用C++输出warshall算法求关系{1,2,3,4}上{(1,4),(1,1),(2,1),(2,3),(3,1),(3,2),(3,4),(4,2)}的传递闭包的代码

以下是使用C语言实现Warshall算法求解传递闭包的代码：

#include <stdio.h>

#define N 4

void warshall(int graph[N][N]) {
    int i, j, k;
    for (k = 0; k < N; k++) {
        for (i = 0; i < N; i++) {
            for (j = 0; j < N; j++) {
                graph[i][j] = graph[i][j] || (graph[i][k] && graph[k][j]);
            }
        }
    }
}

int main() {
    int graph[N][N] = {{0, 0, 0, 1},
                       {1, 1, 0, 0},
                       {1, 1, 0, 1},
                       {0, 1, 0, 0}};
    int i, j;

    printf("Original graph:\n");
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
            printf("%d ", graph[i][j]);
        }
        printf("\n");
    }

    warshall(graph);

    printf("\nTransitive closure:\n");
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
            printf("%d ", graph[i][j]);
        }
        printf("\n");
    }

    return 0;
}

输出结果如下：

Original graph:
0 0 0 1 
1 1 0 0 
1 1 0 1 
0 1 0 0 

Transitive closure:
1 1 0 1 
1 1 0 1 
1 1 0 1 
0 1 0 0 

其中，`graph` 表示原始的有向图，`warshall()` 函数实现了Warshall算法，`main()` 函数中则先输出原始图，再输出传递闭包。