# POJ 3259：Wormholes bellman_ford判定负环

Wormholes

 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 37906 Accepted: 13954

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..NM (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.

Bellman_ford模板题，用来对每一条边都进行松弛，然后看最后结果是否依然能够松弛。如果还能松弛，说明有负环；如果不能松弛了，就是没有负环。

``````#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

struct E{
int s;
int e;
int l;
}edge[5205];

int N,M,W,edge_num;
int dis[505];

{
edge_num++;

edge[edge_num].s=start;
edge[edge_num].e=end;
edge[edge_num].l=len;
}

bool bellman_ford()
{
int i,j;
for(i=1;i<=N-1;i++)
{
int flag=0;
for(j=1;j<=edge_num;j++)
{
if(dis[edge[j].e]>dis[edge[j].s]+edge[j].l)
{
flag=1;
dis[edge[j].e]=dis[edge[j].s]+edge[j].l;
}
}
if(flag==0)
break;
}
for(j=1;j<=edge_num;j++)
{
if(dis[edge[j].e]>dis[edge[j].s]+edge[j].l)
{
return true;
}
}
return false;
}

int main()
{
int i,start,end,len;
int Test;
cin>>Test;

while(Test--)
{
edge_num=0;
memset(dis,0,sizeof(dis));

cin>>N>>M>>W;

for(i=1;i<=M;i++)
{
cin>>start>>end>>len;

}
for(i=1;i<=W;i++)
{
cin>>start>>end>>len;

}
if(bellman_ford())
{
cout<<"YES"<<endl;
}
else
{
cout<<"NO"<<endl;
}
}
return 0;
}``````
原文作者：Bellman - ford算法
原文地址: https://blog.csdn.net/u010885899/article/details/49473187
本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。