spfa首发 POJ 3259

Wormholes

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 backT 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：判断有没有负环

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <math.h>
#include <queue>
#include <algorithm>
using namespace std;
#define N 5210
#define INF 0xfffffff

int cnt, dist[N], Head[N], num[N], vis[N];
int n, m, w;

struct Edge
{
    int v, w, next;
}e[N];

void Add(int u, int v, int w)
{
    e[cnt].v = v;
    e[cnt].w = w;
    e[cnt].next = Head[u];
    Head[u] = cnt++;
}

bool spfa()//spfa模板；
{
    memset(vis, 0, sizeof(vis));
    memset(num, 0, sizeof(num));
    queue<int>Q;
    vis[1] = 1;
    dist[1] = 0;
    Q.push(1);
    num[1]++;
    while(Q.size())
    {
        int p=Q.front();
        Q.pop();
        vis[p] = 0;
        for(int i=Head[p]; i!=-1; i=e[i].next)
        {
            int q = e[i].v;
            if(dist[q] > dist[p] + e[i].w)
            {
                dist[q] = dist[p] + e[i].w;
                if(!vis[q])
                {
                    vis[q] = 1;
                    Q.push(q);
                    num[q] ++;
                    if(num[q]>n)
                        return true;
                }
            }
        }
    }
    return false;
}

int main()
{
    int T, a, b, c;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d %d %d",&n,&m,&w);

        cnt = 0;
        memset(Head, -1, sizeof(Head));
        for(int i=1; i<=n; i++)
            dist[i] = INF;

        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d", &a, &b, &c);
            Add(a, b, c);
            Add(b, a, c);
        }
        for(int i=1; i<=w; i++)
        {
            scanf("%d%d%d", &a, &b, &c);
            Add(a, b, -c);
        }

        if( spfa() )//存在负环；
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}