poj3259Wormholes

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 37403 Accepted: 13757

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.

Source

判断有没有负环，用spfa算法，当然用贝尔曼福特算法也行。

附代码：

#include<stdio.h>#include<algorithm>#include<iostream>using namespace std;#include<queue>#define INF 0x3f3f#include<string.h>#define MAXN 2000#define MAXM 20010struct Edge{ int from,to,val,next;}edge[MAXM];int dist[MAXN],vis[MAXN],head[MAXN],used[MAXN];int n,m,top=0,k,flag;void init() {    memset(vis,0,sizeof(vis));    memset(dist,INF,sizeof(dist));    memset(head,-1,sizeof(head));    memset(used,0,sizeof(used));}void addEdge(int u,int v,int w) {    edge[top].from=u;    edge[top].to=v;    edge[top].val=w;    edge[top].next=head[u];    head[u]=top++;}int spfa(int sx){    flag=0;    queue<int>Q;    Q.push(sx);    dist[sx]=0;    vis[sx]=1;    used[sx]=1;    while(!Q.empty() )    {        int u=Q.front();        Q.pop() ;        vis[u]=0;        for(int i=head[u];i!=-1;i=edge[i].next )        {            int v=edge[i].to ;            if(dist[v]>dist[u]+edge[i].val )            {                dist[v]=dist[u]+edge[i].val ;                if(!vis[v])                {                    vis[v]=1;                    Q.push(v);                     used[v]++;                    if(used[v]>n)                    {                        flag=1;                        return 1;                    }                }            }        }    }    return 0;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        init();        top=0;        int u,v,w;    scanf("%d%d%d",&n,&m,&k);    while(m--)    {        scanf("%d%d%d",&u,&v,&w);         addEdge(u,v,w);        addEdge(v,u,w);    }       while(k--)    {        scanf("%d%d%d",&u,&v,&w);        addEdge(u,v,-w);    }      int s=spfa(1);      if(s)      printf("YES\n");      else      printf("NO\n");    }    return 0;}