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.
题目意思：给一个数组，问一个区间内第K大的数。
解题思路：划分树。划分树是基于快速排序的，首先将原始数组a[]进行排序sorted[]，然后取中位数m，将未排序数组中小于m放在m左边，大于m的放在m右边，并记下原始数列中每个数左边有多少数小于m，用数组to_left[depth][]表示，这就是建树过程。重点在于查询过程，设[L,R]为要查询的区间，[l,r]为当前区间，s 表示[L,R]有多少数放到左子树，ss表示[l,L-1]有多少数被放倒左子树，如果s大于等于K，也就是说第K大的数肯定在左子树里，下一步就查询左子树，但这之前先要更新L,R，新的newl=l+ss,
newr=newl+s-1。如果s小于k，也就是说第k大的数在右子树里，下一步查询右子树，也要先更新L,R，dd表示[l,L-1]中有多少数被放到右子树，d表示[L,R]有多少数被放到右子树，那么newl = m+1+dd，newr=m+d+dd, 这样查询逐渐缩小查询区间，直到最后L==R 返回最后结果就行。
给出一个大牛的图片例子，
代码：
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
#define maxn 100000+100
int val[30][maxn];
int to_left[30][maxn];
int sorted[maxn];
int n;
void build(int l,int r,int d,int rt){
if(l==r) return;
int m = (l+r)>>1;
int lsame = m-l+1;
for(int i=l;i<=r;i++){
if(val[d][i]<sorted[m]) lsame--;
}
int lpos=l,rpos=m+1,same=0;
for(int i=l;i<=r;i++){
if(i==l) to_left[d][i]=0;
else to_left[d][i] = to_left[d][i-1];
if(val[d][i]<sorted[m]){
to_left[d][i]++;
val[d+1][lpos++] = val[d][i];
}else if(val[d][i]>sorted[m]){
val[d+1][rpos++] = val[d][i];
}else{
if(same<lsame){
same++;
to_left[d][i]++;
val[d+1][lpos++] = val[d][i];
}else{
val[d+1][rpos++] = val[d][i];
}
}
}
build(l,m,d+1,rt<<1);
build(m+1,r,d+1,rt<<1|1);
}
void print(){
printf("###\n");
for(int i=0;i<10;i++){
for(int j=1;j<=n;j++){
cout << val[i][j]<<" ";
}
cout << endl;
}
printf("****\n");
for(int i=0;i<10;i++){
for(int j=1;j<=n;j++){
cout << to_left[i][j]<<" ";
}
cout << endl;
}
}
int query(int L,int R,int k,int l,int r,int d,int rt){
if(L==R) return val[d][L];
int s,ss;
if(L==l){
s = to_left[d][R];
ss = 0;
}else{
s = to_left[d][R]-to_left[d][L-1];
ss = to_left[d][L-1];
}
int m = (l+r)>>1;
if(s>=k){
int newl = l+ss;
int newr = newl + s-1;
return query(newl,newr,k,l,m,d+1,rt<<1);
}else{
int bb = (L-1)-l+1-ss;
int b = R-L+1 -s;
int newl = m+1+bb;
int newr = m+bb+b; // 4 5 1 3 2 (2,5,4)
return query(newl,newr,k-s,m+1,r,d+1,rt<<1|1);
}
}
int main(){
int m;
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++){
scanf("%d",&val[0][i]);
sorted[i]=val[0][i];
}
sort(sorted+1,sorted+n+1);
build(1,n,0,1);
// print();
while(m--){
int i,j,k;
scanf("%d %d %d",&i,&j,&k);
printf("%d\n",query(i,j,k,1,n,0,1));
}
return 0;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。