# F - Wormholes

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.

Input

Output

F行，每行代表一个农场 每个农场单独的一行，” YES”表示能满足要求，”NO”表示不能满足要求。

```//
// Created by hanyu on 2019/7/19.
//
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=2e5+7;
#define MAX 0x3f3f3f3f
int book[maxn],cnt[maxn],d[maxn],head[maxn];
int n,m,w;
int pos;
struct Node{
int s;
int e;
int t;
}node[maxn];
void add(int s,int e,int t)
{
node[pos].s=e;
node[pos].e=t;
node[pos].t=head[s];
head[s]=pos++;
}
bool spfa(int start)
{

queue<int>qu;
qu.push(start);
book[start]=1;
d[start]=0;
while(!qu.empty())
{
int now=qu.front();
qu.pop();
book[now]=0;
for(int i=head[now];i!=-1;i=node[i].t)
{
int ss=node[i].s;
int ee=node[i].e;
if(d[ss]>d[now]+ee)
{
d[ss]=d[now]+ee;
if(!book[ss])
{
qu.push(ss);
book[ss]=1;
cnt[ss]++;
if(cnt[ss]>=n)
return true;//判断负环
}
}
}
}
return false;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
memset(book,0,sizeof(book));
memset(cnt,0,sizeof(cnt));
memset(d,MAX,sizeof(d));
memset(head,-1,sizeof(head));
memset(node,0,sizeof(node));
scanf("%d%d%d",&n,&m,&w);
int s,e,t;
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&s,&e,&t);
add(s,e,t);
add(e,s,t);
}
for(int i=1;i<=w;i++)
{
scanf("%d%d%d",&s,&e,&t);
add(s,e,-t);
}
cnt=1;
pos=0;
if(spfa(1))
printf("YES\n");
else
printf("NO\n");
}
return 0;
}```

© 2017 版权所有 鲁ICP备17052893号