# Bellman-Ford算法模板题——POJ 3259

Wormholes
Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
Submit
Status
Practice
POJ 3259

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

``````#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int MAXN=10010;
const int INF=1<<30;
int dis;

struct edge
{
int u,v,t;
}E;

bool Bellman_Ford(int n, int m)
{
int i,j;
for(i=1; i<=n; i++) dis[i]=INF;
dis=0;
for(i=1; i<=n-1; i++){
for(j=0; j<m; j++){
if(dis[E[j].u]+E[j].t < dis[E[j].v]) dis[E[j].v] = dis[E[j].u]+E[j].t;
//cout<<E[j].u<<" "<<E[j].v<<" "<<dis[E[j].u]<<" "<<dis[E[j].v]<<endl;
}
}
for(j=0; j<m; j++){
if(dis[E[j].v] > dis[E[j].u]+E[j].t) return 1;
}
return 0;
}

int main()
{
//freopen("in.txt","r",stdin);
int T;
scanf("%d", &T);
while(T--)
{
int n,m,w;
scanf("%d%d%d", &n,&m,&w);
int size=0;
int i,j;
for(i=1; i<=m; i++){
int u,v,t;
scanf("%d%d%d", &u,&v,&t);
E[size].u=u;
E[size].v=v;
E[size].t=t;
size++;
E[size].v=u;
E[size].u=v;
E[size].t=t;
size++;
}
for(i=1; i<=w; i++){
int u,v,t;
scanf("%d%d%d", &u,&v,&t);
E[size].u=u;
E[size].v=v;
E[size].t=-t;
size++;
}
int ok=Bellman_Ford(n,size);
if(ok) printf("YES\n");
else printf("NO\n");
//cout<<m+w<<" "<<size<<endl;
}
return 0;
}``````

原文作者：Bellman - ford算法
原文地址: https://blog.csdn.net/u013351484/article/details/41678335
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