區(qū)間第k小

金牌導(dǎo)航 整體二分-1

題目大意

給出一個序列，有若干查詢，每次查詢給出l,r,k，讓你求l~r這個區(qū)間的第k大

輸入樣例

7 3 1 5 2 6 3 7 4 2 5 3 4 4 1 1 7 3

輸出樣例

5 6 3

數(shù)據(jù)范圍

$1?n?10^5,1?m?50000,1?|a_i|?10^9$

解題思路

先把原序列按數(shù)值排序
然后對于所有查詢進(jìn)行整體二分
對于每一段的二分，先將mid左側(cè)的數(shù)按原下標(biāo)存進(jìn)樹狀數(shù)組，然后每個查詢判斷它查詢的區(qū)間是否個數(shù)大于k
然后繼續(xù)分治

代碼

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ll long long #define N 100010 #define M 50010 using namespace std; int n, m, c[N], ans[N]; struct node {int v, s;bool operator <(const node &b) const{return s < b.s;} }s[N]; struct nodee {int l, r, k, v; }a[M], b[M]; void add(int x, int y) {for (; x <= n; x += x & -x)c[x] += y;return; } int ask(int x) {int sum = 0;for (; x; x -= x & -x)sum += c[x];return sum; } void solve(int l, int r, int L, int R) {if (l == r){for (int i = L; i <= R; ++i)ans[a[i].v] = s[l].s;//計算結(jié)果return;}int mid = l + r >> 1, L1 = L, R1 = R;for (int i = l; i <= mid; ++i)add(s[i].v, 1);//存數(shù)for (int i = L; i <= R; ++i){int g = ask(a[i].r) - ask(a[i].l - 1);if (a[i].k <= g) b[L1++] = a[i];//整體二分else{b[R1] = a[i];b[R1--].k -= g; }}for (int i = L; i <= R; ++i)a[i] = b[i]; for (int i = l; i <= mid; ++i)add(s[i].v, -1);if (L1 > L) solve(l, mid, L, L1 - 1);if (R1 < R) solve(mid + 1, r, R1 + 1, R); } int main() {scanf("%d%d", &n, &m);for (int i = 1; i <= n; ++i){scanf("%d", &s[i].s);s[i].v = i;}sort(s + 1, s + 1 + n);for (int i = 1; i <= m; ++i){scanf("%d%d%d", &a[i].l, &a[i].r, &a[i].k);a[i].v = i;}solve(1, n, 1, m);for (int i = 1; i <= m; ++i)printf("%d\n", ans[i]);return 0; }

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