洛谷 P2850 [USACO06DEC]虫洞Wormholes

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。

输入输出格式

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

输入样例#1：
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
输出样例#1：
NO
YES
说明

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.

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【分析】
spfa判负环
一个点入队次数>n，那么有负环

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【代码】

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mxn=2505;
bool flag;
int T,n,m,w,cnt;
int head[505],dis[505],hello[505];
bool vis[505];
struct edge {int to,next,dis;} f[mxn*2]; 
inline void add(int u,int v,int d)
{
    f[++cnt].to=v,f[cnt].dis=d,f[cnt].next=head[u],head[u]=cnt;
}
inline void spfa()
{
    memset(dis,0x3f,sizeof dis);
    queue <int> q;
    q.push(1);
    hello[1]++;
    dis[1]=0;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=head[u];i;i=f[i].next)
        {
            int v=f[i].to;
            if(dis[v]>dis[u]+f[i].dis)
            {
                dis[v]=dis[u]+f[i].dis;
                if(!vis[v])
                {
                    vis[v]=1,hello[v]++,q.push(v);
                    if(hello[v]>n) {flag=1;return;} 
                }
            }
        }
    }
}
int main()
{
    int i,j,u,v,d;
    scanf("%d",&T);
    while(T--)
    {
        cnt=flag=0;
        M(head);M(vis);M(hello);
        scanf("%d%d%d",&n,&m,&w);
        fo(i,1,m)
        {
            scanf("%d%d%d",&u,&v,&d);
            add(u,v,d),add(v,u,d);
        }
        fo(i,1,w)
        {
            scanf("%d%d%d",&u,&v,&d);
            add(u,v,-d);
        }
        spfa(); 
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}