..这个东西是当初我弱不会划分树的时候写出来替代的一个玩意..被一小撮别有用心的人取了很奇怪的名字> <
想法是对原序列的每一个前缀[1..i]建立出一颗线段树维护值域上每个数的出现次数，然后发现这样的树是可以减的，然后就没有然后了
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K-th Number
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.
Source
Northeastern Europe 2004, Northern Subregion
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这题是主席树裸题。
但是我却狂T——
#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 (100000+10)
#define MAXM (5000+10)
int n,m,a[MAXN],a2[MAXN];
struct node
{
node *ch[2];
int siz;
node(){ch[0]=ch[1]=NULL;siz=0;}
node(node *_ch0,node *_ch1,int _siz):siz(_siz){ch[0]=_ch0,ch[1]=_ch1;}
void update()
{
if (ch[0]) siz+=ch[0]->siz;
if (ch[1]) siz+=ch[1]->siz;
}
}*null=new node(),*root[MAXN]={NULL};
void make_node(node *&y,node *&x,int l,int r,int t)
{
if (x==NULL) x=null;
y=new node();
int m=(l+r)>>1;
if (l==r)
{
*y=*x;
y->siz++;
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);
}
int main()
{
//freopen("poj2104.in","r",stdin);
null->ch[0]=null; null->ch[1]=null;
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,m)
{
int l,r,k;
scanf("%d%d%d",&l,&r,&k);
find(root[l-1],root[r],1,size,k);
}
return 0;
}
于是在柯老师的教导下，我发现致使我狂T的因素是不断new 结点，于是我将结点事先用数组开好。
#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);
}
int main()
{
//freopen("hdu2665.in","r",stdin);
null->ch[0]=null; null->ch[1]=null;
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,m)
{
int l,r,k;
scanf("%d%d%d",&l,&r,&k);
find(root[l-1],root[r],1,size,k);
}
return 0;
}