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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.
就是求是否存在负环，这样的话一定能够时间倒退。。。
代码如下：
#include<iostream> #include<cstring> #include<cstdio> #include<queue> using namespace std; const int INF=10e8; const int MaxN=510; struct Edge { int v,cost; Edge(int _v,int _cost):v(_v),cost(_cost) {} }; vector <Edge> E[MaxN]; bool vis[MaxN]; int couNode[MaxN]; bool SPFA(int lowcost[],int n,int start) { queue <int> que; int u,v,c; int len; for(int i=1;i<=n;++i) { lowcost[i]=INF; vis[i]=0; couNode[i]=0; } vis[start]=1; couNode[start]=1; lowcost[start]=0; que.push(start); while(!que.empty()) { u=que.front(); que.pop(); vis[u]=0; len=E[u].size(); for(int i=0;i<len;++i) { v=E[u][i].v; c=E[u][i].cost; if(lowcost[u]+c<lowcost[v]) { lowcost[v]=lowcost[u]+c; if(!vis[v]) { vis[v]=1; ++couNode[v]; que.push(v); if(couNode[v]>=n) return 0; } } } } return 1; } inline void addEdge(int u,int v,int c) { E[u].push_back(Edge(v,c)); } int ans[MaxN]; int main() { int T; int N,M,W; int a,b,c; scanf("%d",&T); while(T--) { scanf("%d %d %d",&N,&M,&W); for(int i=1;i<=N;++i) E[i].clear(); for(int i=1;i<=M;++i) { scanf("%d %d %d",&a,&b,&c); addEdge(a,b,c); addEdge(b,a,c); } for(int i=1;i<=W;++i) { scanf("%d %d %d",&a,&b,&c); addEdge(a,b,-c); } if(SPFA(ans,N,1)) printf("NO\n"); else printf("YES\n"); } return 0; }
(简单) POJ 3259 Wormholes，SPFA判断负环。
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原文地址：http://www.cnblogs.com/whywhy/p/4338558.html
