poj2104(主席树)

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.

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).

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

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

5
6
3

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

 

 

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

const int MAXN=100000+100;
const int MAXM=MAXN*20;

int tot,n,m;
int da[MAXN],sDa[MAXN];
int leftChild[MAXM],rightChild[MAXM],wei[MAXM],chairTNode[MAXM];


/**********************************
*参数：待处理元素区间
*功能：建立一棵空线段树
*返回值：返回根节点下标
***********************************/
int Build(int left,int right)
{
	int id=tot++;
	wei[id]=0;
	if(left<right)
	{
		int mid=(left+right)>>1;
		leftChild[id]=Build(left,mid);
		rightChild[id]=Build(mid+1,right);
	}
	return id;
}


int Update(int root,int pos,int val)
{
	int l=1,r=m,mid,newRoot=tot++,retRoot=newRoot;
	wei[newRoot]=wei[root]+val;
	while(l<r)
	{
		mid=(l+r)>>1;
		if(pos<=mid)
		{
			//确定节点孩子节点
			leftChild[newRoot]=tot++;
			rightChild[newRoot]=rightChild[root];

			//确定待跟新节点以及历史版本
			newRoot=leftChild[newRoot];
			root=leftChild[root];

			r=mid;
		}
		else 
		{
			rightChild[newRoot]=tot++;
			leftChild[newRoot]=leftChild[root];
			newRoot=rightChild[newRoot];
			root=rightChild[root];
			l=mid+1;
		}
		wei[newRoot]=wei[root]+val;
	}
	return retRoot;
}

int Query(int leftRoot,int rightRoot,int k)
{
	int  l=1,r=m,mid;
	while(l<r)
	{
		mid=(l+r)>>1;
		if(wei[leftChild[leftRoot]]-wei[leftChild[rightRoot]]>=k)//第k小值在左子树
		{
			//确定查找新区间
			leftRoot=leftChild[leftRoot];
			rightRoot=leftChild[rightRoot];

			r=mid;
		}
		else 
		{
			k-=wei[leftChild[leftRoot]]-wei[leftChild[rightRoot]];
			leftRoot=rightChild[leftRoot];
			rightRoot=rightChild[rightRoot];
			l=mid+1;
		}
	}
	return l;
}

int main()
{
	int q,i;
	int ql,qr,k;
	while(scanf("%d%d",&n,&q)!=EOF)
	{
		m=0;
		tot=0;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&da[i]);
			sDa[i]=da[i];
		}
		sort(sDa+1,sDa+n+1);
		m=unique(sDa+1,sDa+1+n)-sDa-1;
		chairTNode[n+1]=Build(1,m);
		//cout<<"**********"<<endl;

		for(i=n;i>=1;i--)
		{
			int pos=lower_bound(sDa+1,sDa+1+m,da[i])-sDa;
			chairTNode[i]=Update(chairTNode[i+1],pos,1);
		}
		while(q--)
		{
			scanf("%d%d%d",&ql,&qr,&k);
			printf("%d\n",sDa[Query(chairTNode[ql],chairTNode[qr+1],k)]);
		}
	}
	system("pause");
	return 0;
}