poj 2104 K-th Number(划分树)

题目分析：该问题给定一段区间中的值，再给定一段查询区间[ql, qr]，需要给出该查询区间中的值在排序后的第K大的值；

使用划分树即可解决该问题；划分树的建树的复杂度为O(NlogN),查询一个区间的第K大值的复杂度为O(logN)；

代码如下：

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;

const int MAX_N = 100000 + 10;
int sorted[MAX_N];
int to_left[20][MAX_N];
int tree[20][MAX_N];

void Build(int l, int r, int deep)
{
    if (l == r) return;
    int mid = (l + r) >> 1;
    int lc_same = mid - l + 1;
    int l_pos = l, r_pos = mid + 1;

    for (int i = l; i <= r; ++i)
    {
        if (tree[deep][i] < sorted[mid])
            lc_same--;
    }

    for (int i = l; i <= r; ++i)
    {
        if (tree[deep][i] < sorted[mid])
            tree[deep + 1][l_pos++] = tree[deep][i];
        else if (tree[deep][i] == sorted[mid] && lc_same > 0)
        {
            lc_same--;
            tree[deep + 1][l_pos++] = tree[deep][i];
        }
        else
            tree[deep + 1][r_pos++] = tree[deep][i];
        to_left[deep][i] = to_left[deep][l - 1] + l_pos - l;
    }
    Build(l, mid, deep + 1);
    Build(mid + 1, r, deep + 1);
}


int Query(int l, int r, int ql, int qr, int deep, int k)
{
    if (l == r)
        return tree[deep][l];
    int mid = (l + r) >> 1;
    int cnt = to_left[deep][qr] - to_left[deep][ql - 1];

    if (cnt >= k)
    {
        int new_ql = l + to_left[deep][ql - 1] - to_left[deep][l - 1];
        int new_qr = new_ql + cnt - 1;
        return Query(l, mid, new_ql, new_qr, deep + 1, k);
    }
    else
    {
        int new_qr = qr + to_left[deep][r] - to_left[deep][qr];
        int new_ql = new_qr - (qr - ql - cnt);
        return Query(mid + 1, r, new_ql, new_qr, deep + 1, k - cnt);
    }
}


int main()
{
    int n, query_times;

    scanf("%d %d", &n, &query_times);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d", &tree[0][i]);
        sorted[i] = tree[0][i];
    }

    sort(sorted + 1, sorted + n + 1);
    Build(1, n, 0);
    for (int i = 0; i < query_times; ++i)
    {
        int ql, qr, k, ans;

        scanf("%d %d %d", &ql, &qr, &k);
        ans = Query(1, n, ql, qr, 0, k);
        printf("%d
", ans);
    }
    return 0;
}