Wormholes (spfa)

一种路是双向的，路的长度是正值；另一种路是单向的，路的长度是负值；  如果有负环输出YES；否则输出NO；不同的路可能有相同的起点和终点：必须用邻接表

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

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.

Sponsor

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<string>
#define LL long long
#define eps 1e-8
using namespace std;
const int inf = 0x3f3f3f3f;
int n,m,w,a,b,c,t,tot,ans;
struct node{
    int l,r,num,nex;
}e[10010];
int head[550],vis[550],dis[550],sum[550];

void init()
{
    
    memset(head,-1,sizeof(head));
    tot=0;
    ans=0;
}

void add(int l,int r,int num)
{
    e[tot].l=l,e[tot].r=r,e[tot].num=num;
    e[tot].nex=head[l];
    head[l]=tot++;
}

int spfa(int x)
{
    queue<int >q;
    while(!q.empty())
        q.pop();
    memset(dis,inf,sizeof(dis));
    memset(vis,0,sizeof(vis));
    memset(sum,0,sizeof(sum));
    q.push(x);
    vis[x]=1;
    dis[x]=0;
    sum[x]=1;
    while(!q.empty())
    {
        int now=q.front(); q.pop();
        vis[now]=0;
        for(int i=head[now];i!=-1;i=e[i].nex)
        {
            int to=e[i].r;
            if(dis[to]>dis[now]+e[i].num)
            {
                dis[to]=dis[now]+e[i].num;
                if(!vis[to])
                {
                    sum[to]++;
                    if(sum[to]>n) return 1;
                    q.push(to);
                    vis[to]=1;
                }
            }
        }
    }
    return 0;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&w);
        init();
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        while(w--)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,-c);
        }
        for(int i=1;i<=n;i++)
        {
            if(spfa(i)){
                ans=1;
                break;//否则超时 
            } 
        }
        if(ans) printf("YES\n");
        else printf("NO\n");
    }
}