Wormholes POJ

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 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.

解题思路：

本题的本质就是判断一个图是否存在负环

spfa用队列，不用优先队列
还有就是次数的问题：
我们都知道spfa算法是对bellman算法的优化，
那么如何用spfa算法来判断负权回路呢？我们考虑一个节点入队的条件是什么，
只有那些在前一遍松弛中改变了距离估计值的点，才可能引起他们的邻接点的距离估计值的改变。
因此，用一个先进先出的队列来存放被成功松弛的顶点。同样，我们有这样的定理：“两点间如果有最短路，
那么每个结点最多经过一次。也就是说，这条路不超过n-1条边。”
（如果一个结点经过了两次，那么我们走了一个圈。如果这个圈的权为正，显然不划算；如果是负圈，
那么最短路不存在；如果是零圈，去掉不影响最优值）。也就是说，每个点最多入队n-1次
（这里比较难理解，需要仔细体会，n-1只是一种最坏情况，实际中，这样会很大程度上影响程序的效率）。 

 

#include <iostream>
#include <string.h>
#include <queue>
#include <cstdio>
using namespace std;
#define N 50005
#define INF 0x3f3f3f
int dis[N];
int cnt[N];
int n,m,w,k,head[N];
int vis[N];
struct edge
{
    int v,w;
    int next;
}e[N];
void add(int u,int v,int w)
{
    e[k].v=v;
    e[k].w=w;
    e[k].next=head[u];
    head[u]=k;
    k++;
}
int spfa(int u)
{
    memset(vis,0,sizeof(vis));
    memset(cnt,0,sizeof(cnt));
    memset(dis,INF,sizeof(dis));
    queue<int>q;
    dis[u]=0;
    cnt[u]++;
    vis[u]=1;
    q.push(u);
    while(!q.empty())
    {
        int k=q.front();
        q.pop();
        vis[k]=0;
        for(int i=head[k];i!=-1;i=e[i].next)
        {
            int v=e[i].v;
            int w=e[i].w;
            if(dis[k]+w<dis[v])
            {
                dis[v]=dis[k]+w;
                if(!vis[v])
                {
                    vis[v]=1;
                    q.push(v);
                    cnt[v]++;
                    if(cnt[v]>=n)//如果入队次数大于等于n，那么就说明存在负环
                    {
                        return 1;
                    }
                }
            }
        }

    }
    return 0;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int flag=0;
        memset(head,-1,sizeof(head));
        k=0;
        cin>>n>>m>>w;
        for(int i=1; i<=m; i++)
        {
            int u,v,w;
            cin>>u>>v>>w;
            add(u,v,w);
            add(v,u,w);
        }
        for(int i=1;i<=w;i++)
        {
            int u,v,w;
            cin>>u>>v>>w;
            add(u,v,-w);
        }
        for(int i=1;i<=n;i++)
        {
            if(spfa(i)==1)
            {
                flag=1;
                break;
            }
        }
        if(flag==1)
        {
            cout<<"YES"<<endl;
        }else
        {
            cout<<"NO"<<endl;
        }
    }
    return 0;
}