hdu 2545 求当前结点到根节点的距离

求当前结点到根节点的距离

Sample Input
2 1 //n m
1 2
1 2 //询问

5 2
1 2
1 3
3 4
3 5
4 2 //询问
4 5
0 0

Sample Output
lxh
pfz
lxh

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <cmath>
# include <queue>
# define LL long long
using namespace std ;

const int MAXN = 100010;
int F[MAXN];
int dist[MAXN] ;

int find(int x)//找x的祖先结点
{
    if(F[x]==x) return x;
    int t = F[x] ;
    F[x]=find(F[x]);
    dist[x] += dist[t] ;
    return F[x] ;
}

int main()
{
    //freopen("in.txt","r",stdin) ;
    int n , m ;
    while(scanf("%d %d", &n , &m) != EOF)
    {
         if (n == 0 && m == 0)
             break ;
         int u , v ;
         int i ;
         for(i = 1 ; i <= n ; i++)
         {
             F[i] = i ;
             dist[i] = 0 ;
         }
         for(i = 1 ; i < n ; i++)
         {
             scanf("%d %d" , &u , &v) ;
             F[v] = u ;
             dist[v] = 1 ;
         }

        while(m--)
        {
            scanf("%d %d" , &u , &v) ;
            find(u) ;
            find(v) ;
            if (dist[u] <= dist[v])
                printf("lxh
") ;
            else
                printf("pfz
") ;
        }

    }
    return 0;
}