HDU 2665(Kth number-区间第k大[内存限制+重数])

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3155    Accepted Submission(s): 1055

Problem Description

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

 

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 quaere.

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

 

Source

HDU男生专场公开赛——赶在女生之前先过节（From WHU）

 

Recommend

zty

 

这题……不想说什么.

题目还行主要是内存！！

我应该2分内存的……555……

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
#include<cctype>
#include<map>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define RepD(i,n) for(int i=n;i>=0;i--)
#define MAXN (200000+10)
#define MAXM (200000+10)
int n,m,a[MAXN],a2[MAXN];
struct node
{
	node *ch[2];
	int a,siz;
	node(){ch[0]=ch[1]=NULL;siz=a=0;}
	void update()
	{
		siz=a;
		if (ch[0]) siz+=ch[0]->siz;
		if (ch[1]) siz+=ch[1]->siz;
	}
}*null=new node(),*root[MAXN]={NULL},q[MAXN*9];
int q_s;
void make_node(node *&y,node *&x,int l,int r,int t)
{
	if (x==NULL) x=null;
	y=&q[++q_s];
	*y=node();
	int m=(l+r)>>1;
	if (l==r)
	{
		*y=*x;
		y->siz++;y->a++;
		return;
	}
	if (t<=a2[m])
	{
		make_node(y->ch[0],x->ch[0],l,m,t);
		y->ch[1]=x->ch[1];
		y->update();
	}
	else
	{
		make_node(y->ch[1],x->ch[1],m+1,r,t);
		y->ch[0]=x->ch[0];
		y->update();
	}
}
void find(node *&x1,node *&x2,int l,int r,int k)
{
	if (x1==NULL) x1=null;
	if (x2==NULL) x2=null;
	if (l==r) {printf("%d\n",a2[l]);return;}
	int m=(l+r)>>1;
	int ls=0;
	if (x2->ch[0]) ls+=x2->ch[0]->siz;
	if (x1->ch[0]) ls-=x1->ch[0]->siz;
	if (ls>=k) find(x1->ch[0],x2->ch[0],l,m,k);
	else find(x1->ch[1],x2->ch[1],m+1,r,k-ls);
}
void print(node *x,int _x)
{
	cout<<_x<<':'<<x->siz<<' ';
	//cout<<'<';
	if (x->ch[0]!=null) print(x->ch[0],_x*2);
	//cout<<'>';
	if (x->ch[1]!=null) print(x->ch[1],_x*2+1);
	//cout<<'>';
}
int main()
{
	//freopen("hdu2665.in","r",stdin);
	null->ch[0]=null; null->ch[1]=null;
	int t;
	scanf("%d",&t);
	while (t--)
	{
		q_s=0;
		scanf("%d%d",&n,&m);
		For(i,n) scanf("%d",&a[i]),a2[i]=a[i];
		sort(a2+1,a2+1+n);
		int size=unique(a2+1,a2+1+n)-(a2+1);
		For(i,n)
		{
			make_node(root[i],root[i-1],1,size,a[i]);
		}
		//For(i,n) print(root[i],1),cout<<endl;
		For(i,m)
		{
			int l,r,k;
			scanf("%d%d%d",&l,&r,&k);
			find(root[l-1],root[r],1,size,k);
		}
		For(i,n) root[i]=NULL;
	}
	return 0;
}