题目：
| Wormholes| Time Limit: 2000MS | | Memory Limit: 65536K |
| ---------------------------- | - | ------------------------ |
| Total Submissions: 48056 | | Accepted: 17754 |DescriptionWhile 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.InputLine 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.OutputLines 1.. F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).Sample Input2 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 8Sample Output```
NO YES
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思路：
农夫约翰有F个农场，每个农场有N块地，其间有M条路，W条时光隧道（时间倒流）。就是当你过去的时候时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。就是N个点m条边，先给出了一个图，然后有W个虫洞，虫洞有穿越时光的作用，比如从a-->b的权为-c，用folyd找负权回路，容易超时\
代码：
```cpp
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define N 500+20
#define M 100000+20
#define inf 0x3f3f3f3f
using namespace std;
int map[N][N];
int n,m,w;
int floyd()
{
for(int k=1; k<=n; k++)
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
if(map[i][j]>map[i][k]+map[k][j])
map[i][j]=map[i][k]+map[k][j];
if(map[i][i]<0)
return 1;
}
return 0;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int a,b,c;
scanf("%d%d%d",&n,&m,&w);
mem(map,inf);
for(int i=1; i<=n; i++)
map[i][i]=0;
for(int i=1; i<=m; i++)
{
scanf("%d%d%d",&a,&b,&c);
if(c<map[a][b])
map[a][b]=map[b][a]=c;
}
for(int i=1; i<=w; i++)
{
scanf("%d%d%d",&a,&b,&c);
map[a][b]=-c;
}
if(floyd())
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
\
