poj3259 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..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.

题意：N块地。M条路。W个虫洞。推断有没有能够是时间倒流的路径。注意把W次输入的w[i]改成-w[i]。

SPFA:推断有没有入队次数超过N的点。

特别提醒：

不要相信给出的数据范围。这是个大坑！。測试数据绝对有超过非常多的，我开了20000的数组。

#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>
#include<iostream>
#include<list>
#include<set>
#include<cmath>
#define INF 0x3f // 一个大数
#define eps 1e-6
using namespace std;
#define maxn 20000 //数组开大些。
#define Maxn 20000
int n,m,W;
int first[maxn];
int next[maxn];
int u[Maxn];
int v[Maxn];
int w[Maxn];
int dis[maxn];
int go[maxn];
int vist[maxn];
int conut[maxn];
int ttt;
queue<int> Q;
int spfa(int x)
{
    vist[x]=1;
    memset(dis,INF,sizeof(dis));
    dis[x]=0;
    Q.push(x);
    conut[x]++;
    while(!Q.empty())
    {
        int t=Q.front();
        Q.pop();
        vist[t]=0;
        conut[t]++;
        for(int i=first[t];i!=-1;i=next[i])
        {
            if(dis[t]+w[i]<dis[v[i]])
            {
                dis[v[i]]=dis[t]+w[i];
                if(!vist[v[i]])
                {
                    if(conut[t]>=n) //假设某点入队超过n次，则可判定存在负环。。
。。
                      return 1;
                    Q.push(v[i]);
                    ++conut[v[i]];

                }
            }
        }
    }
    return 0;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        memset(vist,0,sizeof(vist));
        memset(conut,0,sizeof(conut));
        while(!Q.empty())
            Q.pop();
          //  cout<<"二货@"<<endl;
        scanf("%d%d%d",&n,&m,&W);
        for(int i=1;i<=n;i++)
            first[i]=-1;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&u[i],&v[i],&w[i]);
            u[i+m]=v[i];
            v[i+m]=u[i];
            w[i+m]=w[i];
        }
        for(int i=m*2+1;i<=2*m+W;i++)
        {
             scanf("%d%d%d",&u[i],&v[i],&w[i]);
             w[i]=-w[i];
        }
        for(int i=1;i<=2*m+W;i++)
           {
               next[i]=first[u[i]];
               first[u[i]]=i;
           }
         int qq=spfa(1);
         if(qq)
            puts("YES");
         else
            puts("NO");
    }
    return 0;
}