POJ-3259.Wormholes(SPFA算法)

我会带着你远行 2022-02-09 08:39 197阅读 0赞

# [3259.Wormholes][] #

### 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.

### spfa算法 ###

这道题是 判断图中是否有负环图，所以spfa算法比较合适的，spfa算法可以检验是否存在负环图，我们设定一个队列，来保存被拓展的点，当有点的入队次数大于等于n时，说明图中存在负环图。

#include <iostream>
    #include <algorithm>
    #include <vector>
    #include <queue>
    using namespace std;
    
    const int maxv = 1000;
    const int INF = 0x3f3f3f3f;
    
    struct Node{  
        int v;
        int dist;
        Node(int v,int d){ 
            this->v = v;
            this->dist = d;
        }
    };
    
    vector<Node> G[maxv]; // 邻接表
    int dist[maxv]; // 记录最短路径
    int ranks[maxv]; // 记录每个点的入队次数
    bool book[maxv]; // 标记被拓展过的点
    
    bool spfa(int s,int n){ 
        fill(dist,dist + n + 1,INF);
        fill(ranks,ranks + n + 1,0);
        fill(book,book + n + 1,false);
        queue<int> que;
        ranks[s] = 1;
        book[s] = true;
        dist[s] = 0;
        que.push(s);
        while(!que.empty()){ 
            int u = que.front();
            que.pop();
            book[u] = false; 
            for(size_t i = 0;i < G[u].size();i++){ 
                Node e = G[u][i];
                if(dist[e.v] > dist[u] + e.dist){ 
                    dist[e.v] = dist[u] + e.dist;
                    if(!book[e.v]){ 
                        que.push(e.v);
                        book[e.v] = true;
                        ranks[e.v]++; // 当有点入队次数等于n时，说明有负环
                        if(ranks[e.v] >= n ) return false; 
                    }
                }
            }
        }
        return true;
    }
    int main(){ 
        //freopen("test.txt","r",stdin);
        int t,n,m,w;
        cin >> t;
        while(t--){ 
            cin >> n >> m >> w;
            for(int i = 0;i <= n; i++) G[i].clear();
            for(int i = 0;i < m; i++){ 
                int u, v, d;
                cin >> u >> v >> d;
                G[u].push_back(Node(v,d));
                G[v].push_back(Node(u,d));
            }
            for(int i = 0; i < w; i++){ 
                int u, v, d;
                cin >> u >> v >> d;
                G[u].push_back(Node(v,-d));
            }
            /* for(int i = 1; i <= n; i++){ */
            /* G[n+1].push_back(Node(i,0)); */
            /* } */
            if(spfa(1,n)) puts("NO");
            else puts("YES");
        }
        return 0;
    }

[3259.Wormholes]: http://poj.org/problem?id=3259