【洛谷 2850】虫洞

题目描述

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.

John在他的农场中闲逛时发现了许多虫洞。虫洞可以看作一条十分奇特的有向边，并可以使你返回到过去的一个时刻（相对你进入虫洞之前）。John的每个农场有M条小路（无向边)连接着N （从1..N标号）块地，并有W个虫洞。其中1<=N<=500,1<=M<=2500,1<=W<=200。 现在John想借助这些虫洞来回到过去（出发时刻之前），请你告诉他能办到吗。 John将向你提供F(1<=F<=5)个农场的地图。没有小路会耗费你超过10000秒的时间，当然也没有虫洞回帮你回到超过10000秒以前。

输入格式

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.

输出格式

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

输入输出样例

输入 #1复制

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复制

NO
YES

说明/提示

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.

 

题解：spfa，负权环简单应用恩。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>
using namespace std;
#define maxn 501
#define min(a,b) (a<b?a:b)
#define I inline
#define R register
#define inf 0x7fffff
using namespace std;
int n,m,x,y,z,w,F;
queue<int> q;
int dis[maxn],cnt[maxn],so[maxn][maxn];
bool f[maxn];
int spfa(int s){
    memset(cnt,0,sizeof(cnt));
    memset(dis,0x3f,sizeof(dis));
    q.push(s); f[s]=true; dis[s]=0;
    while(!q.empty()){
        int x=q.front(); q.pop(); f[x]=false;
        for(int i=1;i<=n;i++){
            if(dis[x]+so[x][i]<dis[i]){
                dis[i]=dis[x]+so[x][i];
                if(!f[i]){
                    q.push(i); cnt[i]++;
                    if(cnt[i]==n)
                        return 1;
                    f[i]=true;
                }
            }
            if(dis[1]<0) return 1;
        }
    }
    return 0;
}
int main(){
    freopen("2850.in","r",stdin);
    freopen("2850.out","w",stdout);
    scanf("%d",&F);
    while(F--){
        scanf("%d %d %d",&n,&m,&w);
        memset(so,0x3f,sizeof(so));
        for(int i=1;i<=m;i++){
            scanf("%d %d %d",&x,&y,&z);
            so[x][y]=min(z,so[x][y]);
            so[y][x]=so[x][y];
        }
        for(int i=1;i<=w;i++){
            scanf("%d%d%d",&x,&y,&z);
            so[x][y]=min(-z,so[x][y]);
        }
        if(spfa(1)) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}