poj3259 Wormholes(spfa)

就是给出一些点边关系， 有虫洞的就是负值。 注意普通道路是双向边。。一开始这里建错图了

之后处理用了spfa ，类似模板的一道题

 

 

题目：

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 26848		Accepted: 9669

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.

Source

USACO 2006 December Gold

 

 

代码：

#include <iostream>
#include <vector>
#include <cstring>
#include <queue>
#include <cstdio>
using namespace std;
#define LEN  500+10
#define INF 10000*501
#define MP(x,y) make_pair(x,y)
typedef pair<int,int> pii;

vector<pii> G[LEN];
int N,M,W;
int dis[LEN];
void init()
{
    cin>>N>>M>>W;
    for(int i=1;i<=N;i++)
        G[i].clear();

    int s,e,t;
    for(int i=0;i<M;i++)
    {
        cin>>s>>e>>t;
        G[s].push_back( MP(e,t));
        G[e].push_back( MP(s,t));
    }
    for(int i=0;i<W;i++)
    {
        cin>>s>>e>>t;
        G[s].push_back( MP(e,-t));
    }
}

bool spfa(int s)
{
    queue<int> q;
    int vis[LEN] = {0}, cnt[LEN] ={0};
    for(int i=1;i<=N;i++)dis[i] = INF;

    dis[s]=0;
    q.push(s);
    vis[s] = 1;
    cnt[s]++;

    while(!q.empty())
    {
        int v = q.front(); q.pop();
        for(int i=0;i<G[v].size();i++)
        {
            int  x = G[v][i].first , d = G[v][i].second;

            if( dis[x] > dis[v]+ d)
            {
                dis[x] = dis[v]+d;
                if(!vis[x])
                {
                    q.push(x);
                    vis[x] = 1;
                    cnt[x] ++;
                    if( cnt[x]>N-1)return true;
                }
            }
        }
        vis[v]= 0;
    }
    return false;
}
int main()
{
    int F;
    cin>>F;
    while(F--)
    {
        init();

        if( spfa(1) || dis[1]<0)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}