poj 2104 K-th Number（划分树）

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K-th Number

Time Limit: 20000MS Memory Limit: 65536KTotal Submissions: 43361 Accepted: 14314Case 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 31 5 2 6 3 7 42 5 34 4 11 7 3

Sample Output

563

给定一个数组 求给定区间内第k大的数

划分树板子。。

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#include <vector>#include <queue>#define MEM(a,x) memset(a,x,sizeof a)#define eps 1e-8#define MOD 10009#define MAXN 100010#define MAXM 100010#define INF 99999999#define ll __int64#define bug cout<<"here"<<endl#define fread freopen("ceshi.txt","r",stdin)#define fwrite freopen("out.txt","w",stdout)using namespace std;int tree[20][MAXN];//表示每层每个位置的值int sorted[MAXN];//已经排序好的数int toleft[20][MAXN];//toleft[p][i]表示第i层从1到i有数分到左边void build(int l,int r,int dep){    if(l==r) return;    int mid=(l+r)/2;    int same=mid-l+1;//表示等于中间值而且被分入左边的个数    for(int i=l;i<=r;i++)        if(tree[dep][i]<sorted[mid])            same--;    int lpos=l;    int rpos=mid+1;    for(int i=l;i<=r;i++)    {        if(tree[dep][i]<sorted[mid])            tree[dep+1][lpos++]=tree[dep][i];        else if(tree[dep][i]==sorted[mid]&&same>0)        {            tree[dep+1][lpos++]=tree[dep][i];            same--;        }        else            tree[dep+1][rpos++]=tree[dep][i];        toleft[dep][i]=toleft[dep][l-1]+lpos-l;    }    build(l,mid,dep+1);    build(mid+1,r,dep+1);}//查询区间第K大的数 [L,R]是大区间 [l,r]是要查询的小空间int query(int L,int R,int l,int r,int dep,int k){    if(l==r) return tree[dep][l];    int mid=(L+R)/2;    int cnt=toleft[dep][r]-toleft[dep][l-1];    if(cnt>=k)    {        int newl=L+toleft[dep][l-1]-toleft[dep][L-1];        int newr=newl+cnt-1;        return query(L,mid,newl,newr,dep+1,k);    }    else    {        int newr=r+toleft[dep][R]-toleft[dep][r];        int newl=newr-(r-l-cnt);        return query(mid+1,R,newl,newr,dep+1,k-cnt);    }}int main(){//    fread;    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        MEM(tree,0);        for(int i=1;i<=n;i++)        {            scanf("%d",&tree[0][i]);            sorted[i]=tree[0][i];        }        sort(sorted+1,sorted+n+1);        build(1,n,0);        int l,r,k;        while(m--)        {            scanf("%d%d%d",&l,&r,&k);            printf("%d\n",query(1,n,l,r,0,k));        }    }    return 0;}