使用最近邻算法,C++ 程序实现旅行商问题
使用最近邻算法实现旅行商问题的 C++ 程序。
必需的函数和伪代码
算法
Begin Initialize c = 0, cost = 1000; Initialize g[][]. function swap() is used to swap two values x and y. function cal_sum() to calculate the cost which take array a[] and size of array as input. Initialize sum = 0. for i = 0 to n compute s+= g[a[i %3]][a[(i+ 1) %3]]; if (cost >s) cost = s function permute() is used to perform permutation: if there is one element in array call cal_sum(). else for j = i to n swap (a+i) with (a + j) cal_sum(a+1,n) swap (a+i) with (a + j) End
示例代码
#include<iostream>
using namespace std;
int c = 0, cost = 1000;
int g[3][3 ]={{1, 2, 3}, {4, 5, 8}, {6, 7, 10}};
void swap(int *x, int *y) {
int t;
t = *x;
*x = *y;
*y = t;
}
void cal_sum(int *a, int n) {
int i, s= 0;
for (i = 0; i <= n; i++) {
s+= g[a[i %3]][a[(i+ 1) %3]];
} if (cost >s) {
cost = s;
}
}
void permute(int *a,int i,int n) {
int j, k;
if (i == n) {
cal_sum (a,n);
} else {
for (j = i; j <= n; j++) {
swap((a + i), (a + j));
cal_sum(a+1,n);
swap((a + i), (a + j));
}
}
}
int main() {
int i, j;
int a[] = {1,2,3};
permute(a, 0,2);
cout << "minimum cost:" << cost << endl;
}结果
Comparing str1 and str2 using ==, Res: 0 Comparing str1 and str3 using ==, Res: 1 Comparing str1 and str2 using compare(), Res: -1024 Comparing str1 and str3 using compare(), Res: 0
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