poj 3259 Wormholes
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 31422 | Accepted: 11429 |
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 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
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 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.
这题就是判存不存在负圈，可以用bellman-ford或者spfa，这里为了练习spfa。注意一下，在题目没有说明图连通的情况下，必须对每个点进行spfa判负圈，实际上可以优化，即每做一次spfa，dist！=inf，这些点不用继续spfa了。可以证明，如果这些点包含负圈，那么在这次spfa中就可以判出存在负圈，既然还要往下做，显然不存在负圈。
代码：
#include<cstdio>
#include<iostream>
#include<cstring>
#define Maxe 6000
#define Maxv 510
using namespace std;
struct line{
int to,w;
line(){}
line(int t,int ww):to(t),w(ww){}
}edge[Maxe];
int head[Maxv],nxt[Maxe],f[Maxv];
int q[Maxv],dist[Maxv],vis[Maxv],cnt[Maxv];
const int inf=1<<30;
bool spfa(int u,int n){
memset(vis,0,sizeof vis);
memset(cnt,0,sizeof cnt);
for(int i=1;i<=n;i++) dist[i]=inf;
dist[u]=0,cnt[u]++;
int s=0,e;
q[e=1]=u;
bool flag=true;
while(s!=e&&flag){
int v=q[s=(s+1)%Maxv];
vis[v]=0;
for(int i=head[v];i!=-1;i=nxt[i]){
int r=edge[i].to,w=edge[i].w;
if(dist[v]+w<dist[r]){
dist[r]=dist[v]+w;
if(++cnt[r]>=n) {flag=false;break;}
if(!vis[r]) {q[e=(e+1)%Maxv]=r;vis[r]=1;}
}
}
}
if(!flag) return false;
for(int i=1;i<=n;i++)
if(dist[i]!=inf) f[i]=true;
return true;
}
int main()
{
int t,n,m,s,fr,to,w;
scanf("%d",&t);
while(t--){
memset(head,-1,sizeof head);
scanf("%d%d%d",&n,&m,&s);
for(int i=0;i<m+s;i++){
scanf("%d%d%d",&fr,&to,&w);
w=i>=m?-w:w;
edge[i+m]=line(to,w),nxt[i+m]=head[fr],head[fr]=i+m;
if(i<m) edge[i]=line(fr,w),nxt[i]=head[to],head[to]=i;
}
memset(f,0,sizeof f);
int k=1;
for(;k<=n;k++)
if(!f[k]){
if(!spfa(k,n)) break;
}
if(k<=n) puts("YES");
else puts("NO");
}
return 0;
}