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Description：
While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ‘s farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.
As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .
To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.
Input：
Output：
Sample Input：
2 3 3 1 1 2 2 1 3 4 2 3 1 3 1 3 3 2 1 1 2 3 2 3 4 3 1 8
Sample Output：
NO YES
Hint：
#include<stdio.h> #include<queue> #include<algorithm> using namespace std; const int N=510; const int INF=0x3f3f3f3f; int G[N][N], dist[N], cou[N], vis[N], n; ///cou数组保存该点被遍历的次数 void Init() { int i, j; for (i = 1; i <= n; i++) { dist[i] = INF; cou[i] = vis[i] = 0; for (j = 1; j <= n; j++) G[i][j] = INF; G[i][i] = 0; } } int Spfa() { int i, v; queue<int>Q; Q.push(1); dist[1] = 0; cou[1]++; while (!Q.empty()) { v = Q.front(); Q.pop(); for (i = 1; i <= n; i++) { if (dist[i] > G[v][i]+dist[v]) { dist[i] = G[v][i]+dist[v]; if (!vis[i]) { Q.push(i); vis[i] = 1; cou[i]++; if (cou[i] >= n && dist[i] < 0) return 1; ///如果存在回路并且并且到达时光隧道的时间比时光隧道穿梭的时间要小 } } } vis[v] = 0; } return 0; } int main () { int T, a, b, c, m, w, ans; scanf("%d", &T); while (T--) { scanf("%d%d%d", &n, &m, &w); Init(); while (m--) { scanf("%d%d%d", &a, &b, &c); G[a][b] = min(G[a][b], c); G[b][a] = G[a][b]; } while (w--) { scanf("%d%d%d", &a, &b, &c); G[a][b] = -c; ///因为时光隧道可以回到c秒前，直接赋值为-c就行了 } ans = Spfa(); if (ans) printf("YES\n"); else printf("NO\n"); } return 0; }
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原文地址：http://www.cnblogs.com/syhandll/p/4812606.html
