poj 3259 Wormholes ([kuangbin带你飞]专题四 最短路练习)

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.

题目大意：

农夫 FJ 有 N 块田地【编号 1...n】 (1<=N<=500)
        田地间有 M 条路径 【双向】(1<= M <= 2500)
        同时有 W 个孔洞,可以回到以前的一个时间点【单向】(1<= W <=200)
        问：FJ 是否能在田地中遇到以前的自己

算法：flod算法

思路： 田地间的双向路径加边,权值为正
        孔洞间的单向路径加边,权值为负【可以回到以前】
        判断有向图是否存在负环
      因为如果存在了负数环,时间就会不停的减少,
      那么 FJ 就可以回到以前更远的地方,肯定能遇到以前的自己的

<pre name="code" class="html">#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn = 510;
const int inf = 9999999;
struct Node{
    int u,v;
    int t;
}E[2500*2+200+10];
int d[maxn];
int c;
int n,m,w;
bool flod(){
    for(int i=1;i<maxn;i++) d[i]=inf;		
    d[1]=0;			//到某个点的时间 
    for(int i=1;i<n;i++){
            bool flag = false;

        for(int j=0;j<c;j++){
            int u=E[j].u;
            int v=E[j].v;
            int t=E[j].t;
            if(d[v]>d[u]+t){		//g跟新每个点时间 
                d[v]=d[u]+t;
                flag = true;
            }
        }
        if(!flag) return false;		//如果不能跟新则不存在负边 
    }
    for(int i=0;i<c;i++){
        if(d[E[i].v]>d[E[i].u]+E[i].t) return true;
    }
    return false;
}



int main(){
    int f;
    scanf("%d",&f);
    while(f--){

        scanf("%d%d%d",&n,&m,&w);
        c=0;
        for(int i=0;i<m;i++){		//m条双向正边 
            int u,v,t;
            scanf("%d%d%d",&u,&v,&t);
            E[c].u=u;
            E[c].v=v;
            E[c].t=t;
            c++;
            E[c].u=v;
            E[c].v=u;
            E[c].t=t;
            c++;
        }
        for(int i=0;i<w;i++){		//w条单向负边 
            int u,v,t;
            scanf("%d%d%d",&u,&v,&t);
            E[c].u=u;
            E[c].v=v;
            E[c].t=-t;
            c++;
        }
        if(flod()){
            printf("YES\n");
        }
        else printf("NO\n");
    }
    return 0;
}