HDU 3887 Counting Offspring

Counting Offspring

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

You are given a tree, it’s root is p, and the node is numbered from 1 to n. Now define f(i) as the number of nodes whose number is less than i in all the succeeding nodes of node i. Now we need to calculate f(i) for any possible i.

 

Input

Multiple cases (no more than 10), for each case:
The first line contains two integers n (0<n<=10^5) and p, representing this tree has n nodes, its root is p.
Following n-1 lines, each line has two integers, representing an edge in this tree.
The input terminates with two zeros.

 

Output

For each test case, output n integer in one line representing f(1), f(2) … f(n), separated by a space.

 

Sample Input

15 7 7 10 7 1 7 9 7 3 7 4 10 14 14 2 14 13 9 11 9 6 6 5 6 8 3 15 3 12 0 0

 

Sample Output

0 0 0 0 0 1 6 0 3 1 0 0 0 2 0

 

Author

bnugong

 

Source

2011 Multi-University Training Contest 5 - Host by BNU

 

   树状数组，父亲节点i 时，统计比 i 小的个数 ，递归回退到 i 时，再次统计比 i 小的个数。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <vector>
#include <set>
#include <algorithm>
#include <vector>
#include <stack>
#include <math.h>
#include <stdlib.h>
#include <fstream>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#pragma comment(linker, "/STACK:16777216")
using namespace std;
typedef long long LL ;
const int size=1000008 ;
struct Edge{
    int v ;
    int next ;
};
Edge edge[size*2] ;
int id ;
int C[size] ;
int vec[size] ;
int ans[size] ;
bool visited[size] ;
struct Me{
    int N ;
    int root ;
    Me(){} ;
    Me(int n,int r):N(n),root(r){
         fill(C,C+1+N,0) ;
         id=0 ;
         fill(vec,vec+1+N,-1) ;
         fill(visited,visited+1+N,0) ;
         fill(ans,ans+1+N,0) ;
    };
    inline void add_edge(int u ,int v){
        edge[id].v=v ;
        edge[id].next=vec[u] ;
        vec[u]=id++ ;
    }
    inline int lowbit(int x){
        return x&(-x) ;
    }
    void insert(int id ,int x){
        while(id<=N){
            C[id]+=x ;
            id+=lowbit(id) ;
        }
    }
    int get_sum(int id){
        int sum=0 ;
        while(id>=1){
            sum+=C[id] ;
            id-=lowbit(id) ;
        }
        return sum ;
    }
    void dfs(int now){
       visited[now]=1 ;
       int less_sum=get_sum(now-1) ;
       insert(now,1) ;
       for(int e=vec[now];e!=-1;e=edge[e].next){
            int son=edge[e].v ;
            if(visited[son])
                continue ;
            dfs(son) ;
       }
       ans[now]=get_sum(now-1)-less_sum ;
    }
    void gao(){
       int u ,v ;
       for(int i=1;i<N;i++){
           scanf("%d%d",&u,&v) ;
           add_edge(u,v) ;
           add_edge(v,u) ;
       }
       dfs(root) ;
       printf("%d",ans[1]) ;
       for(int i=2;i<=N;i++)
          printf(" %d",ans[i]) ;
       puts("") ;
    }
};
int main(){
   int n , p ;
   while(scanf("%d%d",&n,&p)){
       if(n==0&&p==0)
           break  ;
       Me me(n,p) ;
       me.gao() ;
   }
   return 0;
}