poj 3259 spfa

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Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 58013		Accepted: 21697

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：从位置1到位置1若存在负环就输出YES，否则输出NO。

题解：跑一遍spfa，若到点1的距离为负就退出输出YES，否则输出NO，spfa模板题。

AC代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>

using namespace std;

const int maxn = 510;
const int INF = 0x3f3f3f3f;

typedef pair <int,int> P;
vector <P> E[maxn];
int n,m,w;
int inq[maxn],d[maxn];

void init()
{
    for(int i=0;i<maxn;i++) E[i].clear();
    for(int i=0;i<maxn;i++) inq[i]=0;
    for(int i=0;i<maxn;i++) d[i]=INF;
}

int spfa()
{
    queue <int> que;
    que.push(1);
    d[1]=0; inq[1]=1;
    while(!que.empty())
    {
        int now=que.front(); que.pop();
        inq[now]=0;
        for(int i=0;i<E[now].size();i++)
        {
            int v=E[now][i].first;
            if(d[v]>d[now]+E[now][i].second)
            {
                d[v]=d[now]+E[now][i].second;
                if(d[1]<0) return d[1];
                if(inq[v]==1) continue;
                inq[v]=1;
                que.push(v);
            }
        }
    }
    return d[1];
}

int main()
{
    int t;
    int u,v,a;
    scanf("%d",&t);
    while(t--)
    {
        init();
        scanf("%d%d%d",&n,&m,&w);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&u,&v,&a);
            E[u].push_back(P(v,a));
            E[v].push_back(P(u,a));
        }
        for(int i=1;i<=w;i++)
        {
            scanf("%d%d%d",&u,&v,&a);
            E[u].push_back(P(v,-a));
        }
        if(spfa()>=0) puts("NO");
        else puts("YES");
    }
    return 0;
}