# poj 3259 bellman-ford算法 判断是否存在负权回路

Wormholes

 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 41838 Accepted: 15364

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.

Source

USACO 2006 December Gold

bellman-ford的模版～

``````#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;

#define INF 0x3f3f3f3f
#define N 1100
int n,m,w,tol;
struct node
{
int s,e,t;
}p;

int bellman()
{
int i,j,dis[N];
for(i=1;i<=n;i++)
{
dis[i]=(i==1)?0:INF;
}

for(i=0;i<=n-1;i++)
{
int f=0;
for(j=0;j<tol;j++)
{
if(dis[p[j].e]>dis[p[j].s]+p[j].t)
{
dis[p[j].e]=dis[p[j].s]+p[j].t;
f=1;
}
}
if(f==0)
break;
}

for(i=0;i<tol;i++)
{
if(dis[p[i].e]>dis[p[i].s]+p[i].t)
return 1;
}
return 0;
}

int main()
{
int T,a,b,c,i;
scanf("%d",&T);
while(T--)
{
tol=0;
scanf("%d%d%d",&n,&m,&w);
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&a,&b,&c);
p[tol].s=a;
p[tol].e=b;
p[tol].t=c;
tol++;
p[tol].s=b;
p[tol].e=a;
p[tol].t=c;
tol++;
}
for(i=1;i<=w;i++)
{
scanf("%d%d%d",&a,&b,&c);
p[tol].s=a;
p[tol].e=b;
p[tol].t=-c;
tol++;
}
if(bellman())
printf("YES\n");
else
printf("NO\n");
}
return 0;
}``````

原文作者：Bellman - ford算法
原文地址: https://blog.csdn.net/qq_33406883/article/details/51548153
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