# poj Wormholes（Bellman_ford寻找负权环）

Wormholes

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代码：

``````#include<stdio.h>
#include<string.h>

#define maxn 600+10
#define maxv 6000+10
const int inf=0x3f3f3f3f;
int d[maxn];
struct node
{
int u,v,w;
} E[maxv];
int n,m,s,len;

void Bellman_ford()
{
for(int i=1; i<=n; i++)d[i]=inf;
int flag=1,cnt=0;
while(flag)
{
flag=0;
if(++cnt>n)
{
printf("YES\n");
return ;
}
for(int j=1; j<len; j++)
{
int x=E[j].u,y=E[j].v;
if(d[y]>d[x]+E[j].w)
d[y]=d[x]+E[j].w,flag=1;
}
}
printf("NO\n");
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&s);
int i,x,y,z;
len=1;
for(i=1; i<=m; i++)
{
scanf("%d%d%d",&x,&y,&z);
E[len].u=x,E[len].v=y,E[len].w=z,len++;
E[len].u=y,E[len].v=x,E[len].w=z,len++;
}
for(i=1; i<=s; i++)
{
scanf("%d%d%d",&x,&y,&z);
E[len].u=x,E[len].v=y,E[len].w=-z,len++;
}
Bellman_ford();
}
return 0;
}``````

spfa代码：

``````#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;

#define maxn 500+10
#define maxv 6000+10
const int inf=0x3f3f3f3f;
struct node
{
int u,v,w;
}E[maxv];
int n,m,s,len;
int d[maxn];
int first[maxn],next[maxv];
bool inq[maxn];
int vis[maxn];

bool spfa(int st)
{
memset(inq,0,sizeof(inq));
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)d[i]=inf;
d[st]=0,inq[st]=1;
queue<int>q;
q.push(st);
while(!q.empty())
{
int now=q.front();
q.pop();
inq[now]=0;
for(int i=first[now];i!=-1;i=next[i])
{
int x=E[i].v,y=E[i].w;
if(d[x]>d[now]+y)
{
d[x]=d[now]+y;
if(!inq[x])
{
inq[x]=1;
vis[x]++;
q.push(x);
if(vis[x]>n)
return true;
}
}
}
if(d[st]<0)
return true;
}
return false;
}

{
E[len].u=u,E[len].v=v,E[len].w=w;
next[len]=first[u];
first[u]=len++;
}
void solve()
{
for(int i=1;i<=n;i++)
{
if(spfa(i))
{
printf("YES\n");
return ;
}
}
printf("NO\n");
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(first,-1,sizeof(first));
scanf("%d%d%d",&n,&m,&s);
int u,v,w,i;
len=1;
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&u,&v,&w);
}
for(i=1;i<=s;i++)
{
scanf("%d%d%d",&u,&v,&w);
}
solve();
}
return 0;
}

``````

ps：此题spfa比较慢，还是Bellman_ford比较好，但是测试数据好像比较弱，直接把起点看成1也能过 0.0

原文作者：Bellman - ford算法
原文地址: https://blog.csdn.net/blessLZH0108/article/details/54585913
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