某售货员要到若干城市去推销商品,已知各城市之间的路程(或旅费)。他要选定一条从驻地出发,经过每个城市一次,最后回到驻地的路线,使总的路程(或总旅费)最小。用分支与限界法通过C语言实现
时间: 2024-06-14 07:05:19 浏览: 26
以下是使用分支与限界法通过C语言实现某售货员旅行问题的示例代码:
```c
#include <stdio.h>
#include <stdbool.h>
#define N 4 // 城市数量
int minCost = INT_MAX; // 最小路程(或旅费)
int minPath[N+1]; // 最小路线
// 计算当前路径的总路程(或旅费)
int calculateCost(int graph[N][N], int path[N+1]) {
int cost = 0;
for (int i = 0; i < N; i++) {
cost += graph[path[i]][path[i+1]];
}
return cost;
}
// 分支与限界法求解最小路程(或旅费)
void tsp(int graph[N][N], int path[N+1], bool visited[N], int level, int cost) {
if (level == N) {
cost += graph[path[level-1]][path[0]]; // 回到起点
if (cost < minCost) {
minCost = cost;
for (int i = 0; i <= N; i++) {
minPath[i] = path[i];
}
}
return;
}
for (int i = 0; i < N; i++) {
if (!visited[i]) {
path[level] = i;
visited[i] = true;
int newCost = cost + graph[path[level-1]][i];
if (newCost < minCost) {
tsp(graph, path, visited, level+1, newCost);
}
visited[i] = false;
}
}
}
int main() {
int graph[N][N] = {
{0, 10, 15, 20},
{10, 0, 35, 25},
{15, 35, 0, 30},
{20, 25, 30, 0}
};
int path[N+1];
bool visited[N] = {false};
path[0] = 0; // 起点为0号城市
visited[0] = true;
tsp(graph, path, visited, 1, 0);
printf("最小路程(或旅费)为:%d\n", minCost);
printf("最小路线为:");
for (int i = 0; i <= N; i++) {
printf("%d ", minPath[i]);
}
printf("\n");
return 0;
}
```