UVA125 - Numbering Paths(floyd)
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UVA125 - Numbering Paths(floyd)
UVA125 - Numbering Paths
题目大意:
给m条有方向的边,然后要求你给出N * N的矩阵,矩阵G【i】【j】代表的是i到j之间的总路径数,如果i到j之间存在着环,那么G【i】【j】 = -1.
解题思路:
i到j的路径数目等于i到k乘以k到j(经过k到达的话)。用floyd可以求出i到j之间的所有的路径数目,G【i】【j】 += G【i】【k】 * G【k】【j】;如果i,j之间存在着环,那么必然会存在一点i使得G【i】【i】 != 0.
代码:
#include <cstdio>#include <cstring>const int maxn = 35;int N, M;int G[maxn][maxn];int max (int a, int b) { return a > b ? a: b;}void floyd () { for (int k = 0; k <= N; k++) for (int i = 0; i <= N; i++) for (int j = 0; j <= N; j++) if (G[i][k] && G[k][j]) G[i][j] += G[i][k] * G[k][j]; for (int k = 0; k <= N; k++) if (G[k][k]) { for (int i = 0; i <= N; i++) for (int j = 0; j <= N; j++) if (G[i][k] && G[k][j]) G[i][j] = -1; }}int main () { int u, v; int T = 0; while (scanf("%d", &M) != EOF) { memset(G, 0, sizeof(G)); N = 0; for (int i = 0; i < M; i++) { scanf ("%d%d", &u, &v); G[u][v] = 1; N = max(N, max(u, v)); } printf ("matrix for city %d\n", T++); floyd(); for (int i = 0; i <= N; i++) for (int j = 0; j <= N; j++) { if (j != N) printf ("%d ", G[i][j]); else printf ("%d\n", G[i][j]); } } return 0;}
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