划分树

这篇文章有些酱油。。。

首先我觉得划分树是一种很搞笑的数据结构：静态意味着不能修改，被树套树完爆，功能也不如主席树等强大，只是在求解区间 k 大值上有一个很优秀的 O(logn) per operation。。。。

构造的时间空间复杂度都是 O(nlogn) 的。

其核心思想类似快速排序。。。对于每个区间选取中位数，把小于之的按原顺序放到左侧，大于之到右侧，然后记一下每一位放至左侧的数量。。接着递归处理。。。

然后查找时就可以发现，刚才记的数量是可减的。。。于是便可以非递归 O(logn) 求出区间 k 大值了。。。

然后记几个 tip : 区间左端点数量的初值很坑。。。然后查找的可减性还是推一下更好。。。

hdu2665 code :

#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#include <ctime>#include <cctype>#include <map>#include <set>#include <string>#include <queue>#include <algorithm>#include <fstream>#include <iostream>using namespace std;#ifdef WIN32#define fmt64 "%I64d"#else#define fmt64 "%lld"#endif#define PI M_PI#define oo 0x13131313#define PB push_back#define PO pop_back#define MP make_pair#define iter iterator#define fst first#define snd second#define pque priority_queue#define cstr(a) (a).c_str()#define FOR(i, j, k) for (i = (j); i <= (k); ++i)#define ROF(i, j, k) for (i = (j); i >= (k); --i)#define FER(e, d, u) for (e = d[u]; e; e = e->n)#define FRE(i, a) for (i = (a).begin(); i != (a).end(); ++i)typedef unsigned int uint;typedef long long int64;typedef unsigned long long uint64;typedef long double real;template<class T> inline bool minfy(T &a, const T &b) {return b < a ? a = b, 1 : 0;}template<class T> inline bool maxfy(T &a, const T &b) {return b > a ? a = b, 1 : 0;}template<class T> inline T sqr(const T &a) {return a * a;}#define maxn 100005int n, m, A, B, k;int a[maxn], b[maxn];struct node {int w[maxn], num[maxn];};node null[20];int *u, *v, *num, x, ls, rs, same;void build(node *p, int l, int r){u = p->w, v = (p + 1)->w, num = p->num;int mid = (l + r) >> 1, x = b[mid];int i;ls = l - 1, rs = mid, same = mid - l + 1;FOR (i, l, r) if (a[u[i]] < a[x]) --same;FOR (i, l, r) {num[i] = i == l ? 0 : num[i - 1];if (a[u[i]] < a[x]) ++num[i], v[++ls] = u[i];else if (a[u[i]] > a[x]) v[++rs] = u[i];else if (same) --same, ++num[i], v[++ls] = u[i];else v[++rs] = u[i];}if (l < mid) build(p + 1, l, mid);if (mid + 1 < r) build(p + 1, mid + 1, r);}int query(){int l = 1, r = n; node *p = null;for (; l < r; ++p) {int mid = (l + r) >> 1;ls = A == l ? 0 : p->num[A - 1], rs = p->num[B];if (rs - ls >= k) {A = l + ls, B = l + rs - 1, r = mid;} else {A = mid + A - l - ls + 1;B = mid + B - l + 1 - rs;k -= rs - ls, l = mid + 1;}}return p->w[l];}bool cmp(int i, int j) {return a[i] < a[j];}void show(int *v){int i;FOR (i, 1, n) cerr << a[v[i]] << " ";cerr << endl;}int main(){freopen("2665.in", "r", stdin);freopen("2665.out", "w", stdout);int i, T;for (scanf("%d", &T); T--; ) {scanf("%d%d", &n, &m);FOR (i, 1, n) scanf("%d", a + i), b[i] = null->w[i] = i;sort(b + 1, b + n + 1, cmp);build(null, 1, n);for (; m--; ) {scanf("%d%d%d", &A, &B, &k);printf("%d\n", a[query()]);}}return 0;}