BZOJ[Usaco2017 Jan]Promotion Counting——线段树合并

题目描述

The cows have once again tried to form a startup company, failing to remember from past experience t

hat cows make terrible managers!The cows, conveniently numbered 1…N1…N (1≤N≤100,000), organize t

he company as a tree, with cow 1 as the president (the root of the tree). Each cow except the presid

ent has a single manager (its "parent" in the tree). Each cow ii has a distinct proficiency rating, 

p(i), which describes how good she is at her job. If cow ii is an ancestor (e.g., a manager of a man

ager of a manager) of cow jj, then we say jj is a subordinate of ii.

Unfortunately, the cows find that it is often the case that a manager has less proficiency than seve

ral of her subordinates, in which case the manager should consider promoting some of her subordinate

s. Your task is to help the cows figure out when this is happening. For each cow ii in the company, 

please count the number of subordinates jj where p(j)>p(i).

n只奶牛构成了一个树形的公司，每个奶牛有一个能力值pi，1号奶牛为树根。
问对于每个奶牛来说，它的子树中有几个能力值比它大的。

输入

The first line of input contains N

The next N lines of input contain the proficiency ratings p(1)…p(N) 

for the cows. Each is a distinct integer in the range 1…1,000,000,000

The next N-1 lines describe the manager (parent) for cows 2…N 

Recall that cow 1 has no manager, being the president.

n，表示有几只奶牛 n<=100000
接下来n行为1-n号奶牛的能力值pi
接下来n-1行为2-n号奶牛的经理（树中的父亲）

输出

Please print N lines of output. The ith line of output should tell the number of 

subordinates of cow ii with higher proficiency than cow i.

共n行，每行输出奶牛i的下属中有几个能力值比i大

样例输入

5
804289384
846930887
681692778
714636916
957747794
1
1
2
3

样例输出

2
0
1
0
0

 

线段树合并练习题。

其实这道题直接用线段树查询子树区间就能做，但为了练一下线段树合并，所以写了一下线段树合并。

首先因为权值太大要离散化。

每个点动态开点建一棵权值线段树，每个点维护区间权值个数，从根开始dfs，回溯时合并每个点和子节点的线段树，都合并完之后单点查询即可。

#include<set>
#include<map>
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m;
int x;
int tot;
int cnt;
char ch[2];
int v[100010];
int h[100010];
int to[100010];
int ans[100010];
int ls[2000010];
int rs[2000010];
int sum[2000010];
int next[100010];
int head[100010];
int root[100010];
void add(int x,int y)
{
    tot++;
    next[tot]=head[x];
    head[x]=tot;
    to[tot]=y;
}
void insert(int &rt,int l,int r,int k)
{
    if(!rt)
    {
        rt=++cnt;
    }
    if(l==r)
    {
        sum[rt]++;
        return ;
    }
    int mid=(l+r)>>1;
    if(k<=mid)
    {
        insert(ls[rt],l,mid,k);
    }
    else
    {
        insert(rs[rt],mid+1,r,k);
    }
    sum[rt]=sum[ls[rt]]+sum[rs[rt]];
}
void merge(int &x,int y)
{
    if(!x||!y)
    {
        x=x+y;
        return ;
    }
    sum[x]=sum[x]+sum[y];
    merge(ls[x],ls[y]);
    merge(rs[x],rs[y]);
}
int query(int rt,int l,int r,int k)
{
    if(l==r)
    {
        return 0;
    }
    int mid=(l+r)>>1;
    if(k<=mid)
    {
        return query(ls[rt],l,mid,k)+sum[rs[rt]];
    }
    else
    {
        return query(rs[rt],mid+1,r,k);
    }
}
void dfs(int x)
{
    for(int i=head[x];i;i=next[i])
    {
        dfs(to[i]);
        merge(root[x],root[to[i]]);
    }
    ans[x]=query(root[x],1,m,v[x]);
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&v[i]);
        h[i]=v[i];
    }
    sort(h+1,h+n+1);
    m=unique(h+1,h+n+1)-h-1;
    for(int i=2;i<=n;i++)
    {
        scanf("%d",&x);
        add(x,i);
    }
    for(int i=1;i<=n;i++)
    {
        v[i]=lower_bound(h+1,h+1+m,v[i])-h;
        insert(root[i],1,m,v[i]);
    }
    dfs(1);
    for(int i=1;i<=n;i++)
    {
        printf("%d\n",ans[i]);
    }
}