bzoj 3110 [Zjoi2013]K大数查询 整体二分

题面

题目传送门

解法

树套树比较苟，考虑整体二分

因为要求第K大，那么我们在二分的时候把\(v\)大于\(mid\)的放在右边并修改

修改直接用线段树区间加区间求和即可

时间复杂度：\(O(m\ log^2\ n)\)

代码

#include <bits/stdc++.h>
#define int long long
#define N 50010
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Node {
	int op, l, r, v, id;
} a[N], tx[N], ty[N];
struct SegmentTree {
	struct Node {
		int l, r, sum, del;
	} t[N * 4];
	void build(int k, int l, int r) {
		t[k] = (Node) {l, r, 0, 0};
		if (l == r) return;
		int mid = (l + r) >> 1;
		build(k << 1, l, mid);
		build(k << 1 | 1, mid + 1, r);
	}
	void pushdown(int k) {
		int x = t[k].del; t[k].del = 0;
		int lc = k << 1, rc = k << 1 | 1;
		t[lc].del += x, t[rc].del += x;
		t[lc].sum += (t[lc].r - t[lc].l + 1) * x;
		t[rc].sum += (t[rc].r - t[rc].l + 1) * x; 
	}
	void modify(int k, int L, int R, int v) {
		int l = t[k].l, r = t[k].r;
		if (L <= l && r <= R) {
			t[k].del += v, t[k].sum += (r - l + 1) * v;
			return;
		}
		if (t[k].del) pushdown(k);
		int mid = (l + r) >> 1;
		if (L <= mid && mid < R)
			modify(k << 1, L, mid, v), modify(k << 1 | 1, mid + 1, R, v);
		if (R <= mid) modify(k << 1, L, R, v);
		if (L > mid) modify(k << 1 | 1, L, R, v);
		t[k].sum = t[k << 1].sum + t[k << 1 | 1].sum;
	}
	int query(int k, int L, int R) {
		int l = t[k].l, r = t[k].r;
		if (L <= l && r <= R) return t[k].sum;
		if (t[k].del) pushdown(k);
		int mid = (l + r) >> 1;
		if (R <= mid) return query(k << 1, L, R);
		if (L > mid) return query(k << 1 | 1, L, R);
		return query(k << 1, L, mid) + query(k << 1 | 1, mid + 1, R);
	}
} T;
int ans[N];
void solve(int l, int r, int L, int R) {
	if (l > r) return;
	if (L == R) {
		for (int i = l; i <= r; i++)
			if (a[i].op == 2) ans[a[i].id] = L;
		return;
	}
	int mid = (L + R) >> 1, fx = 0, fy = 0, lx = 0, ly = 0;
	for (int i = l; i <= r; i++)
		if (a[i].op == 1) {
			if (a[i].v > mid) T.modify(1, a[i].l, a[i].r, 1), ty[++ly] = a[i];
				else tx[++lx] = a[i];
		} else {
			int tmp = T.query(1, a[i].l, a[i].r);
			if (a[i].v > tmp) a[i].v -= tmp, tx[++lx] = a[i], fx = 1;
				else ty[++ly] = a[i], fy = 1;
		}
	for (int i = l; i <= r; i++)
		if (a[i].op == 1 && a[i].v > mid) T.modify(1, a[i].l, a[i].r, -1);
	for (int i = 1; i <= lx; i++) a[i + l - 1] = tx[i];
	for (int i = 1; i <= ly; i++) a[i + l + lx - 1] = ty[i];
	if (fx) solve(l, l + lx - 1, L, mid);
	if (fy) solve(l + lx, r, mid + 1, R);
}
main() {
	int n, m, q = 0; read(n), read(m);
	for (int i = 1; i <= m; i++) {
		read(a[i].op), read(a[i].l), read(a[i].r), read(a[i].v);
		if (a[i].op == 2) a[i].id = ++q;
	}
	T.build(1, 1, n), solve(1, m, 1, n);
	for (int i = 1; i <= q; i++) cout << ans[i] << "\n";
	return 0;
}