旅行商问题分支限界法C语言

以下是使用分支限界法解决旅行商问题的 C 语言代码示例：

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

#define MAX_N 100 // 最大城市数量

int n; // 城市数量
int city[MAX_N][MAX_N]; // 城市间距离矩阵
int visited[MAX_N]; // 标记城市是否已经被访问
int best_path[MAX_N]; // 记录最优路径
int best_dist = INT_MAX; // 记录最优路径长度

// 节点结构体，用于存储搜索状态
typedef struct {
    int level; // 当前搜索深度
    int bound; // 当前搜索路径长度下界
    int path[MAX_N]; // 当前搜索路径
} Node;

// 计算当前路径长度
int get_path_dist(int path[]) {
    int dist = 0;
    for (int i = 0; i < n-1; i++) {
        dist += city[path[i]][path[i+1]];
    }
    dist += city[path[n-1]][path[0]];
    return dist;
}

// 计算当前节点的下界
int get_bound(Node node) {
    int bound = 0;
    int path[MAX_N];
    for (int i = 0; i < n; i++) {
        path[i] = node.path[i];
    }
    // 对未访问的城市进行排序
    for (int i = node.level+1; i < n; i++) {
        int min_dist = INT_MAX;
        int min_index = -1;
        for (int j = i; j < n; j++) {
            if (!visited[j] && city[path[i-1]][j] < min_dist) {
                min_dist = city[path[i-1]][j];
                min_index = j;
            }
        }
        if (min_index != -1) {
            int temp = path[i];
            path[i] = path[min_index];
            path[min_index] = temp;
        }
    }
    // 计算下界
    for (int i = 0; i < n-1; i++) {
        if (!visited[path[i]]) {
            bound += city[path[i]][path[i+1]];
        }
    }
    bound += city[path[n-1]][path[0]];
    return bound;
}

// 搜索函数
void search(Node node) {
    if (node.level == n-1) {
        int dist = get_path_dist(node.path);
        if (dist < best_dist) {
            best_dist = dist;
            for (int i = 0; i < n; i++) {
                best_path[i] = node.path[i];
            }
        }
        return;
    }
    for (int i = node.level+1; i < n; i++) {
        if (!visited[node.path[i]]) {
            Node child;
            child.level = node.level + 1;
            child.path[child.level] = node.path[i];
            for (int j = 0; j <= node.level; j++) {
                child.path[j] = node.path[j];
            }
            visited[node.path[i]] = 1;
            child.bound = node.bound - city[node.path[node.level]][node.path[i]] + city[node.path[node.level]][node.path[node.level+1]];
            child.bound += get_bound(child);
            if (child.bound < best_dist) {
                search(child);
            }
            visited[node.path[i]] = 0;
        }
    }
}

int main() {
    // 读入城市数量和城市间距离矩阵
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            scanf("%d", &city[i][j]);
        }
    }
    // 初始化搜索状态
    Node initial;
    initial.level = -1;
    initial.bound = 0;
    for (int i = 0; i < n; i++) {
        initial.path[i] = i;
        visited[i] = 0;
    }
    visited[0] = 1;
    // 计算下界
    initial.bound = get_bound(initial);
    // 开始搜索
    search(initial);
    // 输出结果
    printf("最短路径长度为：%d\n", best_dist);
    printf("最短路径为：");
    for (int i = 0; i < n; i++) {
        printf("%d ", best_path[i]);
    }
    printf("\n");
    return 0;
}

该代码使用了分支限界法求解 TSP 问题，其中 `get_bound` 函数计算了当前节点的下界，`search` 函数进行搜索，`main` 函数处理输入和输出。