poj2761(划分树)

Feed the dogs

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 14029		Accepted: 4273

Description

Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, but not the dogs, so Jiajia use a special way to feed the dogs. At lunchtime, the dogs will stand on one line, numbered from 1 to n, the leftmost one is 1, the second one is 2, and so on. In each feeding, Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of course Jiajia has his own way of deciding the pretty value of each dog. It should be noted that Jiajia do not want to feed any position too much, because it may cause some death of dogs. If so, Wind will be angry and the aftereffect will be serious. Hence any feeding inteval will not contain another completely, though the intervals may intersect with each other.

Your task is to help Jiajia calculate which dog ate the food after each feeding.

Input

The first line contains n and m, indicates the number of dogs and the number of feedings.

The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier.

Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding.

You can assume that n<100001 and m<50001.

Output

Output file has m lines. The i-th line should contain the pretty value of the dog who got the food in the i-th feeding.

Sample Input

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

Sample Output

3
2

Source

POJ Monthly--2006.02.26,zgl & twb

 

        本题给定一个数的序列，然后给定若干查询区间，要求区间中的第k小值。

        这是道典型的划分树模板题。建立一颗二叉树然后直接查询即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

//*****************************************************************
const int MAXN=100001+100;
int tree[30][MAXN];//表示每层每个位置的值
int sorted[MAXN];//已经排序的数
int toleft[30][MAXN];//toleft[p][i]表示第p层从1到i有多少个数分入左边

//创建划分树
//时间复杂度为O(N*log(N))
void build(int l,int r,int dep)
{
	int i;
    if(l==r)return;
    int mid=(l+r)>>1;
    int same=mid-l+1;//表示等于中间值而且被分入左边的个数
    for(i=l;i<=r;i++)
      if(tree[dep][i]<sorted[mid])
         same--;
    int lpos=l;
    int rpos=mid+1;
    for(i=l;i<=r;i++)
    {
        if(tree[dep][i]<sorted[mid])//比中间的数小，分入左边
             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  //比中间值大分入右边
            tree[dep+1][rpos++]=tree[dep][i];
        toleft[dep][i]=toleft[dep][l-1]+lpos-l;//从1到i放左边的个数

    }
    build(l,mid,dep+1);
    build(mid+1,r,dep+1);
}

//查询区间第k小的数，[L,R]是大区间，[l,r]是要查询的小区间
//时间复杂度为O(log(N))
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)>>1;
    int cnt=toleft[dep][r]-toleft[dep][l-1];//[l,r]中位于左边的个数
    if(cnt>=k)
    {
        //L+要查询的区间前被放在左边的个数
        int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
        //左端点加上查询区间会被放在左边的个数
        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()
{
	int n,m,i,r,l,k;
	while(~scanf("%d%d",&n,&m))
	{
		memset(toleft,0,sizeof(toleft));
		for(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);
		while(m--)
		{
			scanf("%d%d%d",&l,&r,&k);
			printf("%d\n",query(1,n,l,r,0,k));
		}
	}
	return 0;
}