2831.   
Wormholes
Time Limit: 1.0 Seconds   
Memory Limit: 65536K

Total Runs: 1271   
Accepted Runs: 404    
Multiple test files

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 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

Input Details

Two farm maps. The first has three paths and one wormhole, and the second has two paths and one wormhole.

Output Details

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.

题意是FJ有好几个农场，农场中有不同的虫洞，在不同虫洞之间王往来有N条路，各有自己的cost，而通过虫洞可以回到过去，使时间减少。他想是否可以时空穿越从一个点出发再回到同一个点上。这道题就成了判断个图中是否有负权环，利用bellmamnford，再加上spfa优化，然后要注意这里并不能保证每个点之间都是互相联通的，所以需要以每个未标记vis[]的点为起点再跑一遍。

#include <iostream>
#include<stdio.h>
#include<cstring>
#define inf 0x3f3f3f3f
using namespace std;

int n,m,w,head[500],nc,dis[500];

struct edge{
    int to,nex,cost;
};
edge e[50005];
int addedge(int a,int b,int c)
{
    e[nc].to=b;
    e[nc].cost=c;
    e[nc].nex=head[a];
    head[a]=nc++;
    return a;
}

bool spfa(int s)
{
    int x,y,l[1000],top,tail,used[520];
    bool mark[520];
    memset(mark,false,sizeof(mark));
    memset(used,0,sizeof(used));
    top=tail=0;
    l[top++]=s;
    used[s]++;
    mark[s]=1;
    dis[s]=0;
    while(tail!=top)
    {
        x=l[tail++];
        mark[x]=0;
        if(tail==999)
            tail=0;
        for(int i=head[x];i!=-1;i=e[i].nex)
        {
            y=e[i].to;
            if(dis[y]>e[i].cost+dis[x])
            {
                dis[y]=e[i].cost+dis[x];
                used[y]++;
                if(used[y]>=n)
                    return false;
                if(!mark[y])
                {
                    mark[y]=1;
                    l[top++]=y;
                    if(top==999)
                        top=0;
                }
            }

        }
    }
    return true;
}

int main()
{
    int num,a,b,c;
    scanf("%d",&num);
    while(num--)
    {
        nc=0;
        scanf("%d%d%d",&n,&m,&w);
        memset(head,-1,sizeof(head));
        memset(dis,inf,sizeof(dis));
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            addedge(a,b,c);
            addedge(b,a,c);
        }
        for(int i=0;i<w;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            addedge(a,b,-c);
        }
        int i;
        for(i=1;i<=n;i++)
        {
            if(dis[i]==inf)
                if(!spfa(i))
            {
                i=n+2;
                break;
            }
        }
         if(i==n+1) printf("NO\n");
         else printf("YES\n");
    }
    return 0;
}