, 0] + [90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95] + [0, 0, 0, 0] # fmt: on ) source = 0 sink = 13 tasks = 4 # Define an array of supplies at each node. supplies = [tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks]
C++
// Define the directed graph for the flow. const std::vector<int64_t> team_A = {1, 3, 5}; const std::vector<int64_t> team_B = {2, 4, 6}; const std::vector<int64_t> start_nodes = { 0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10}; const std::vector<int64_t> end_nodes = { 11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13}; const std::vector<int64_t> capacities = {2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}; const std::vector<int64_t> unit_costs = { 0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0}; const int64_t source = 0; const int64_t sink = 13; const int64_t tasks = 4; // Define an array of supplies at each node. const std::vector<int64_t> supplies = {tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};
Java
// Define the directed graph for the flow. // int[] teamA = new int[] {1, 3, 5}; // int[] teamB = new int[] {2, 4, 6}; int[] startNodes = new int[] {0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10}; int[] endNodes = new int[] {11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13}; int[] capacities = new int[] {2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}; int[] unitCosts = new int[] {0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0}; int source = 0; int sink = 13; int tasks = 4; // Define an array of supplies at each node. int[] supplies = new int[] {tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};
C#
// Define the directed graph for the flow. int[] teamA = { 1, 3, 5 }; int[] teamB = { 2, 4, 6 }; // Define four parallel arrays: sources, destinations, capacities, and unit costs // between each pair. int[] startNodes = { 0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10 }; int[] endNodes = { 11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13 }; int[] capacities = { 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 }; int[] unitCosts = { 0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0 }; int source = 0; int sink = 13; int tasks = 4; // Define an array of supplies at each node. int[] supplies = { tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks };
वर्कर, नोड 1 - 6 के बारे में हैं. टीम A में 1, 3, और 5 कर्मचारी शामिल हैं. और टीम B में 2, 4, और 6 कर्मचारी हैं. टास्क की संख्या 7 से लेकर 10 तक है.
सोर्स और वर्कर के बीच, दो नए नोड 11 और 12 हैं. नोड 11 टीम A के लिए नोड से कनेक्ट होता है और नोड 12, टीम B, जिसमें चापों की क्षमता 1 है. नीचे दिया गया ग्राफ़, सोर्स से लेकर वर्कर तक के सिर्फ़ नोड और चाप दिखाता है.
वर्कलोड के बीच संतुलन बनाने के लिए सबसे ज़रूरी है कि सोर्स 0, नोड 11 से जुड़ा हो और 12 x की आर्क ऑफ़ कपैसिटी 2. इसका मतलब है कि नोड 11 और 12 (इसलिए, जो टीम A और B) का फ़्लो ज़्यादा से ज़्यादा दो हो सकता है. नतीजे के तौर पर, हर टीम ज़्यादा से ज़्यादा दो काम कर सकती है.
कंस्ट्रेंट बनाएं
Python
# Add each arc. for i in range(0, len(start_nodes)): smcf.add_arc_with_capacity_and_unit_cost( start_nodes[i], end_nodes[i], capacities[i], costs[i] ) # Add node supplies. for i in range(0, len(supplies)): smcf.set_node_supply(i, supplies[i])
C++
// Add each arc. for (int i = 0; i < start_nodes.size(); ++i) { int arc = min_cost_flow.AddArcWithCapacityAndUnitCost( start_nodes[i], end_nodes[i], capacities[i], unit_costs[i]); if (arc != i) LOG(FATAL) << "Internal error"; } // Add node supplies. for (int i = 0; i < supplies.size(); ++i) { min_cost_flow.SetNodeSupply(i, supplies[i]); }
Java
// Add each arc. for (int i = 0; i < startNodes.length; ++i) { int arc = minCostFlow.addArcWithCapacityAndUnitCost( startNodes[i], endNodes[i], capacities[i], unitCosts[i]); if (arc != i) { throw new Exception("Internal error"); } } // Add node supplies. for (int i = 0; i < supplies.length; ++i) { minCostFlow.setNodeSupply(i, supplies[i]); }
C#
// Add each arc. for (int i = 0; i < startNodes.Length; ++i) { int arc = minCostFlow.AddArcWithCapacityAndUnitCost(startNodes[i], endNodes[i], capacities[i], unitCosts[i]); if (arc != i) throw new Exception("Internal error"); } // Add node supplies. for (int i = 0; i < supplies.Length; ++i) { minCostFlow.SetNodeSupply(i, supplies[i]); }
सॉल्वर आज़माएं
Python
# Find the minimum cost flow between node 0 and node 10. status = smcf.solve()
C++
// Find the min cost flow. int status = min_cost_flow.Solve();
Java
// Find the min cost flow. MinCostFlowBase.Status status = minCostFlow.solve();
C#
// Find the min cost flow. MinCostFlow.Status status = minCostFlow.Solve();
प्रोग्राम का आउटपुट
Python
if status == smcf.OPTIMAL: print("Total cost = ", smcf.optimal_cost()) print() for arc in range(smcf.num_arcs()): # Can ignore arcs leading out of source or intermediate, or into sink. if ( smcf.tail(arc) != source and smcf.tail(arc) != 11 and smcf.tail(arc) != 12 and smcf.head(arc) != sink ): # Arcs in the solution will have a flow value of 1. # There start and end nodes give an assignment of worker to task. if smcf.flow(arc) > 0: print( "Worker %d assigned to task %d. Cost = %d" % (smcf.tail(arc), smcf.head(arc), smcf.unit_cost(arc)) ) else: print("There was an issue with the min cost flow input.") print(f"Status: {status}")
C++
if (status == MinCostFlow::OPTIMAL) { LOG(INFO) << "Total cost: " << min_cost_flow.OptimalCost(); LOG(INFO) << ""; for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) { // Can ignore arcs leading out of source or intermediate nodes, or into // sink. if (min_cost_flow.Tail(i) != source && min_cost_flow.Tail(i) != 11 && min_cost_flow.Tail(i) != 12 && min_cost_flow.Head(i) != sink) { // Arcs in the solution have a flow value of 1. Their start and end // nodes give an assignment of worker to task. if (min_cost_flow.Flow(i) > 0) { LOG(INFO) << "Worker " << min_cost_flow.Tail(i) << " assigned to task " << min_cost_flow.Head(i) << " Cost: " << min_cost_flow.UnitCost(i); } } } } else { LOG(INFO) << "Solving the min cost flow problem failed."; LOG(INFO) << "Solver status: " << status; }
Java
if (status == MinCostFlow.Status.OPTIMAL) { System.out.println("Total cost: " + minCostFlow.getOptimalCost()); System.out.println(); for (int i = 0; i < minCostFlow.getNumArcs(); ++i) { // Can ignore arcs leading out of source or intermediate nodes, or into sink. if (minCostFlow.getTail(i) != source && minCostFlow.getTail(i) != 11 && minCostFlow.getTail(i) != 12 && minCostFlow.getHead(i) != sink) { // Arcs in the solution have a flow value of 1. Their start and end nodes // give an assignment of worker to task. if (minCostFlow.getFlow(i) > 0) { System.out.println("Worker " + minCostFlow.getTail(i) + " assigned to task " + minCostFlow.getHead(i) + " Cost: " + minCostFlow.getUnitCost(i)); } } } } else { System.out.println("Solving the min cost flow problem failed."); System.out.println("Solver status: " + status); }
C#
if (status == MinCostFlow.Status.OPTIMAL) { Console.WriteLine("Total cost: " + minCostFlow.OptimalCost()); Console.WriteLine(""); for (int i = 0; i < minCostFlow.NumArcs(); ++i) { // Can ignore arcs leading out of source or into sink. if (minCostFlow.Tail(i) != source && minCostFlow.Tail(i) != 11 && minCostFlow.Tail(i) != 12 && minCostFlow.Head(i) != sink) { // Arcs in the solution have a flow value of 1. Their start and end nodes // give an assignment of worker to task. if (minCostFlow.Flow(i) > 0) { Console.WriteLine("Worker " + minCostFlow.Tail(i) + " assigned to task " + minCostFlow.Head(i) + " Cost: " + minCostFlow.UnitCost(i)); } } } } else { Console.WriteLine("Solving the min cost flow problem failed."); Console.WriteLine("Solver status: " + status); }
प्रोग्राम का आउटपुट नीचे दिखता है.
Total cost = 250 Worker 1 assigned to task 9. Cost = 75 Worker 2 assigned to task 7. Cost = 35 Worker 5 assigned to task 10. Cost = 75 Worker 6 assigned to task 8. Cost = 65 Time = 0.00031 seconds
टीम A को टास्क 9 और 10 असाइन किए गए हैं. वहीं, टीम B को टास्क 7 और 8 असाइन किए गए हैं.
ध्यान दें कि इस सवाल के लिए कम से कम कॉस्ट फ़्लो सॉल्वर, एमआईपी सॉल्वर, जिसमें करीब 0.006 सेकंड लगते हैं.
पूरा प्रोग्राम
पूरा प्रोग्राम नीचे दिखाया गया है.
Python
"""Assignment with teams of workers.""" from ortools.graph.python import min_cost_flow def main(): """Solving an Assignment with teams of worker.""" smcf = min_cost_flow.SimpleMinCostFlow() # Define the directed graph for the flow. team_a = [1, 3, 5] team_b = [2, 4, 6] start_nodes = ( # fmt: off [0, 0] + [11, 11, 11] + [12, 12, 12] + [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6] + [7, 8, 9, 10] # fmt: on ) end_nodes = ( # fmt: off [11, 12] + team_a + team_b + [7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10] + [13, 13, 13, 13] # fmt: on ) capacities = ( # fmt: off [2, 2] + [1, 1, 1] + [1, 1, 1] + [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] + [1, 1, 1, 1] # fmt: on ) costs = ( # fmt: off [0, 0] + [0, 0, 0] + [0, 0, 0] + [90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95] + [0, 0, 0, 0] # fmt: on ) source = 0 sink = 13 tasks = 4 # Define an array of supplies at each node. supplies = [tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks] # Add each arc. for i in range(0, len(start_nodes)): smcf.add_arc_with_capacity_and_unit_cost( start_nodes[i], end_nodes[i], capacities[i], costs[i] ) # Add node supplies. for i in range(0, len(supplies)): smcf.set_node_supply(i, supplies[i]) # Find the minimum cost flow between node 0 and node 10. status = smcf.solve() if status == smcf.OPTIMAL: print("Total cost = ", smcf.optimal_cost()) print() for arc in range(smcf.num_arcs()): # Can ignore arcs leading out of source or intermediate, or into sink. if ( smcf.tail(arc) != source and smcf.tail(arc) != 11 and smcf.tail(arc) != 12 and smcf.head(arc) != sink ): # Arcs in the solution will have a flow value of 1. # There start and end nodes give an assignment of worker to task. if smcf.flow(arc) > 0: print( "Worker %d assigned to task %d. Cost = %d" % (smcf.tail(arc), smcf.head(arc), smcf.unit_cost(arc)) ) else: print("There was an issue with the min cost flow input.") print(f"Status: {status}") if __name__ == "__main__": main()
C++
#include <cstdint> #include <vector> #include "ortools/graph/min_cost_flow.h" namespace operations_research { // MinCostFlow simple interface example. void BalanceMinFlow() { // Instantiate a SimpleMinCostFlow solver. SimpleMinCostFlow min_cost_flow; // Define the directed graph for the flow. const std::vector<int64_t> team_A = {1, 3, 5}; const std::vector<int64_t> team_B = {2, 4, 6}; const std::vector<int64_t> start_nodes = { 0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10}; const std::vector<int64_t> end_nodes = { 11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13}; const std::vector<int64_t> capacities = {2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}; const std::vector<int64_t> unit_costs = { 0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0}; const int64_t source = 0; const int64_t sink = 13; const int64_t tasks = 4; // Define an array of supplies at each node. const std::vector<int64_t> supplies = {tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks}; // Add each arc. for (int i = 0; i < start_nodes.size(); ++i) { int arc = min_cost_flow.AddArcWithCapacityAndUnitCost( start_nodes[i], end_nodes[i], capacities[i], unit_costs[i]); if (arc != i) LOG(FATAL) << "Internal error"; } // Add node supplies. for (int i = 0; i < supplies.size(); ++i) { min_cost_flow.SetNodeSupply(i, supplies[i]); } // Find the min cost flow. int status = min_cost_flow.Solve(); if (status == MinCostFlow::OPTIMAL) { LOG(INFO) << "Total cost: " << min_cost_flow.OptimalCost(); LOG(INFO) << ""; for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) { // Can ignore arcs leading out of source or intermediate nodes, or into // sink. if (min_cost_flow.Tail(i) != source && min_cost_flow.Tail(i) != 11 && min_cost_flow.Tail(i) != 12 && min_cost_flow.Head(i) != sink) { // Arcs in the solution have a flow value of 1. Their start and end // nodes give an assignment of worker to task. if (min_cost_flow.Flow(i) > 0) { LOG(INFO) << "Worker " << min_cost_flow.Tail(i) << " assigned to task " << min_cost_flow.Head(i) << " Cost: " << min_cost_flow.UnitCost(i); } } } } else { LOG(INFO) << "Solving the min cost flow problem failed."; LOG(INFO) << "Solver status: " << status; } } } // namespace operations_research int main() { operations_research::BalanceMinFlow(); return EXIT_SUCCESS; }
Java
package com.google.ortools.graph.samples; import com.google.ortools.Loader; import com.google.ortools.graph.MinCostFlow; import com.google.ortools.graph.MinCostFlowBase; /** Minimal Assignment Min Flow. */ public class BalanceMinFlow { public static void main(String[] args) throws Exception { Loader.loadNativeLibraries(); // Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow(); // Define the directed graph for the flow. // int[] teamA = new int[] {1, 3, 5}; // int[] teamB = new int[] {2, 4, 6}; int[] startNodes = new int[] {0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10}; int[] endNodes = new int[] {11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13}; int[] capacities = new int[] {2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}; int[] unitCosts = new int[] {0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0}; int source = 0; int sink = 13; int tasks = 4; // Define an array of supplies at each node. int[] supplies = new int[] {tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks}; // Add each arc. for (int i = 0; i < startNodes.length; ++i) { int arc = minCostFlow.addArcWithCapacityAndUnitCost( startNodes[i], endNodes[i], capacities[i], unitCosts[i]); if (arc != i) { throw new Exception("Internal error"); } } // Add node supplies. for (int i = 0; i < supplies.length; ++i) { minCostFlow.setNodeSupply(i, supplies[i]); } // Find the min cost flow. MinCostFlowBase.Status status = minCostFlow.solve(); if (status == MinCostFlow.Status.OPTIMAL) { System.out.println("Total cost: " + minCostFlow.getOptimalCost()); System.out.println(); for (int i = 0; i < minCostFlow.getNumArcs(); ++i) { // Can ignore arcs leading out of source or intermediate nodes, or into sink. if (minCostFlow.getTail(i) != source && minCostFlow.getTail(i) != 11 && minCostFlow.getTail(i) != 12 && minCostFlow.getHead(i) != sink) { // Arcs in the solution have a flow value of 1. Their start and end nodes // give an assignment of worker to task. if (minCostFlow.getFlow(i) > 0) { System.out.println("Worker " + minCostFlow.getTail(i) + " assigned to task " + minCostFlow.getHead(i) + " Cost: " + minCostFlow.getUnitCost(i)); } } } } else { System.out.println("Solving the min cost flow problem failed."); System.out.println("Solver status: " + status); } } private BalanceMinFlow() {} }
C#
using System; using Google.OrTools.Graph; public class BalanceMinFlow { static void Main() { // Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow(); // Define the directed graph for the flow. int[] teamA = { 1, 3, 5 }; int[] teamB = { 2, 4, 6 }; // Define four parallel arrays: sources, destinations, capacities, and unit costs // between each pair. int[] startNodes = { 0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10 }; int[] endNodes = { 11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13 }; int[] capacities = { 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 }; int[] unitCosts = { 0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0 }; int source = 0; int sink = 13; int tasks = 4; // Define an array of supplies at each node. int[] supplies = { tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks }; // Add each arc. for (int i = 0; i < startNodes.Length; ++i) { int arc = minCostFlow.AddArcWithCapacityAndUnitCost(startNodes[i], endNodes[i], capacities[i], unitCosts[i]); if (arc != i) throw new Exception("Internal error"); } // Add node supplies. for (int i = 0; i < supplies.Length; ++i) { minCostFlow.SetNodeSupply(i, supplies[i]); } // Find the min cost flow. MinCostFlow.Status status = minCostFlow.Solve(); if (status == MinCostFlow.Status.OPTIMAL) { Console.WriteLine("Total cost: " + minCostFlow.OptimalCost()); Console.WriteLine(""); for (int i = 0; i < minCostFlow.NumArcs(); ++i) { // Can ignore arcs leading out of source or into sink. if (minCostFlow.Tail(i) != source && minCostFlow.Tail(i) != 11 && minCostFlow.Tail(i) != 12 && minCostFlow.Head(i) != sink) { // Arcs in the solution have a flow value of 1. Their start and end nodes // give an assignment of worker to task. if (minCostFlow.Flow(i) > 0) { Console.WriteLine("Worker " + minCostFlow.Tail(i) + " assigned to task " + minCostFlow.Head(i) + " Cost: " + minCostFlow.UnitCost(i)); } } } } else { Console.WriteLine("Solving the min cost flow problem failed."); Console.WriteLine("Solver status: " + status); } } }