POJ3259 Wormholes(SPFA判断负环)

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.

题意:农夫约翰沉迷于守望屁股,现在他的农场里有一些双向道路和单向虫洞,穿过一条道路需要Ti的时间,进入虫洞则可以带你回到Ti时间之前,请问约翰是否可以在他出发之前回到他出发的农场(默认从1出发???),守望到他的屁股?

题解:很明显的一道spfa判负环

然后也没什么好说的了,WA到哭只因忘清空vector

代码如下:

#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;

vector< pair<int,int> > g[];
int d[],vis[],cnt[];
int ttt,n,m,w;

int spfa()
{
    memset(vis,,sizeof(vis));
    memset(cnt,,sizeof(cnt));
    d[]=;
    queue<int> q;
    q.push();
    cnt[]++;
    vis[]=;
    while(!q.empty())
    {
        int x=q.front();
        vis[x]=;
        q.pop();
        int sz=g[x].size();
        for(int i=; i<sz; i++)
        {
            int y=g[x][i].first;
            int w=g[x][i].second;
            if(d[x]+w<d[y])
            {
                d[y]=d[x]+w;
                if(!vis[y])
                {
                    q.push(y);
                    vis[y]=;
                    cnt[y]++;
                    if(cnt[y]>n)
                    {
                        return ;
                    }
                }
            }
        }
    }
    return ;
}

int main()
{
    scanf("%d",&ttt);
    while(ttt--)
    {
        scanf("%d%d%d",&n,&m,&w);
        for(int i=;i<=n;i++)
        {
            g[i].clear();
        }
        for(int i=; i<=m; i++)
        {
            int f,t,c;
            scanf("%d%d%d",&f,&t,&c);
            g[f].push_back(make_pair(t,c));
            g[t].push_back(make_pair(f,c));
        }
        for(int i=; i<=w; i++)
        {
            int f,t,c;
            scanf("%d%d%d",&f,&t,&c);
            g[f].push_back(make_pair(t,-c));
        }
        for(int i=; i<=n; i++)
        {
            d[i]=inf;
        }
        int ans=spfa();
        if(ans)
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
}