HDU 3887 Counting Offspring(DFS序求子树权值和）

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

题意：求以x为根节点的子树权值和。
题解：求得dfs序之后，就是简单的求和问题，BIT/线段树均可。
因为题目是求比当前x小的，所以我们要从大的开始更新。不想手动模拟请加栈

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int INF = 0x3f3f3f3f;
const int maxn=1e5+100;
int L[maxn],R[maxn];
int c[maxn*2+10];
vector<int>v[maxn];
int ans[maxn];
int n,rt,dfn;
void dfs(int u,int pre)
{
    L[u]=++dfn;
    for(int i=0;i<v[u].size();i++)
    {
        int to=v[u][i];
        if(to==pre) continue;
        dfs(to,u);
    }
    R[u]=++dfn;
}
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int val)
{
    while(x<=dfn)
    {
        c[x]+=val;
        x+=lowbit(x);
    }
}
int query(int x)
{
    int sum=0;
    while(x>0)
    {
        sum+=c[x];
        x-=lowbit(x);
    }
    return sum;
}
int main()
{
    int x,y;
    while(~scanf("%d%d",&n,&rt),n+rt)
    {
        CLEAR(L,0);
        REP(i,n+1)
          v[i].clear();
        for(int i=0;i<n-1;i++)
        {
            scanf("%d%d",&x,&y);
            v[x].push_back(y);
            v[y].push_back(x);
        }
        dfn=0;
        dfs(rt,-1);
        CLEAR(c,0);
        for(int i=1;i<=dfn;i++)
            update(i,1);
        for(int i=n;i>=1;i--)
        {
            ans[i]=(query(R[i]-1)-query(L[i]))/2;
            update(R[i],-1);
            update(L[i],-1);
        }
        REPF(i,1,n)
            printf(i==n?"%d\n":"%d ",ans[i]);
    }
    return 0;
}