hdu 2665 Kth number

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Description

Give you a sequence and ask you the kth big number of a interval.

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 queries.
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

Solution

发现我一直不会主席树,于是今天找了个板子题来学习一下...

• 主席树是可持久化线段树,而对区间查询第(k)大问题,更准确的说,是可持久化权值线段树.所以需要有权值线段树的前置技能.
• 考虑对区间([1,1],[1,2],[1,3]...[1,n])都各自维护一颗权值线段树.那么询问([L,R])时,某个数的出现次数就应该是这个数在([1,R])这颗权值线段树中的出现次数减去在([1,L-1])这颗权值线段树中的出现次数.
• 这就是其他博客中所说的(T[R]-T[L-1])的含义.我们在查询时,就可以当做是在一颗权值线段树里查询第(k)大,这里就是基本操作了.
• 然而,如果真的去建(n)颗线段树,空间无法承受.注意到后一颗线段树比前一颗线段树,只有修改的位置到根节点的路径有变化,最多(logn)个节点.所以我们新建线段树的时候,可以直接复制上颗树对应节点的信息.递归后链上的信息会自动被修改(参见代码).

#include<bits/stdc++.h>
using namespace std;
typedef long long LoveLive;
inline int read()
{
	int out=0,fh=1;
	char jp=getchar();
	while ((jp>'9'||jp<'0')&&jp!='-')
		jp=getchar();
	if (jp=='-')
		{
			fh=-1;
			jp=getchar();
		}
	while (jp>='0'&&jp<='9')
		{
			out=out*10+jp-'0';
			jp=getchar();
		}
	return out*fh;
}
const int MAXN=1e5+10;
int a[MAXN],b[MAXN];
int n,m;
int rt[MAXN];
struct PreSegTree{
	struct node{
		int lson,rson,cnt;
	}Tree[MAXN*20];
	int idx,siz;
	PreSegTree()
		{
			idx=0;
			Tree[0].lson=Tree[0].rson=Tree[0].cnt=0;
		}
	int BuildTree(int l,int r)
		{
			int cur=++idx;
			Tree[cur].cnt=0;
			if(l==r)
				return cur;
			int mid=(l+r)>>1;
			Tree[cur].lson=BuildTree(l,mid);
			Tree[cur].rson=BuildTree(mid+1,r);
			return cur;
		}
	void update(int &cur,int last,int l,int r,int pos)
		{
			cur=++idx;
			Tree[cur]=Tree[last];
			++Tree[cur].cnt;
			if(l==r)
				return;
			int mid=(l+r)>>1;
			if(pos<=mid)
				update(Tree[cur].lson,Tree[last].lson,l,mid,pos);
			else
				update(Tree[cur].rson,Tree[last].rson,mid+1,r,pos);
		}
	int query(int L,int R,int l,int r,int k)
		{
			if(l==r)
				return l;
			int p=Tree[Tree[R].lson].cnt-Tree[Tree[L].lson].cnt;
			int mid=(l+r)>>1;
			if(p>=k)
				return query(Tree[L].lson,Tree[R].lson,l,mid,k);
			else
				return query(Tree[L].rson,Tree[R].rson,mid+1,r,k-p);
		}
}T;
int main()
{
	int cases=read();
	while(cases--)
		{
			n=read(),m=read();
			for(int i=1;i<=n;++i)
				b[i]=a[i]=read();
			sort(b+1,b+1+n);
			int siz=unique(b+1,b+1+n)-(b+1);
			T.idx=0;
			rt[0]=T.BuildTree(1,siz);
			for(int i=1;i<=n;++i)
				{
					a[i]=lower_bound(b+1,b+1+siz,a[i])-b;
					T.update(rt[i],rt[i-1],1,siz,a[i]);
				}
			for(int i=1;i<=m;++i)
				{
					int L=read(),R=read(),k=read();
					int ans=T.query(rt[L-1],rt[R],1,siz,k);
					printf("%d
",b[ans]);
				}
		}
	return 0;
}