分支界限法 旅行商问题C语言

分支界限法是一种求解问题的通用算法，其中包括解决旅行商问题的算法。以下是一个使用 C 语言实现的分支界限法求解旅行商问题的示例代码：

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

#define N 4

int graph[N][N] = {
    {0, 10, 15, 20},
    {10, 0, 35, 25},
    {15, 35, 0, 30},
    {20, 25, 30, 0}
};

typedef struct {
    int path[N];
    int level;
    int bound;
} Node;

int min(int a, int b) {
    return a < b ? a : b;
}

void swap(int *a, int *b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

int calculateBound(Node node) {
    int sum = 0;

    for (int i = 0; i < N - 1; i++) {
        sum += graph[node.path[i]][node.path[i + 1]];
    }

    sum += graph[node.path[N - 1]][node.path[0]];

    for (int i = 0; i < N; i++) {
        if (node.path[i] == -1) {
            int min1 = 999, min2 = 999;

            for (int j = 0; j < N; j++) {
                if (i != j && node.path[j] == -1) {
                    if (graph[i][j] < min1) {
                        min2 = min1;
                        min1 = graph[i][j];
                    } else if (graph[i][j] < min2) {
                        min2 = graph[i][j];
                    }
                }
            }

            sum += min1 + min2;
        }
    }

    return sum;
}

void copyPath(int *dest, int *src) {
    for (int i = 0; i < N; i++) {
        dest[i] = src[i];
    }
}

void printPath(int *path) {
    for (int i = 0; i < N; i++) {
        printf("%d ", path[i]);
    }

    printf("\n");
}

void solve() {
    Node queue[100];
    int head = 0, tail = 0;
    int bestPath[N], bestCost = 999;

    queue[tail].level = 0;
    queue[tail].path[0] = 0;
    queue[tail].bound = calculateBound(queue[tail]);
    tail++;

    while (head != tail) {
        Node node = queue[head];
        head++;

        if (node.level == N - 1) {
            int cost = calculateBound(node);

            if (cost < bestCost) {
                bestCost = cost;
                copyPath(bestPath, node.path);
            }
        } else {
            for (int i = node.level + 1; i < N; i++) {
                Node childNode = node;
                swap(&childNode.path[node.level + 1], &childNode.path[i]);
                childNode.level++;
                childNode.bound = calculateBound(childNode);

                if (childNode.bound < bestCost) {
                    queue[tail] = childNode;
                    tail++;
                }
            }
        }
    }

    printf("Best path: ");
    printPath(bestPath);
    printf("Best cost: %d\n", bestCost);
}

int main() {
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            if (graph[i][j] == 0) {
                graph[i][j] = 999;
            }
        }
    }

    int path[N];

    for (int i = 0; i < N; i++) {
        path[i] = -1;
    }

    path[0] = 0;

    solve();

    return 0;
}

在这个示例代码中，我们定义了一个 `Node` 结构体来表示分支界限法中的节点，其中包括路径、级别和界限。我们还定义了一些辅助函数来计算路径的界限、复制路径、打印路径等。最后我们在 `main` 函数中初始化路径，然后调用 `solve` 函数来求解旅行商问题。