poj 3259 Wormholes

题意：就是有N个，M条道路，W条时间隧道，问是否能从某一点出经过一些道路和时间隧道后再次到达该顶点并且回到该点过去

思路：最短路，只要有负权回路即可判断能回到过去

注意，道路是双向的所以有2500*2条，再加上单向的W200条，贡献一次RE

/*
    spfa()
    2011-8-16
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <climits>
#include <algorithm>
#include <functional>
#include <cstdlib>
#include <queue>
#include <vector>
#include <stack>
#include <cmath>
#define nMax 505
#define mMax 5505
#define INF INT_MAX
using namespace std;
struct Edge
{
    int u, v, w;
}edge[mMax];
int head[nMax], dist[nMax], cnt[nMax];
bool used[nMax];
int e, n, m, w;

void BuildGraph(int from, int to, int weight)
{
    edge[e].v=to;
    edge[e].w=weight;
    edge[e].u=head[from];
    head[from]=e++;
}

void spfa()
{
    memset(used, 0, sizeof(used));
    memset(cnt, 0, sizeof(cnt));
    queue<int> que;

    for(int i=2; i<=n; i++) dist[i]=INF;
    dist[1]=0;
    used[1]=1;

    que.push(1);
    while(!que.empty())
    {
        int u=que.front();
        que.pop();
        used[u]=0;

        for(int i=head[u]; i!=-1; i=edge[i].u)
        {
            int v=edge[i].v;
            if(edge[i].w+dist[u]<dist[v])
            {
                dist[v]=edge[i].w+dist[u];
                if(!used[v])
                {
                    used[v]=1;
                    que.push(v);
                    cnt[v]++;
                    if(cnt[v]>n)
                    {
                        puts("YES");
                        return ;
                    }
                }
            }
        }
    }
    puts("NO");
}

int main()
{
 int ntime;
 scanf("%d", &ntime);
 while(ntime--)
 {
     int s, o, t;
     e=0;
     memset(head, -1, sizeof(head));
     scanf("%d%d%d", &n, &m, &w);
     //cout<<" "<<n<<" "<<m<<" "<<w<<endl;
     for(int i=0; i<m+w; i++)
     {
         scanf("%d%d%d", &s, &o, &t);
         if(i>=m) BuildGraph(s, o, -t);
         else
         {
             BuildGraph(s, o, t);
             BuildGraph(o, s, t);
         }
     }
     spfa();
 }
 return 0;
}