poj2104(主席樹)

K-th Number Time Limit: 20000MS Memory Limit: 65536K Total Submissions: 35704 Accepted: 11396 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. 題解： 本題要求給定區間的第k小元素，可以用劃分樹，在我的博客poj2104劃分樹解法有劃分樹的解法，至於劃分樹可以參考劃分樹 本題要用到的主席樹相關知識在主席樹讀書筆記可以找到。主席樹的每個節點對應一顆線段樹，每個節點的意義為離散後原序列的某個後綴。在每棵線段樹中若元素出現則標記為1，問題可以轉化為求區間的前k個數，第k個數即為第k小的元素，在區間上操作是線段樹的特色。 貼段代碼：

#include
#include
#include
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];


/**********************************
*參數：待處理元素區間
*功能：建立一棵空線段樹
*返回值：返回根節點下標
***********************************/
int Build(int left,int right)
{
	int id=tot++;
	wei[id]=0;
	if(left>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>1;
		if(pos<=mid)
		{
			//確定節點孩子節點
			leftChild[newRoot]=tot++;
			rightChild[newRoot]=rightChild[root];

			//確定待跟新節點以及歷史版本
			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>1;
		if(wei[leftChild[leftRoot]]-wei[leftChild[rightRoot]]>=k)//第k小值在左子樹
		{
			//確定查找新區間
			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<<"**********"<=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;
}