POJ-3259 Wormholes （SPFA 判定负环）

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描述

传送门：POJ-3259 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.

输入描述

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.

输出描述

Lines 1..F: For each farm, output “YES” if FJ can achieve his goal, otherwise output “NO” (do not include the quotes).

示例

输入

1
2
3
4
5
6
7
8
9
10
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

输出

1
2
NO
YES

题解

题目大意

给定F(1<=F<=5)组数据，对应与一个farm，每组数据对应一个结果。
对于其中一组数据，给定 N , M , W ，N为点数，M为无向边数(普通路径，通过之后时间前进)，W为有向边数(虫洞，通过虫洞之后时间倒退)，求每组数据是否存在一条回路能使时间倒退。

思路

若存在这样一条回路，则图中肯定存在负环。只要判断是否有负环就行了。

代码

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#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <cstdio>
#include <queue>
#include <cmath>
const int MAXN = 5*1e3, INF = 0x3f3f3f3f;
using namespace std;
int n, m, w, x, y, z;
int vis[MAXN], cnt[MAXN], dis[MAXN];
vector<pair<int, int> >E[MAXN];

bool SPFA(){
    queue<int> que;
    que.push(1);
    dis[1] = 0;
    vis[1] = 1;
    cnt[1]++;
    while(!que.empty()){
        int now = que.front();
        que.pop();
        vis[now] = 0;
        for(int i = 0; i < E[now].size(); i++){
            int v = E[now][i].first;
            if(dis[v] > dis[now] + E[now][i].second){
                dis[v] = dis[now] + E[now][i].second;
                if(!vis[v]){
                    que.push(v);
                    vis[v] = 1;
                    cnt[v]++;
                    if(cnt[v] >= n){
                        return true;
                    }
                }
            }
        }
    }
    return false;
}

int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        memset(vis, 0, sizeof(vis));
        memset(cnt, 0, sizeof(cnt));
        memset(dis, INF, sizeof(dis));
        for(int i = 0; i <= MAXN; i++) E[i].clear();
        scanf("%d %d %d", &n, &m, &w);
        for(int i = 0; i < m; i++){
            scanf("%d %d %d", &x, &y, &z);
            E[x].push_back(make_pair(y, z));
            E[y].push_back(make_pair(x, z));
        }
        for(int i = 0; i < w; i++){
            scanf("%d %d %d", &x, &y, &z);
            E[x].push_back(make_pair(y, -z));
        }
        if(SPFA()) printf("YES\n");
        else printf("NO\n");
    }
}