F - Wormholes POJ - 3259--SPFA 算法判断负环是否存在

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.

思路以及原理；

在 SPFA算法中，每次松弛的时候，会吧初始点的访问下表变为0，如果图里面存在环的话，SPFA算法是无法结束的，利用这个思维，在一个有n个点的图中，如果不存在自身环下某一个点顶多被所有其他的点相连，这样的话，这个点顶多进队列n-1次，如果存在环，这个点进队的次数一定会大于n-1，利用这个原理，进行一遍SPFA就可以得到答案了

算法 如下

#include <iostream>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
# define inf 0x3f3f3f3f
using namespace std;
int vis[1200];
int dist[1200];
int num[1200];
int n,m,w;
struct node
{
    int id;
    int wi;
} a;
vector<node>v[1200];
int SPFA()
{
    memset(vis,0,sizeof(vis));
    memset(dist,inf,sizeof(dist));
    memset(num,0,sizeof(num));  // 桶排计数的原理
    vis[2]=0;
    dist[2]=0;
    queue<int>q;
    q.push(2);
    num[2]++;
    while(!q.empty())
    {
        int b=q.front();
          if(num[b]>=n)
                {
                    return 1;
                }
        q.pop();
        vis[b]=0;
        for(int i=0; i<v[b].size(); i++)
        {
            int id=v[b][i].id;
            int wi=v[b][i].wi;
            if(dist[id]>dist[b]+wi)
            {
                dist[id]=dist[b]+wi;
                if(!vis[id])
                {
                    vis[id]=1;
                    q.push(id);
                    num[id]++;
                }
            }
        }
    }
    return 0;
}
int main ()
{
    int t;
    int x,y,z;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d %d",&n,&m,&w);
        for(int i=0;i<=n;i++)
        v[i].clear();
        for(int i=0; i<m; i++)
        {
            scanf("%d %d %d",&x,&y,&z);
            a.id=y;
            a.wi=z;
            v[x].push_back(a);
            a.id=x;
            a.wi=z;
            v[y].push_back(a);
        }
        for(int i=0; i<w; i++)
        {
            scanf("%d %d %d",&x,&y, &z);
            a.id=y;
            a.wi=-z;
            v[x].push_back(a);
        }
        int flag=SPFA();
        if(flag) printf("YES\n");
        else  printf("NO\n");
    }
}