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  • POJ Wormholes (SPFA)

    http://poj.org/problem?id=3259

    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..NM (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, FF farm descriptions follow. 
    Line 1 of each farm: Three space-separated integers respectively: NM, and W
    Lines 2.. M+1 of each farm: Three space-separated numbers ( SET) 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.. MW+1 of each farm: Three space-separated numbers ( SET) 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.
     
     

    SPFA判断是否有负权,如果一个点进入队列的次数达到总点数则说明有负权

    dist[i]数组记录源点到i的最短路径,与Dijsktar不同的是dist[i]多次更新

    use[i]记录i点进入队列的次数,即dist[i]被更新的次数;

    vis[i]标记i点是否进入队列

    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<math.h>
    #include<vector>
    #include<queue>
    #define INF 0xffffff
    #define N 520
    using namespace std;
    
    struct node
    {
        int e, w;
    };
    
    vector<node>G[N];
    int n, use[N], dist[N];
    bool vis[N];
    
    void Init()
    {
        int i;
        memset(vis, false, sizeof(vis));
        memset(use, 0, sizeof(use));
        for(i = 0 ; i <= n ; i++)
        {
            G[i].clear();
            dist[i] = INF;
        }
    }
    
    int SPFA(int s)
    {
        queue<node>Q;
        node now, next;
        int i, len;
        now.e = s;
        now.w = 0;
        dist[s] = 0;
        Q.push(now);
        vis[s] = true;
        use[now.e]++;
        while(!Q.empty())
        {
            now = Q.front();
            Q.pop();
            vis[now.e] = false;
    
            len = G[now.e].size();
            for(i = 0 ; i < len ; i++)
            {
                next = G[now.e][i];
                if(dist[next.e] > dist[now.e] + next.w)
                {
                    dist[next.e] = dist[now.e] + next.w;
                    use[next.e]++;
                    if(use[next.e] >= n)
                        return 1;
                    if(!vis[next.e])
                    {
                        vis[next.e] = true;
                        Q.push(next);
                    }
                }
            }
        }
        return 0;
    }
    
    int main()
    {
        int T, m, w, s, e, t, i;
        node p;
        scanf("%d", &T);
        while(T--)
        {
    
            scanf("%d%d%d", &n, &m, &w);
            Init();
            for(i = 1 ; i <= m ; i++)
            {
                scanf("%d%d%d", &s, &e, &t);
                p.w = t;
                p.e = s;
                G[e].push_back(p);
                p.e = e;
                G[s].push_back(p);
            }
            for(i = 1 ; i <= w ; i++)
            {
                scanf("%d%d%d", &s, &e, &t);
                p.w = -t;
                p.e = e;
                G[s].push_back(p);
            }
            if(SPFA(1))
                printf("YES
    ");
            else
                printf("NO
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/qq2424260747/p/4663564.html
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