POJ 3259 Wormholes （floyd或者spfa）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 58636		Accepted: 21921

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

 

题意有点神经病，就是一个人有f个农场，农场里有n个坑，m个正常坑，w个虫洞，虫洞可以让牛回到t秒前，问这个农场存不存在可以让牛无限回到过去的情况（就是负环）。那么说白了判负环是否存在。最直接的思路就是spfa跑一遍，松弛的点大于等于n就特判跳出。（有人floyd也ac了，一看时间限制2s，确实可以，但是跑出来比spfa慢很多）。

//#include<bits/stdc++.h>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cmath>
#include<iostream>
using namespace std;
#define maxn 3005
#define INF 99999999
typedef pair<int,int> pii;
vector<pii> e[maxn];
int d[maxn],vis[maxn],cnt[maxn];
int f,n,m,w,ans;

void init()
{
    for(int i=0; i<maxn; i++) e[i].clear();
    for(int i=0; i<maxn; i++) d[i]=INF;
    memset(vis,0,sizeof(vis));
    memset(cnt,0,sizeof(cnt));
}

void addedge(int x,int y,int z)
{
    e[x].push_back(make_pair(y,z));
    // e[y].push_back(make_pair(x,z));
}

void spfa(int s)
{
    queue<int> q;
    d[s]=0;
    vis[s]=1;
    cnt[s]++;
    q.push(s);
    while(!q.empty())
    {
        int now=q.front();
        q.pop();
        vis[now]=0;
        for(int i=0; i<e[now].size(); i++)
        {
            int v=e[now][i].first;
            int w=e[now][i].second;
            if(d[v]>d[now]+w)
            {
                d[v]=d[now]+w;
                if(!vis[v])
                {
                    vis[v]=1;
                    cnt[v]++;
                    q.push(v);
                    if(cnt[v]>=n)
                    {
                        ans=1;
                        return;
                    }
                }
            }
        }
    }
}



int main()
{
    scanf("%d",&f);
    while(f--)
    {
        init();
        ans=0;
        scanf("%d %d %d",&n,&m,&w);
        while(m--)
        {
            int x,y,z;
            scanf("%d %d %d",&x,&y,&z);
            addedge(x,y,z);
            addedge(y,x,z);
        }
        while(w--)
        {
            int x,y,z;
            scanf("%d %d %d",&x,&y,&z);
            addedge(x,y,-z);
        }
        spfa(1);
        if(ans)
            printf("YES
");
        else
            printf("NO
");
    }


    return 0;
}