Wormholes


Time Limit: 2000MS Memory Limit: 65536K


Total Submissions: 32369 Accepted: 11762


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 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.



Source


USACO 2006 December Gold


题目大意：农场有N个点，M条双向边，每条边信息为——从S到E花费的时间为T，同理从T到S花费的时间

也为T。现在农场出现了W个虫洞，每个虫洞意味着：从S到E花费的时间为-T(时间倒流，但是虫洞是单向边)。

问：从N个点中的某点开始走，能否使时间倒流。

思路：构建一个图，添加M条双向边，再添加W个从S到E权值为-T的负权边，求最长路径，看能否出现负权

回路，若出现负权回路，则说明能使时间倒流，否则则不能使时间倒流。



#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 550;
const int MAXM = 5500;
const int INF = 0xffffff0;
struct EdgeNode
{
    int to;
    int w;
    int next;
}Edges[MAXM];
int Head[MAXN],Dist[MAXN];

bool BellmanFord(int N,int M)
{
    Dist[1] = 0;
    for(int i = 1; i < N; ++i)
    {
        for(int j = 1; j <= N; ++j)
        {
            if(Dist[j] == INF)
                continue;
            for(int k = Head[j]; k != -1; k = Edges[k].next)
            {
                if(Edges[k].w != INF && Dist[Edges[k].to] > Dist[j] + Edges[k].w)
                    Dist[Edges[k].to] = Dist[j] + Edges[k].w;
            }
        }
    }

    for(int j = 1; j <= N; ++j)
    {
        if(Dist[j] == INF)
            continue;
        for(int k = Head[j]; k != -1; k = Edges[k].next)
        {
            if(Edges[k].w != INF && Dist[Edges[k].to] > Dist[j] + Edges[k].w)
                return true;
        }
    }
    return false;
}
int main()
{
    int F,N,M,W,S,E,T;
    scanf("%d",&F);
    while(F--)
    {
        memset(Dist,INF,sizeof(Dist));
        memset(Head,-1,sizeof(Head));
        memset(Edges,0,sizeof(Edges));
        scanf("%d%d%d",&N,&M,&W);
        int id = 0;
        for(int i = 0; i < M; ++i)
        {
            scanf("%d%d%d",&S,&E,&T);
            Edges[id].to = E;
            Edges[id].w = T;
            Edges[id].next = Head[S];
            Head[S] = id++;
            Edges[id].to = S;
            Edges[id].w = T;
            Edges[id].next = Head[E];
            Head[E] = id++;
        }
        for(int i = 0; i < W; ++i)
        {
            scanf("%d%d%d",&S,&E,&T);
            Edges[id].to = E;
            Edges[id].w = -T;
            Edges[id].next = Head[S];
            Head[S] = id++;
        }
        if(BellmanFord(N,M))
            printf("YES\n");
        else
            printf("NO\n");
    }

    return 0;
}