POJ3259-Wormholes(SPFA 判负环)

Wormholes
Time Limit: 2000MS       Memory Limit: 65536K

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..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.

题解

英文题目我就不翻译了，都是学霸。

题目大意：有一张图，包含负权，判断有无负环。

直接跑SPFA判负环。

代码

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int MAXM=25000+10;
const int MAXN=5000+10;
struct edge{int to,next,v;}e[MAXM*2];
int n,m,x,head[MAXN],cnt,dis[MAXN],num[MAXN];
bool vis[MAXN];
inline void add(int u,int v,int w)
{
cnt++;
e[cnt].to=v;
e[cnt].v=w;
e[cnt].next=head[u];
head[u]=cnt;
}
bool spfa(int s)
{
memset(dis,0x3f,sizeof(dis));
memset(vis,0,sizeof(vis));
memset(num,0,sizeof(num));
queue<int>q;
while(!q.empty()) q.pop();
dis[s]=0;
vis[s]=true;
num[s]=1;
q.push(s);
while(!q.empty())
{
int t=q.front();
q.pop();
vis[t]=false;
for(int i=head[t];i;i=e[i].next)
{
int temp=e[i].to;
if(dis[temp]>dis[t]+e[i].v)
{
dis[temp]=dis[t]+e[i].v;
if(!vis[temp])
{
num[temp]++;
if(num[temp]>=n) return true;
vis[temp]=true;
q.push(temp);
}
}
}
}
return false;
}
int main()
{
int _;
scanf("%d",&_);
while(_--)
{
cnt=0;
memset(head,0,sizeof(head));
scanf("%d%d%d",&n,&m,&x);
for(int i=1;i<=m;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);add(v,u,w);
}
for(int i=1;i<=x;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w*-1);
}
bool flag=false;
for(int i=1;i<=n;i++)
{
if(spfa(i))
{
flag=true;
break;
}
}
if(flag) printf("YES\n");
else printf("NO\n");
}
return 0;
}