Wormholes

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

大意：有F个test，共N个编号，M个正向的u,v,t，即从u到v要花t秒，可双向，W个虫洞反向的，只单向，要花-t秒，问最后能不能到达出发之前。

就有if(d[edge.v]>d[edge.u]+edge.t){

flag = 0;

d[edge.v] = d[edge.u]+edge.t;

}

意为如果虫洞返回后的时间前于出发的时间，那么满足，经过N此循环都是满足的话，那么就是满足的。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 22222222;
int F,N,M,W,cnt;
int d[6000];
struct edge
{
    int u;
    int v;
    int t;
}edge[6000];

bool bellman_ford()
{
    for(int i = 1; i <= N;i++)
        d[i] = MAX;
    d[1] = 0;
    bool flag;
    for(int i = 1; i <= N;i++){
            flag = 1;
       for(int j = 0; j < cnt; j++){
            if(d[edge[j].v]>d[edge[j].u]+edge[j].t){
                    flag = 0;
              d[edge[j].v] = d[edge[j].u]+edge[j].t;
         }
      }
    if(flag)
        return true;
   }
   return false;
}

int main()
{
    int u,v,t;
    scanf("%d",&F);
    while(F--){
            cnt = 0;
            scanf("%d%d%d",&N,&M,&W);
            for(int i = 1; i <= M;i++){
                scanf("%d%d%d",&u,&v,&t);
                edge[cnt].u = u; edge[cnt].v = v; edge[cnt].t = t;
                cnt++;
                edge[cnt].u = v; edge[cnt].v = u; edge[cnt].t = t;
                cnt++;
            }
            for(int i = 1; i <= W;i++){
                    scanf("%d%d%d",&u,&v,&t);
                    edge[cnt].u = u;edge[cnt].v = v;edge[cnt].t = -t;
                    cnt++;
            }
        if(bellman_ford())
            printf("NO\n");
        else printf("YES\n");
    }
    return 0;
}