旅行推销员问题用于计算覆盖所有城市的最短路线,然后返回到原始城市。此方法用于查找覆盖图形所有节点的最短路径。
这是查找未加权图的最短路径的程序。
Begin Define a variable vr = 4 universally. Declare an integer function TSP to implement Travelling salesman Problem. Declare a graph grph[][] as a 2D matrix and variable p to the integer datatype. Pass them as a parameter. Declare variable ver to the vector datatype. for (int i = 0; i < vr; i++) if (i != p) then Call push_back(i) function to store the value of all vertex except source vertex. Initialize m_p = INT_MAX to store minimum weight of a graph. do Declare cur_pth, k to the integer datatype. initialize cur_pth = 0. initialize k = p. for (int i = 0; i < ver.size(); i++) cur_pth += grph[k][ver[i]]. k = ver[i]. cur_pth += grph[k][p]. m_p = min(m_p, cur_pth) to update the value of minimum weight. while (next_permutation(ver.begin(), ver.end())). Return m_p. Declare a graph grph[][] as a 2D matrix to the integer datatype. Initialize values of grph[][] graph. Declare variable p to the integer datatype. Initialize p = 0. Print “The result is: ”. Print the return value of TSP() function. End.
#include <bits/stdc++.h> using namespace std; #define vr 4 int TSP(int grph[][vr], int p) // implement traveling Salesman Problem. { vector<int> ver; // for (int i = 0; i < vr; i++) if (i != p) ver.push_back(i); int m_p = INT_MAX; // store minimum weight of a graph do { int cur_pth = 0; int k = p; for (int i = 0; i < ver.size(); i++) { cur_pth += grph[k][ver[i]]; k = ver[i]; } cur_pth += grph[k][p]; m_p = min(m_p, cur_pth); // to update the value of minimum weight } while (next_permutation(ver.begin(), ver.end())); return m_p; } int main() { int grph[][vr] = { { 0, 5, 10, 15 }, //values of a graph in a form of matrix { 5, 0, 20, 30 }, { 10, 20, 0, 35 }, { 15, 30, 35, 0 } }; int p = 0; cout<< "\n The result is: "<< TSP(grph, p) << endl; return 0; }
输出结果
The result is: 75