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

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[20000];
int d[20000],vis[20000],cnt[20000];
int ttt,n,m,w;

int spfa()
{
    memset(vis,0,sizeof(vis));
    memset(cnt,0,sizeof(cnt));
    d[1]=0;
    queue<int> q;
    q.push(1);
    cnt[1]++;
    vis[1]=1;
    while(!q.empty())
    {
        int x=q.front();    
        vis[x]=0;
        q.pop();
        int sz=g[x].size();
        for(int i=0; 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]=1;
                    cnt[y]++;
                    if(cnt[y]>n)
                    {
                        return 1;
                    }
                }
            }
        }
    }
    return 0;
}

int main()
{
    scanf("%d",&ttt);
    while(ttt--)
    {
        scanf("%d%d%d",&n,&m,&w);
        for(int i=1;i<=n;i++)
        {
            g[i].clear();
        }
        for(int i=1; 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=1; 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=1; i<=n; i++)
        {
            d[i]=inf;
        }
        int ans=spfa();
        if(ans)
        {
            printf("YES
");
        }
        else
        {
            printf("NO
");
        }
    }
}