Wormholes

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 toF (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 fromS to E that also moves the traveler back T 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.

就是来判断负数环。也就是回到了这个点之前的存在。

注意这个边的申请多少个，正反。

#include <cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;typedef long long ll;const int maxn=5500;const int INF=0x3f3f3f3f;int N,M,W;int cnt;int time[510];int head[maxn];struct Edge{  int v,w;  int next;}edge[maxn];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 dis[510];int vis[510];int spfa(int st){    memset(dis,INF,sizeof(dis));    memset(time,0,sizeof(time));    memset(vis,0,sizeof(vis));    dis[st]=0;    queue<int> q;    q.push(st);    int flag=0;    time[st]++;    while(q.size()){        int now=q.front();        q.pop();        vis[now]=0;        for(int i=head[now];i!=-1;i=edge[i].next){            int v=edge[i].v,w=edge[i].w;          // printf("st:%d u:%d v:%d w:%d\n",st,now,v,w);            if(dis[v]>dis[now]+w){                dis[v]=dis[now]+w;                if(!vis[v]){                    vis[v]=1;                    time[v]++;                    if(time[v]>N){                        flag=1;                          break;                    }                    q.push(v);                }            }        }    }    if(flag)        return 1;    return 0;}int main(){    int T;    scanf("%d",&T);    while(T--){        cnt=0;        memset(head,-1,sizeof(head));        scanf("%d %d %d",&N,&M,&W);        for(int i=1;i<=M;i++){            int u,v,w;            scanf("%d %d %d",&u,&v,&w);            add_edge(u,v,w);            add_edge(v,u,w);        }        for(int i=1;i<=W;i++){            int u,v,w;            scanf("%d %d %d",&u,&v,&w);            add_edge(u,v,-w);        }        int i;        for(i=1;i<=N;i++){            if(spfa(i)){                printf("YES\n");                break;            }            else                continue;        }        if(i==N+1)            printf("NO\n");    }    return 0;}