POJ2104主席树模板题

完成新成就——B站上看了算法https://www.bilibili.com/video/av4619406/?from=search&seid=17909472848554781180#page=2

K-th Number

Description

You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment. 
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?" 
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

Input

The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000). 
The second line contains n different integer numbers not exceeding 10⁹ by their absolute values --- the array for which the answers should be given. 
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).

Output

For each question output the answer to it --- the k-th number in sorted a[i...j] segment.

Sample Input

7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

Sample Output

5
6
3

Hint

This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

题意很简单，求n个数，m个操作，以及区间第K大数。

我们考虑1个算法，建n颗权值线段树，第i颗表示前i个数中的所有取值个数（离散化一下），这样我们可以用第r棵子树减去第l-1颗子树，得到l~r中各个值的取值情况。这是我们就可以二分第k大值在哪个区间了，如果左边子数的size比k不小于的话往左找，否则往右。

关于多棵树的问题就交给主席树了。

顺便空间开40倍

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#define maxn 100006
using namespace std;
int n,m,cnt,a[maxn],root[maxn],x,y,k;
struct note{
    int l,r,sum;
}T[maxn*40];
vector<int> v;
int getid(int x){return lower_bound(v.begin(),v.end(),x)-v.begin()+1;}

void update(int l,int r,int &x,int y,int pos)
{
    T[++cnt]=T[y];T[cnt].sum++;x=cnt;
    if(l==r) return;
    int mid=(l+r)>>1;
    if(pos<=mid) update(l,mid,T[x].l,T[y].l,pos);else
    update(mid+1,r,T[x].r,T[y].r,pos);
}
int query(int l,int r,int x,int y,int k)
{
    if(l==r) return l;
    int mid=(l+r)/2;
    int sum=T[T[y].l].sum-T[T[x].l].sum;
    if(sum<k)return query(mid+1,r,T[x].r,T[y].r,k-sum);else
    return query(l,mid,T[x].l,T[y].l,k);
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]),v.push_back(a[i]);
    sort(v.begin(),v.end()),v.erase(unique(v.begin(),v.end()),v.end());
    for(int i=1;i<=n;i++) update(1,n,root[i],root[i-1],getid(a[i]));
    for(int i=1;i<=m;i++)
    {
        scanf("%d %d %d",&x,&y,&k);
        printf("%d\n",v[query(1,n,root[x-1],root[y],k)-1]);
    }
    return 0;
 }