[NBUT 1458 Teemo]區間第k大問題，劃分樹

裸的區間第k大問題，劃分樹搞起。

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#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;/** 過程：快排的過程，通過記錄進入左子區間的個數的前綴和來解決區間第k大問題 **/
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); }
};/** 下標從1開始 */PartitionTree pt;
int a[maxn];int main() {
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGEint 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\n", pt.querykth(l, r, k));}}return 0;
}