Easy Tree Problem

A tree structure is very special structure, there is exactly one path connecting each pair of nodes.
Now, given a tree structure, which has N nodes(conveniently labeled from 1 to N). And the root of the tree is always labeled with 1. You are to write a program to figure out that, for every node V in the tree, how many nodes there are in the sub-tree rooted by V and it’s label number is larger than the label number of V.



For the example above:
Ans[1] = 6,Ans[2] = 1,Ans[3] = 2,Ans[4] = 0,
Ans[5] = 0, Ans[6] = 0,Ans[7] = 0

 

Input

 

There are multiple cases.The first line is an integer T representing the number of test cases.The following lines every tow lines representing a test case. For each case there are exactly two lines:The first line with a single integer N(1 <= N <= 200000), representing the size of tree.The second line with N – 1 numbers: P[2], P[3], ……P[n]. (1 <= P[i] <= N),Which mean the father node of node i is P[i].It is guaranteed that the input data is a tree structure and has 1 as root.

 

Output

 

For each test case, output a line of N numbers in the following format:
Case #C: Ans[1] Ans[2] Ans[3] …… Ans[N]

 

Sample Input

 

2
7
1 1 3 2 1 3
4
1 2 3

 

Sample Output

 

Case #1: 6 1 2 0 0 0 0
Case #2: 3 2 1 0

 

Source

 

The 5th ACM Programming Contest of HUST

// 对每个父亲节点 给他左边界标号 len ， 有边界 为len+n ;n 为他的儿子数
// 对于他的儿子标号为len+1,len+2...len+n
// 然后从最大节点开始查找答案，然后更新树状数组
#include<cmath>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define maxn 50003
#define LL long long
using namespace std ;
int head[maxn] , next1[maxn] , to[maxn] ;
int ans[maxn] , len ,xt[maxn+10];
int L[maxn] , R[maxn] ;
int low( int x )
{
  return x&(-x) ;
}
void update( int x )
{
    while( x <= maxn-2 )
    {
        xt[x]++ ;
        x += low(x) ;
    }
}
int sum( int x )
{
    int ans = 0;
    while( x > 0 )
    {
        ans += xt[x] ;
        x -= low(x) ;
    }
    return ans ;
}
void dfs(int u  )
{
	int i , v ;
	L[u] = ++len ;
	for( i = head[u] ; i != -1 ;i = next1[i])
	{
       dfs(to[i]) ;
	}
	R[u] = len ;
}
int main(){
    int i , j , case1 = 0, T , top ;
	int  n ;
   cin >> T ;
   while(T--)
   {
        scanf("%d" , &n ) ;
		memset(ans,0,sizeof(ans)) ;
		memset(head,-1,sizeof(head)) ;
		top = len = 0 ;
		for( i = 2 ; i <= n ;i++ )
		{
			scanf("%d" , &j ) ;
			next1[top] = head[j] ;
			to[top] = i ;head[j] = top++ ;
		}
		memset(xt,0,sizeof(xt)) ;
		dfs(1) ;
		for( i = n ; i >= 1 ;i-- )
		{
			ans[i] = sum(R[i]) - sum(L[i]-1) ;
			update(L[i]) ;
		}
		printf("Case #%d:",++case1 ) ;
		for( i = 1 ; i <= n ;i++ )
			printf(" %d",ans[i]) ;
		puts("") ;
   }
}