POJ3259 Wormholes[SPFA模版]

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 20567		Accepted: 7314

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：
John的农场里N块地，M条路连接两块地，W个虫洞；路是双向的，虫洞是一条单向路，会在你离开之前把你传送到目的地，
就是当你过去的时候时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。
简化下，就是看图中有没有负权环。

 

 

code:

 

#include<iostream>
#include<queue>
using namespace std;

#define MAXN 510
#define intmax 0x3fffffff

int map[MAXN][MAXN];
int cnt[MAXN];
int vst[MAXN];
int dis[MAXN];
int f,n,m,w;

bool spfa(int st)
{
    int i;
    for(i=0;i<=n;i++)
    {
        vst[i]=0;
        dis[i]=intmax;
        cnt[i]=0;
    }
    int temp;
    queue<int>Que;
    Que.push(st);
    dis[st]=0;
    vst[st]=1;
    cnt[st]=1;
    while(!Que.empty())
    {
        temp=Que.front();
        Que.pop();
        vst[temp]=0;
        for(i=1;i<=n;i++)
        {
            if(dis[i]>dis[temp]+map[temp][i])
            {
                dis[i]=dis[temp]+map[temp][i];
                if(!vst[i])
                {
                    cnt[i]++;
                    Que.push(i);
                    vst[i]=1;
                    if(cnt[i]>=n)
                        return false;
                }
            }
        }
    }
    return true;
}

int main()
{
    int i,j;
    int s,e,t;
    scanf("%d",&f);
    while(f--)
    {
        scanf("%d%d%d",&n,&m,&w);
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                map[i][j]=intmax;
        for(i=1;i<=m;i++)
        {
            scanf("%d%d%d",&s,&e,&t);
            if(t<map[s][e])
            {
                map[s][e]=t;
                map[e][s]=t;
            }
        }
        for(i=1;i<=w;i++)
        {
            scanf("%d%d%d",&s,&e,&t);
            map[s][e]=-t;
        }
        if(spfa(1))
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}