| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 22566 | Accepted: 8056 |
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 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
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 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
思路：本题就是判断是否有出现负权回路。用SPFA，看是否有顶点入队数超过顶点数，有则说明有负权回路。
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int nn=510;
const int mm=3100;
const int oo=1e9;
class node
{
public:int v,c;
};
vector<node >e[nn];
int n,m,w,dis[nn],id[nn];
bool vis[nn];
queue<int>q;
bool spfa(int x)
{
memset(id,0,sizeof(id));
memset(vis,0,sizeof(vis));
for(int i=0;i<n;i++)
dis[i]=oo;
dis[x]=0;vis[x]=1;++id[x];
q.push(x);int z;
while(!q.empty())
{
z=q.front();q.pop();vis[z]=0;
for(int i=0;i<e[z].size();i++)
{
if(dis[e[z][i].v]>dis[z]+e[z][i].c)
{
dis[e[z][i].v]=dis[z]+e[z][i].c;
if(!vis[e[z][i].v])
{
vis[e[z][i].v]=1;++id[e[z][i].v];
q.push(e[z][i].v);
if(id[e[z][i].v]>n)
return 1;
}
}
}
}
return 0;
}
int main()
{
int cas;
while(cin>>cas)
{
while(cas--)
{ memset(e,0,sizeof(e));
cin>>n>>m>>w;
int a,b,c;node z;
for(int i=0;i<m;i++)
{
cin>>a>>b>>c;a--;b--;
z.v=a;z.c=c;
e[b].push_back(z);z.v=b;
e[a].push_back(z);
}
for(int i=0;i<w;i++)
{
cin>>a>>b>>c;a--;b--;
z.v=b;z.c=-c;
e[a].push_back(z);
}
bool flag=0;
for(int i=0;i<n;i++)
if(spfa(i))
{
flag=1;break;
}
if(flag)cout<<"YES\n";
else cout<<"NO\n";
}
}
}