区间第k小

金牌导航 整体二分-1

题目大意

给出一个序列，有若干查询，每次查询给出l,r,k，让你求l~r这个区间的第k大

输入样例

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

输出样例

5
6
3

数据范围

$1⩽n⩽10^5,1⩽m⩽50000,1⩽|a_i|⩽10^9$

解题思路

先把原序列按数值排序
然后对于所有查询进行整体二分
对于每一段的二分，先将mid左侧的数按原下标存进树状数组，然后每个查询判断它查询的区间是否个数大于k
然后继续分治

代码

#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;//计算结果return;}int mid = l + r >> 1, L1 = L, R1 = R;for (int i = l; i <= mid; ++i)add(s[i].v, 1);//存数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];//整体二分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;
}