poj 3259 Wormholes

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal 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: 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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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.

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 510using 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;}