Wormholes

 

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36782		Accepted: 13467

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 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 toE 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

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

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.

【思路】

spfa判断负环

一个点是否加入队列n次以上

【code】

#include<queue>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m,w,x,y,z,t;
struct Edge {
    int x,y,z,nxt;
    Edge(int x=0,int y=0,int z=0,int nxt=0):
        x(x),y(y),z(z),nxt(nxt) {}
} edge[2500<<1];
int head[501],sumedge,dis[501],vis[501],cnt[501];
void fir() {
    sumedge=0;
    memset(head,0,sizeof(head));
    memset(dis,127/2,sizeof(dis));
    memset(vis,0,sizeof(vis));
}
int add(int x,int y,int z) {
    edge[++sumedge]=Edge(x,y,z,head[x]);
    return head[x]=sumedge;
}
int spfa() {
    queue <int>q;
    dis[1]=0;
    vis[1]=1;
    q.push(1);
    while(!q.empty()) {
        int now=q.front();
        q.pop();
        vis[now]=0;
        ++cnt[now];
        if(cnt[now]>n)return 1;
        for(int i=head[now]; i; i=edge[i].nxt) {
            int to=edge[i].y;
            if(dis[to]>dis[now]+edge[i].z) {
                dis[to]=dis[now]+edge[i].z;
                if(!vis[to]) {
                    vis[to]=1;
                    q.push(to);
                }
            }
        }
    }
    return 0;
}
int main() {
    scanf("%d",&t);
    while(t--) {
        fir();
        scanf("%d%d%d",&n,&m,&w);
        for(int i=1; i<=m; i++) {
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);
            add(y,x,z);
        }
        for(int i=1; i<=w; i++) {
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,-z);
        }
        if(spfa())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}