# POJ 3259 虫洞 （BellMan-Ford）

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

``````#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int inf = 0x3f3f3f3f;
struct P
{
int from,to,c;
} a[2555][3];
int d[555];
int vis[555];
bool bm(int n,int m,int w)
{
int i,j,p;
fill(d,d+n+1,inf);
memset(vis,0,sizeof(vis));
vis[1]=1;
d[1]=0;
for(i=1; i<=n; i++)
{
p=1;
for(j=1; j<=m+w; j++)
{
if(j>m) p=2;
P e = a[j][p];
if(vis[e.from] && d[e.to]>d[e.from]+e.c)
{
vis[e.to]=1;
d[e.to]=d[e.from]+e.c;
if(i==n) return true;
}
if(p==1)
{
if(vis[e.to] && d[e.from]>d[e.to]+e.c)
{
vis[e.from]=1;
d[e.from]=d[e.to]+e.c;
if(i==n) return true;
}
}
}
}
return false;
}
int main()
{
int T,n,m,w;
int i,j;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&w);
for(i=1; i<=m; i++) scanf("%d%d%d",&a[i][1].from,&a[i][1].to,&a[i][1].c);
for(; i<=w+m; i++) scanf("%d%d%d",&a[i][2].from,&a[i][2].to,&a[i][2].c), a[i][2].c=-a[i][2].c;
if(bm(n,m,w)) 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/h1021456873/article/details/65033012
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