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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
/* 题目大意：虫洞问题，现在有n个点，m条边， 代表现在可以走的通路，比如从a到b和从b到a需要花费c时间， 现在在地上出现了w个虫洞， 虫洞的意义就是你从a到b花费的时间是-c(时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能回到从前 其实就是判断是否存在负环 */ #include <cstdio> #include <cstring> #include <queue> #define MAXN 550 #define MAXM 10000 #define INF 0x3f3f3f3f using namespace std; struct Edge { int u, v, w; int next;//下一个结构体变量的下标 }edge[MAXM]; int head[MAXN];//下标为起点u，值为对应结构体下标 int vis[MAXN];//判断是否加入队列了 int used[MAXN]; int num; int N; int low[MAXN];//存最短路径 void Add_Edge(int u, int v, int w)//加边 { Edge E={u, v, w, head[u]};//初始化结构体 edge[num]=E;//直接赋值 head[u]=num++; } bool SPFA(int s) { int i, j; queue<int> Q; memset(low, INF, sizeof(low)); memset(vis, 0, sizeof(vis)); memset(used,0,sizeof(used)); vis[s] = 1; low[s]=0; Q.push(s); used[s]++; while(!Q.empty()) { int u=Q.front(); Q.pop(); vis[u]=0;//出队列了，不在队列就变成0 for(j = head[u]; j != -1; j = edge[j].next) { int v = edge[j].v; if(low[v] > low[u] + edge[j].w) { low[v] = low[u] + edge[j].w; if(!vis[v] ) { vis[v]=1; Q.push(v); used[v]++; if(used[v]>N) return 0; } } } } return 1; } int main() { int u, v, w; int T; int M, W; scanf("%d",&T); while(T--) { scanf("%d%d%d", &N, &M, &W); memset(head, -1, sizeof(head)); num=0; while(M--) { scanf("%d%d%d", &u, &v, &w); Add_Edge(u, v, w); Add_Edge(v, u, w);//无向边 } while(W--) { scanf("%d%d%d", &u, &v, &w); Add_Edge(u, v, -w); } if(!SPFA(1)) printf("YES\n"); else printf("NO\n"); } return 0; }
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原文地址：http://blog.csdn.net/yuzhiwei1995/article/details/47806003
