[NBUT 1458 Teemo]区间第k大问题，划分树

裸的区间第k大问题，划分树搞起。

#pragma comment(linker, "/STACK:10240000")
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define fillarray(a, b)     memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}

const double PI = acos(-1.0);
const int INF = 1e9 + 7;
const double EPS = 1e-12;

/* -------------------------------------------------------------------------------- */

const int maxn = 1e5 + 7;


/** 过程：快排的过程，通过记录进入左子区间的个数的前缀和来解决区间第k大问题 **/
class PartitionTree {
    int cnt[20][maxn], val[20][maxn], buf[maxn];
    int n;

    void init(int a[], int n) {
        this->n = n;
        fillchar(cnt, 0);
        fillchar(val, 0);
        fillarray(val[0], a);
        fillarray(buf, a);
        sort(buf, buf + n);
    }

    void build(int l, int r, int dep) {
        if (l == r) return ;
        int m = (l + r) >> 1, c = 0, small = 0;
        for (int i = l; i <= r; i ++) small += val[dep][i] < buf[m];
        for (int i = l; i <= r; i ++) {
            if (c < m - l + 1) {
                if (val[dep][i] < buf[m] || val[dep][i] == buf[m] && small < m - l + 1) {
                    cnt[dep][i] = 1;
                    val[dep + 1][l + c ++] = val[dep][i];
                    small += val[dep][i] == buf[m];
                }
            }
            else break;
        }
        for (int i = l; i <= r; i ++) {
            if (!cnt[dep][i]) val[dep + 1][l + c ++] = val[dep][i];
        }
        build(l, m, dep + 1);
        build(m + 1, r, dep + 1);
    }
    /** 第k小 */
    int querykth(int L, int R, int k, int l, int r, int dep) {
        if (k <= 0 || k > R - L + 1) return - 1;
        if (L == R) return val[dep][L];
        int m = (l + r) >> 1, cl = cnt[dep][L - 1] - cnt[dep][l - 1], cr = cnt[dep][R] - cnt[dep][l - 1];
        if (cr - cl >= k) return querykth(l + cl, l + cr - 1, k, l, m, dep + 1);
        return querykth(m + 1 + L - l - cl, m + R - l + 1 - cr, k - cr + cl, m + 1, r, dep + 1);
    }
public:
    void build(int a[], int n) {
        init(a, n);
        build(1, n, 0);
        for (int i = 0; i < 20; i ++) {
            for (int j = 2; j <= n; j ++) {
                cnt[i][j] += cnt[i][j - 1];
            }
        }
    }
    int querykth(int L, int R, int k) { return querykth(L, R, k, 1, n, 0); }
};/** 下标从1开始 */

PartitionTree pt;
int a[maxn];

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int n, m;
    while (cin >> n >> m) {
        for (int i = 1; i <= n; i ++) {
            scanf("%d", a + i);
        }
        pt.build(a, n);
        int l, r, k;
        for (int i = 0; i < m; i ++) {
            scanf("%d%d%d", &l, &r, &k);
            printf("%d
", pt.querykth(l, r, k));
        }
    }
    return 0;
}