poj3259 Wormholes flord判负环

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 46744		Accepted: 17292

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.

Source

USACO 2006 December Gold

题意：少年在农场里发现了虫洞，虫洞就是能让人穿越回到过去。少年很兴奋，他想通过穿越虫洞来遇见过去的自己，题目会给我们m条路，每条路参数a、b、c代表从a点到b点是连通的并且从a点走到b点花费c秒时间。w条虫洞，参数a、b、c代表点a到点b为虫洞并且可以往前回溯c秒，我们需要判断少年是否能遇见过去的自己

看完题目的时候我还以为时间回溯到过去的时候少年所在的点也是要相同的才行，但是其实没有这么复杂，只需要时间是负的即可。那么就是在这个图中寻找负环的存在

为什么会把判断dis【i】【i】写在 j 循环后面 i 循环里面呢？是应该在更新自身到自身的点后来判断是否存在负环吗？

还不熟悉，多写一点吧。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
using namespace std;
const int maxn = 510;
const int INF = 0x3f3f3f3f;
int n,m,w;
int dis[maxn][maxn];
void init(){
    int i,j;
    for(i=1;i<=n;i++){
        for(j=1;j<=n;j++){
            if(i==j){
                dis[i][j] = 0;
            }else{
                dis[i][j] = INF;
            }
        }
    }
}
bool floyd(){
    int i,j,k;
    for(k=1;k<=n;k++){
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                if(dis[i][j]>dis[i][k]+dis[k][j]){
                    dis[i][j] = dis[i][k]+dis[k][j];
                    //cout<<i<<" "<<j<<" "<<dis[i][j]<<endl;
                }
            }
            if(dis[i][i]<0){
                //flag = true;
                return true;
            }
        }
    }
    return false;
}
int main(){
    int t,i,j;
    scanf("%d",&t);
    while (t--) {
        scanf("%d%d%d",&n,&m,&w);
        init();
        int a,b,c;
        for(i=1;i<=m;i++){
            scanf("%d%d%d",&a,&b,&c);
            dis[a][b] = min(dis[a][b],c);
            dis[b][a] = min(dis[b][a],c);
        }
        for(i=1;i<=w;i++){
            scanf("%d%d%d",&a,&b,&c);
            dis[a][b] = -c;
        }
        
        if(floyd()){
            printf("YES\n");
        }else{
            printf("NO\n");
        }
        
    }
    return 0;
}