| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 30169 | Accepted: 10914 |
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
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
Source
AC代码：
#include<iostream>
using namespace std;
struct Point{
int s,e,t;
}a[10000];
int se;
int n,m,w;
int bell_man(int start){
int dis[10000];
for(int i=1;i<=n;i++)
dis[i]=999999;
dis[start]=0;
for(int i=1;i<n;i++)
for(int j=0;j<se;j++)
dis[a[j].e] = dis[a[j].e] > dis[a[j].s] + a[j].t ? dis[a[j].s] + a[j].t : dis[a[j].e];
for(int i=0;i<se;i++){
if(dis[a[i].e] > dis[a[i].s] + a[i].t)
return 1;
}
return 0;
}
int main(){
int T; cin>>T;
while(T--){
se=0;
cin>>n>>m>>w;
for(int i=0;i<m;i++){
int s,e,t;
cin>>s>>e>>t;
a[se].s=s; a[se].e=e; a[se++].t=t;
a[se].s=e; a[se].e=s; a[se++].t=t;
}
for(int i=0;i<w;i++){
int s,e,t;
cin>>s>>e>>t;
a[se].s=s; a[se].e=e; a[se++].t=-t;
}
//int k;
//for(k=1;k<=n;k++){ //其实正确的起点应该要历遍所有点，但是这样的超时了
//这个题目只要1点就可以了，算是题目的一个很大漏洞吧，数据太水了
if(bell_man(1)){
cout<<"YES"<<endl;
//break;
}
//}
//if(k>n)
else
cout<<"NO"<<endl;
}
return 0;
}
poj 3259(bellman最短路径),布布扣,bubuko.com
原文：http://blog.csdn.net/my_acm/article/details/37996185