HDU 2665 Kth number(划分树）

Kth number

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

Total Submission(s): 37 Accepted Submission(s): 25

Problem Description Give you a sequence and ask you the kth big number of a inteval.

Input The first line is the number of the test cases.  For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere.  The second line contains n integers, describe the sequence.  Each of following m lines contains three integers s, t, k.  [s, t] indicates the interval and k indicates the kth big number in interval [s, t]

Output For each test case, output m lines. Each line contains the kth big number.

Sample Input 1 10 1 1 4 2 3 5 6 7 8 9 0 1 3 2

Sample Output 2

Source HDU男生专场公开赛——赶在女生之前先过节（From WHU）

Recommend zty

/*----------------------------------------------
File:
Date: 2017/6/9 21:40:22
Author: LyuCheng
----------------------------------------------*/

#include <bits/stdc++.h>
#define MAXN 100005

using namespace std;

int t,n;
int q,l,r,k;
/************************************划分树模板************************************/
int a[MAXN];       //原数组
int sorted[MAXN];  //排序好的数组
//是一棵树，但把同一层的放在一个数组里。
int num[][MAXN];   //num[i] 表示i前面有多少个点进入左孩子
int val[][MAXN];   //20层,每一层元素排放，0层就是原数组
void build(int l,int r,int ceng)
{
    if(l==r) return ;
    int mid=(l+r)/,isame=mid-l+;  //isame保存有多少和sorted[mid]一样大的数进入左孩子
    for(int i=l;i<=r;i++) if(val[ceng][i]<sorted[mid]) isame--;
    int ln=l,rn=mid+;   //本结点两个孩子结点的开头，ln左
    for(int i=l;i<=r;i++)
    {
        if(i==l) num[ceng][i]=;
        else num[ceng][i]=num[ceng][i-];
        if(val[ceng][i]<sorted[mid] || val[ceng][i]==sorted[mid]&&isame>)
        {
            val[ceng+][ln++]=val[ceng][i];
            num[ceng][i]++;
            if(val[ceng][i]==sorted[mid]) isame--;
        }
        else
        {
            val[ceng+][rn++]=val[ceng][i];
        }
    }
    build(l,mid,ceng+);
    build(mid+,r,ceng+);
}

int look(int ceng,int sl,int sr,int l,int r,int k)
{
    if(sl==sr) return val[ceng][sl];
    int ly;  //ly 表示l 前面有多少元素进入左孩子
    if(l==sl) ly=;  //和左端点重合时
    else ly=num[ceng][l-];
    int tolef=num[ceng][r]-ly;  //这一层l到r之间进入左子树的有tolef个
    if(tolef>=k)
    {
        return look(ceng+,sl,(sl+sr)/,sl+ly,sl+num[ceng][r]-,k);
    }
    else
    {
        // l-sl 表示l前面有多少数，再减ly 表示这些数中去右子树的有多少个
        int lr = (sl+sr)/ +  + (l-sl-ly);  //l-r 去右边的开头位置
        // r-l+1 表示l到r有多少数，减去去左边的，剩下是去右边的，去右边1个，下标就是lr，所以减1
        return look(ceng+,(sl+sr)/+,sr,lr,lr+r-l+-tolef-,k-tolef);
    }
}
/************************************划分树模板************************************/

int main(int argc, char *argv[]){

    // freopen("in.txt","r",stdin);
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&q);
        for(int i=;i<=n;i++){
            scanf("%d",&val[][i]);
            sorted[i]=val[][i];
        }
        sort(sorted+,sorted+n+);
        build(,n,);
        while(q--){
            scanf("%d%d%d",&l,&r,&k);
            printf("%d\n",look(,,n,l,r,k));
        }
    }
    return ;
}