55 lines
1.6 KiB
C++
55 lines
1.6 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 30;
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int m, n, t;
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int dp[N][N];
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void floyd() {
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for (int k = 0; k < n; k++)
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for (int i = 0; i < n; i++)
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for (int j = 0; j < n; j++)
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dp[i][j] += dp[i][k] * dp[k][j]; //记录边数
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}
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int main() {
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while (scanf("%d", &m) != EOF) {
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int a, b;
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n = 0;
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memset(dp, 0, sizeof(dp));
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while (m--) {
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scanf("%d%d", &a, &b);
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dp[a][b] = 1; //单向边，同时边权为1
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n = max({n, a, b}); //最大节点号
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}
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//注意有0节点，所以矩阵边长为n+1。
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n++;
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floyd();
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printf("matrix for city %d\n", t++);
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//结论：某个节点到自己的距离不为0，就是有环
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for (int k = 0; k < n; k++)
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if (dp[k][k]) { //如果存在环的话，需要特殊处理
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for (int i = 0; i < n; i++)
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for (int j = 0; j < n; j++)
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if (dp[i][k] && dp[k][j]) //如果有其它点i,j利用过k，则i->k,k->j都是无数条路径
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dp[i][j] = -1;
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dp[k][k] = -1; //标识k->k也有无数条路径
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}
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//严格控制格式，输出矩阵
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < n; j++)
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if (j == 0)
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printf("%d", dp[i][j]);
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else
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printf(" %d", dp[i][j]);
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puts("");
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}
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}
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return 0;
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} |