# POJ3259 Wormholes SPFA 或者 bellman_ford

Wormholes

 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 37539 Accepted: 13818

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.

``````#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
const int INF = (1<<30);
using namespace std;

struct Node
{
int v,w,nt;
}f[100009];

int cnt;
int used[100009];
int inq[100009];
int dis[100009];
queue<int>q;

{
f[cnt].v=u;
f[cnt].w=w;
}

int spfa(int n)
{
for(int i=1;i<=n;i++)
dis[i]=INF;
dis[1]=0;

q.push(1);
used[1]++;

while(!q.empty())
{
int x=q.front(); q.pop();
inq[x]=0;
{
if(dis[f[u].v] > dis[x]+f[u].w)//如果可松弛则进行操作，否则不动
{
dis[f[u].v] = dis[x]+f[u].w;
if(!inq[f[u].v])
{
inq[f[u].v]=1;
if(++used[f[u].v]>n-1) return 0;
q.push(f[u].v);
}
}
}
}
return 1;
}

void init(int n)
{
cnt=0;
for(int i=1;i<=n;i++)
{
used[i]=0;
inq[i]=0;
}
}

int main()
{
int T;
int u,v,w;
int n,m,l;

scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&l);

init(n);

for(int i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
}

for(int i=0;i<l;i++)
{
scanf("%d%d%d",&u,&v,&w);
}

if(!spfa(n)) puts("YES");
else puts("NO");
}

return 0;
}

``````

``````#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
const int INF = (1<<26);

struct Node
{
int u,v,w;
}f[10009];

int cnt;
int dis[10009];

{
f[cnt].u=u;
f[cnt].v=v;
f[cnt++].w=w;
}

int bellman_ford(int n)
{
for(int i=1;i<=n;i++) dis[i]=INF;

dis[1]=0;
int u,v,w;

for(int i=0;i<n-1;i++)
for(int j=0;j<cnt;j++)
{
u=f[j].u;
v=f[j].v;
w=f[j].w;

if(dis[u]<INF && dis[v]>dis[u]+w)
{
dis[v]=dis[u]+w;
}
}

for(int j=0;j<cnt;j++)
{
u=f[j].u;
v=f[j].v;
w=f[j].w;

if(dis[u]<INF && dis[v]>dis[u]+w)
{
return 0;
}
}
return 1;
}

int main()
{
int T;
scanf("%d",&T);
int n,m,l;
while(T--)
{
scanf("%d%d%d",&n,&m,&l);
cnt=0;
int u,v,w;

for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&u,&v,&w);
}

for(int i=0;i<l;i++)
{
scanf("%d%d%d",&u,&v,&w);
}

if(bellman_ford(n)==0) puts("YES");
else puts("NO");

}
return 0;
}
``````

原文作者：Bellman - ford算法
原文地址: https://blog.csdn.net/wust_ZJX/article/details/48916355
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