Wormholes

用spfa判断是否有负环，其次存边的数组要开的大一些（不仅有正常的双向边，还有单向的虫洞）

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

Line 1: A single integer, F. F farm descriptions follow. 
Line 1 of each farm: Three space-separated integers respectively: N, M, and W 
Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path. 
Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler backT seconds.

Output

Lines 1.. F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

Hint

For farm 1, FJ cannot travel back in time. 
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

#include<stdio.h>#include<queue>#include<string.h>#define maxn 505#define maxm 2505#define inf 0x3f3f3f3fusing namespace std;int head[maxn],time[maxn],dis[maxn],cnt;bool vis[maxn];struct Edge{int v,w,next;}edge[3*maxm];void add_edge(int u,int v,int w){edge[cnt].v=v;edge[cnt].w=w;edge[cnt].next=head[u];head[u]=cnt++;}int spfa(int s,int tol){memset(dis,inf,sizeof(dis));memset(vis,0,sizeof(vis));memset(time,0,sizeof(time));queue<int> q;dis[s]=0;vis[s]=true;time[s]++;q.push(s);while(!q.empty()){int Q=q.front();q.pop();vis[Q]=false;for(int i=head[Q];i!=-1;i=edge[i].next){int to=edge[i].v;if(dis[to]>dis[Q]+edge[i].w){dis[to]=dis[Q]+edge[i].w;if(!vis[to]){vis[to]=true;q.push(to);time[to]++;if(time[to]==tol)return 1;}}}} return 0;}int main(){int f;scanf("%d",&f);while(f--){int n,m,w;cnt=0;memset(head,-1,sizeof(head));memset(edge,0,sizeof(edge));scanf("%d%d%d",&n,&m,&w);for(int i=0;i<m;i++){int s,e,t;scanf("%d%d%d",&s,&e,&t);add_edge(s,e,t);add_edge(e,s,t);}for(int i=0;i<w;i++){int s,e,t;scanf("%d%d%d",&s,&e,&t);add_edge(s,e,-t);}if(spfa(1,n))printf("YES\n");elseprintf("NO\n");}return 0;}