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K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 69053		Accepted: 24471
Case Time Limit: 2000MS

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.

Source

Northeastern Europe 2004, Northern Subregion

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　　分析：　　

　　没错，这是主席树的模板，但是这里我们也可以用整体二分来做（shui）。

　　关于整体二分，这个博主也才刚学，知识点什么的也不多讲，只放模板代码，关于知识点可以参考一下这个julao的博客。

　　Code：

//It is made by HolseLee on 5th Oct 2018
//POJ2104
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;

const int N=2e5+7;
int n,m,ans[N],cnt,c[N];
struct Node {
    int x,y,k,pos,type;
    Node() {}
    Node(const int _x,const int _y,const int _k,const int _p,const int _t):
        x(_x), y(_y), k(_k), pos(_p), type(_t) {}
}q[N],q1[N],q2[N];

inline int read()
{
    char ch=getchar(); int num=0; bool flag=false;
    while( ch<'0' || ch>'9' ) {
        if( ch=='-' ) flag=true; ch=getchar();
    }    
    while( ch>='0' && ch<='9' ) {
        num=num*10+ch-'0'; ch=getchar();
    }
    return flag ? -num : num;
}

inline int lowbit(int x)
{
    return x&(-x);
}

inline void add(int pos,int x)
{
    for(; pos<=n; pos+=lowbit(pos)) c[pos]+=x;
}

inline int quary(int pos)
{
    int ret=0;
    for(; pos>0; pos-=lowbit(pos)) ret+=c[pos];
    return ret;
}

void solve(int l,int r,int L,int R)
{
    if( l>r || L>R ) return;
    if( l==r ) {
        for(int i=L; i<=R; ++i) 
        if( q[i].type ) ans[q[i].pos]=l;
        return ;
    }
    int mid=(l+r)>>1, cnt1=0, cnt2=0;
    for(int i=L; i<=R; ++i) 
    if( q[i].type ) {
        int tmp=quary(q[i].y)-quary(q[i].x-1);
        if( tmp>=q[i].k ) q1[++cnt1]=q[i];
        else q[i].k-=tmp, q2[++cnt2]=q[i];
    } else {
        if( q[i].x<=mid ) q1[++cnt1]=q[i], add(q[i].pos,1);
        else q2[++cnt2]=q[i];
    }
    for(int i=1; i<=cnt1; ++i) 
    if(!q1[i].type) add(q1[i].pos,-1);
    for(int i=1; i<=cnt1; ++i) q[L+i-1]=q1[i];
    for(int i=1; i<=cnt2; ++i) q[L+cnt1+i-1]=q2[i];
    solve(l,mid,L,L+cnt1-1), solve(mid+1,r,L+cnt1,R);
}

int main()
{
    n=read(), m=read();
    int x,y,z;
    for(int i=1; i<=n; ++i) 
        x=read(), q[++cnt]=Node(x,1,0,i,0);
    for(int i=1; i<=m; ++i) {
        x=read(), y=read(), z=read();
        q[++cnt]=Node(x,y,z,i,1);
    }
    solve(-inf,inf,1,cnt);
    for(int i=1; i<=m; ++i) 
        printf("%d
",ans[i]);
    return 0;
}