POJ3259 Wormholes（链式前向星+SPFA判断负环）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 83995		Accepted: 31141

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

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.

题目大意：

有n(1 ≤ n ≤ 500)个农场，m(1 ≤ m ≤ 2500)条路, w(1 ≤ w ≤ 200)个虫洞，虫洞是单向的，而路径始终是双向的（例如有(5, 6), 则必有(6, 5)）。

问农夫是否能够通过一些路与虫洞，使得农夫能够看到以前的自己（时间倒流）。

思路：把每个顶点都入一次队，判断图中是否存在负环即可。

之前把重复的边也处理了，导致了WA多发（太惨了

删掉以后就A了

代码：

#include <iostream>
#include <queue>
#include <cstring>
#include <vector>
#include <set>
#include <algorithm>
#include <string>
#include <cstdio>
#include <cstdlib>
using namespace std;
#define MAXN 2503
#define INF 0x3f3f3f3f

int dis[MAXN];
int n, m, wei;

int h[4 * MAXN], w[4 * MAXN], e[4 * MAXN], ne[4 * MAXN], cnt[MAXN], idx;
bool vis[MAXN];

inline int read() {
    int s = 0, w = 1;
    char ch = getchar();
    while (ch<'0' || ch>'9') { if (ch == '-')w = -1; ch = getchar(); }
    while (ch >= '0'&&ch <= '9') s = s * 10 + ch - '0', ch = getchar();
    return s * w;
}
void add(int u, int v, int wei){
    e[idx] = v; w[idx] = wei; ne[idx] = h[u]; h[u] = idx ++;
    
}


bool spfa(){
    memset(cnt, 0, sizeof cnt);
    memset(vis, 0, sizeof vis);
    int cur;
    queue<int> q;
    for(int i = 1; i <= n; ++ i){
        q.push(i);
        vis[i] = true;
        cnt[i] ++;
    }
    while(!q.empty()){
        cur = q.front();
        q.pop();
        vis[cur] = false;
        for(int i = h[cur]; i != -1; i = ne[i]){
            int j = e[i];
            if(dis[j] > dis[cur] + w[i]){
                dis[j] = dis[cur] + w[i];
                
                cnt[j] ++;
                if(cnt[j] > n)
                    return true;
                if(!vis[j]){
                    q.push(j);
                    vis[j] = true;
                }
            }
        }
    }
    return false;
}

int main(){
    int t, S, E, T, W;
    for(cin >> t; t --;){
        idx = 0;
        memset(h, -1, sizeof h);
        cin >> n >> m >> W;
        for(int i = 0; i < m; ++ i){
            S = read();
            E = read();
            wei = read();
            add(S, E, wei);
            add(E, S, wei);
        }
        for(int i = 0; i < W; ++ i){
            S = read();
            E = read();
            wei = read();
            add(S, E, -wei);
        }
        
        cout << (spfa()?"YES":"NO") << endl;
                                                                                                                                                                            
    }
    return 0;
}