编辑代码

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

int main() {
    int N, M;
    scanf("%d %d", &N, &M);

    // 创建邻接矩阵表示电话线连接关系
    int* graph = (int*)calloc((N + 1) * (N + 1), sizeof(int));
    for (int i = 0; i <= N; i++) {
        graph[i] = (int*)calloc((N + 1), sizeof(int));
    }

    // 读取电话线连接关系并构建邻接矩阵
    for (int i = 0; i < M; i++) {
        int u, v;
        scanf("%d %d", &u, &v);
        graph[u][v] = 1;
        graph[v][u] = 1;
    }

    int count = 0;

    // 枚举所有可能的爆破方案
    for (int i = 1; i <= N; i++) {
        for (int j = i + 1; j <= N; j++) {
            if (graph[i][j]) {
                // 假设爆破第i和第j条电话线
                graph[i][j] = 0;
                graph[j][i] = 0;

                // 使用DFS遍历所有与第i个阵地相连的阵地
                int visited[N + 1];
                for (int k = 1; k <= N; k++) {
                    visited[k] = 0;
                }

                visited[i] = 1;
                int stack[N];
                int top = -1;
                stack[++top] = i;

                while (top >= 0) {
                    int node = stack[top--];

                    for (int k = 1; k <= N; k++) {
                        if (graph[node][k] && !visited[k]) {
                            visited[k] = 1;
                            stack[++top] = k;
                        }
                    }
                }

                // 检查是否存在通信盲区
                int blindSpot = 0;
                for (int k = 1; k <= N; k++) {
                    if (!visited[k]) {
                        blindSpot = 1;
                        break;
                    }
                }

                // 如果不存在通信盲区，增加方案计数
                if (!blindSpot) {
                    count++;
                }

                // 恢复第i和第j条电话线连接关系
                graph[i][j] = 1;
                graph[j][i] = 1;
            }
        }
    }

    // 释放内存
    for (int i = 0; i <= N; i++) {
        free(graph[i]);
    }
    free(graph);

    printf("%d\n", count);

    return 0;
}