poj 2104

K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 76200		Accepted: 27393
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.

Source

Northeastern Europe 2004, Northern Subregion

 

题目大意：给你n个数和m个询问，每个询问都是关于区间第k小的数是多少。 

 

思路：  这个题目的方法很多。  

1. 分块

2.整体二分

3. 主席数（可持久化线段树)

4. 区间线段树套平衡树

5. 权值线段树套平衡树

6. 线段树套线段树

7. 替罪羊树套权值线段树。 

 

附整体二分代码： 

#include<iostream>
#include<cstring>
#include<cmath>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<ctime>
using namespace std;
#define re(i,n)     for(int i=0;i<n;i++)
#define RE(i,n)     for(int i=n-1;i;i--)
#define REP(i,n)    for(int i=n;i;i--)
#define rep(i,n)    for(int i=1;i<=n;i++)
#define repx(i,a,b) for(int i=a;i<=b;i++)
#define repy(i,a,b) for(int i=a;i>=b;i--)
#define ms(i,a)     memset(a,i,sizeof(a))
#define sz(x)       (x.size())
#define pb          push_back
#define sqr(x)      ((x)*(x))
#define vi          vector<int>
#define LL           long long
#define pi          pair<int,int>
#define mp          make_pair
#define lch         (o<<1)
#define rch         (o<<1|1)
#define mid         (l+r)/2
inline void  read(int &x){
    int w=0; char ch=0; x=0;
    while(ch<'0' || ch>'9') {w|=ch=='-';ch=getchar();}
    while(ch>='0' && ch<='9') x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
    if (w) x=-x;
}

inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
int const maxn=100003;
int const inf=1e9;
struct node{
    int x,num;
    bool operator < ( const node &rhs) const{
        return x<rhs.x;
    }
}a[maxn];

struct query{
    int x,y,id,cnt,k;
}q[maxn],tmp[maxn];
int n,m,ans[maxn],t[maxn];
void calc(int l,int r,int al,int Mid){
    int L=1, R=n+1;
    while (L<R){
        int M=(L+R)/2;
        if (a[M].x>=al) R=M;
        else L=M+1;
    }
    for(int i=R;i<=n && a[i].x<=Mid;i++)
        for(int j=a[i].num;j<=n; j+= (j&-j) ) t[j]++;
    for(int i=l;i<=r;i++){
        q[i].cnt=0;
        for(int j=q[i].y; j; j-=(j&-j)) q[i].cnt+=t[j];
        for(int j=q[i].x-1;j;j-=(j&-j)) q[i].cnt-=t[j];
    }
    for(int i=R;i<=n && a[i].x<=Mid;i++)
        for(int j=a[i].num;j<=n;j+=(j&-j)) t[j]--;
}
void solve(int  l,int  r,LL  al,LL  ar){
    if(al==ar){
        repx(i,l,r) ans[q[i].id]=al; return;
    }
    int  Mid=al+(ar-al)/2;
    calc(l,r,al,Mid);
    int p1=l,p2=r;
    for(int i=l;i<=r;i++)
        if (q[i].cnt>=q[i].k) tmp[p1++]=q[i];
        else q[i].k-=q[i].cnt,tmp[p2--]=q[i];
    for(int i=l;i<=r;i++) q[i]=tmp[i];
    if(l<=p1-1) solve(l,p1-1,al,Mid);
    if(p2+1<=r) solve(p2+1,r,Mid+1,ar);
}

int main(){
    scanf("%d%d",&n,&m);
    rep(i,n){
        scanf("%d",&a[i].x);
        a[i].num=i;
    }
    a[n+1].x=2*inf;
    sort(a+1,a+n+1);
    rep(i,m){
        scanf("%d%d%d",&q[i].x,&q[i].y,&q[i].k);
        q[i].id=i;
    }
    solve(1,m,-inf,inf);
    rep(i,m) printf("%d\n",ans[i]);
    return 0;
}