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

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;
}

View Code