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http://poj.org/problem?id=3259
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
SPFA判断是否有负权，如果一个点进入队列的次数达到总点数则说明有负权
dist[i]数组记录源点到i的最短路径，与Dijsktar不同的是dist[i]多次更新
use[i]记录i点进入队列的次数，即dist[i]被更新的次数；
vis[i]标记i点是否进入队列
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<vector> #include<queue> #define INF 0xffffff #define N 520 using namespace std; struct node { int e, w; }; vector<node>G[N]; int n, use[N], dist[N]; bool vis[N]; void Init() { int i; memset(vis, false, sizeof(vis)); memset(use, 0, sizeof(use)); for(i = 0 ; i <= n ; i++) { G[i].clear(); dist[i] = INF; } } int SPFA(int s) { queue<node>Q; node now, next; int i, len; now.e = s; now.w = 0; dist[s] = 0; Q.push(now); vis[s] = true; use[now.e]++; while(!Q.empty()) { now = Q.front(); Q.pop(); vis[now.e] = false; len = G[now.e].size(); for(i = 0 ; i < len ; i++) { next = G[now.e][i]; if(dist[next.e] > dist[now.e] + next.w) { dist[next.e] = dist[now.e] + next.w; use[next.e]++; if(use[next.e] >= n) return 1; if(!vis[next.e]) { vis[next.e] = true; Q.push(next); } } } } return 0; } int main() { int T, m, w, s, e, t, i; node p; scanf("%d", &T); while(T--) { scanf("%d%d%d", &n, &m, &w); Init(); for(i = 1 ; i <= m ; i++) { scanf("%d%d%d", &s, &e, &t); p.w = t; p.e = s; G[e].push_back(p); p.e = e; G[s].push_back(p); } for(i = 1 ; i <= w ; i++) { scanf("%d%d%d", &s, &e, &t); p.w = -t; p.e = e; G[s].push_back(p); } if(SPFA(1)) printf("YES\n"); else printf("NO\n"); } return 0; }
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原文地址：http://www.cnblogs.com/qq2424260747/p/4663564.html
