python调用gurobi求解最短路问题
文章目录
- 最短路问题Shortest Path Problem
- 一、符号说明
- 二、数学模型
- 三、python 调用gurobi求解
-
- 问题数据
- 代码
- 结果展示
最短路问题Shortest Path Problem
最短路径问题是图论研究中的一个经典算法问题,旨在寻找图(由结点和路径组成的)中两结点之间的最短路径。
copytright@baidu
一、符号说明
| 符号 | 说明 |
|---|---|
| c i j c_{ij} cij | 弧ij上的权重，可以理解为距离，流量，时间等 |
| x i j x_{ij} xij | 弧ij是否被选择通过，是binary变量 |
| V V V | 图中的点集合 |
二、数学模型
min ∑ ( i , j ) ∈ A c i j x i j ∑ ( j , i ) ∈ A x i j − ∑ ( i , j ) ∈ A x j i = b i , ∀ i ∈ V , b i = { − 1 , i f i = s , 0 , i f i ≠ s a n d i ≠ t , 1 , i f i = t , \min \sum_{\left( i,j \right) \in A}{c_{ij}x_{ij}} \\ \sum_{\left( j,i \right) \in A}{x_{ij}}-\sum_{\left( i,j \right) \in A}{x_{ji}}=b_i,\forall i\in V, \\ b_i=\begin{cases} -1, if\,\,i=s,\\ 0, if\,\,i\ne s\,\,and\,\,i\ne t,\\ 1, if\,\,i=t,\\ \end{cases} min(i,j)∈A∑cijxij(j,i)∈A∑xij−(i,j)∈A∑xji=bi,∀i∈V,bi=⎩⎪⎨⎪⎧−1,ifi=s,0,ifi=sandi=t,1,ifi=t,
三、python 调用gurobi求解
问题数据
这里我们使用距离矩阵来表示图。
copyright@https://jingyan.baidu.com/article/359911f58636ad57ff030645.html
代码
代码如下（示例）：
# _*_coding:utf-8 _*_
from __future__ import print_function
from __future__ import division, print_function
from gurobipy import *
import numpy as np
import copy
import time
starttime = time.time()
#返回最优的路线
def reportMIP(model, Routes):
if model.status == GRB.OPTIMAL:
print("Best MIP Solution: ", model.objVal, "\n")
var = model.getVars()
for i in range(model.numVars):
if (var[i].x > 0):
print(var[i].varName, " = ", var[i].x)
print("Optimal route:", Routes[i])
def getValue(var_dict, nodeNum):
x_value = np.zeros([nodeNum, nodeNum])
for key in var_dict.keys():
a = key[0]
b = key[1]
x_value[a][b] = var_dict[key].x
return x_value
def getRoute(x_value):
x = copy.deepcopy(x_value)
# route_temp.append(0)
previousPoint = 0
route_temp = [previousPoint]
count = 0
while (count != len(x_value)-1):
# print('previousPoint: ', previousPoint )
if (x[previousPoint][count] > 0):
previousPoint = count
route_temp.append(previousPoint)
count = 0
continue
else:
count += 1
route_temp.append(len(x_value)-1)
return route_temp
if __name__ == "__main__":
nodeNum =11
#距离矩阵
cost =[[0,2,8,1,1000,1000,1000,1000,1000,1000,1000],[2,0,6,1000,1,1000,1000,1000,1000,1000,1000]
,[8,6,0,7,5,1,2,1000,1000,1000,1000],[1,1000,7,0,1000,1000,9,100,100,100,100],[100,1,5,100,0,3,100,2,100,100,100]
,[100,100,1,100,3,0,4,100,6,100,100],[100,100,2,9,100,4,0,100,3,1,100],[100,100,100,100,2,100,100,0,7,100,9]
,[100,100,100,100,100,6,3,7,0,1,2],[100,100,100,100,100,100,1,100,1,0,4],[100,100,100,100,100,100,100,100,9,2,4,0]]
print("cost", cost)
model = Model('TSP')
# creat decision variables，决策变量
X = {}
for i in range(nodeNum):
for j in range(nodeNum):
if (i != j):
X[i, j] = model.addVar(vtype=GRB.BINARY
, name='x_' + str(i) + '_' + str(j)
)
# set objective function，目标函数
obj = LinExpr(0)
for key in X.keys():
i = key[0]
j = key[1]
obj.addTerms(cost[key[0]][key[1]], X[key])
model.setObjective(obj, GRB.MINIMIZE)
# add constraints 出发点的流量约束
lhs_1 =LinExpr(0)
lhs_2 =LinExpr(0)
for j in range(1, nodeNum-1):
for i in range(0,nodeNum):
if i == 0:
lhs_1.addTerms(1, X[i, j])
model.addConstr(lhs_1 == 1, name='visit_' + str(i)+"start")
#终点的流量约束
for i in range(1,nodeNum-1):
for j in range(1, nodeNum):
if j == nodeNum-1:
lhs_2.addTerms(1, X[i, j])
model.addConstr(lhs_2 == 1, name='visit_' + str(j) + "end")
#其余点的流量约束
for j in range(1,nodeNum-1):
lhs3 = LinExpr(0)
for i in range(0, nodeNum-1):
if i != j:
lhs3.addTerms(1, X[i, j])
for i in range(1, nodeNum):
if i != j:
lhs3.addTerms(-1, X[j, i])
model.addConstr(lhs3 == 0, name='visit_' + str(j)+'balance_flow')
model.write('modelshortestpath.lp')
model.setParam(GRB.Param.MIPGap, 0)
model.setParam(GRB.Param.TimeLimit, 60)
model.optimize()
# 可以输出求解的决策变量
print(model.ObjVal)
X_NEW={}
for var in model.getVars():
if (var.x > 0):
print(var.varName, '\t', var.x)
X_NEW[var.varName]=var.x
x_value = getValue(X, nodeNum)
route = getRoute(x_value)
print('optimal route:', route)
结果展示
最短距离为13
最短路径为[0, 1, 4, 5, 2, 6, 9, 8, 10]
作者：王基光，清华大学，清华伯克利深圳学院（硕士在读）
刘兴禄，清华大学，清华伯克利深圳学院（博士在读）