Description
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 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
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。
划分树是基于高速排序的，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一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。假设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。假设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 返回最后结果即可。
给出一个大牛的图片样例。
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("###
");
for(int i=0;i<10;i++){
for(int j=1;j<=n;j++){
cout << val[i][j]<<" ";
}
cout << endl;
}
printf("****
");
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
",query(i,j,k,1,n,0,1));
}
return 0;
}