bzoj 2683 简单题 cdq分治

题面

题目传送门

解法

可以离线，那么就是非常简单的cdq分治了

只要把询问拆成4个，然后就变成了一个三维偏序问题

时间复杂度：\(O(q\ log^2\ n)\)

代码

#include <bits/stdc++.h>
#define int long long
#define N 1000010
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, tim, x, y, v;
	bool operator < (const Node &a) const {
		return x < a.x;
	}
} a[N], t[N];
int n, f[N], ans[N];
int lowbit(int x) {return x & -x;}
void modify(int x, int v) {
	for (int i = x; i <= n; i += lowbit(i))
		f[i] += v; 
}
int query(int x) {
	int ret = 0;
	for (int i = x; i; i -= lowbit(i))
		ret += f[i];
	return ret;
}
void cdq(int l, int r) {
	if (l == r) return;
	int mid = (l + r) >> 1, tx = l, ty = mid + 1;
	for (int i = l; i <= r; i++)
		if (a[i].tim <= mid) t[tx++] = a[i];
			else t[ty++] = a[i];
	for (int i = l; i <= r; i++) a[i] = t[i];
	tx = l, ty = mid + 1;
	while (ty <= r) {
		while (tx <= mid && a[tx].x <= a[ty].x) {
			if (a[tx].op == 1) modify(a[tx].y, a[tx].v);
			tx++;
		}
		if (a[ty].op == 2) ans[a[ty].v] += query(a[ty].y);
		if (a[ty].op == 3) ans[a[ty].v] -= query(a[ty].y);
		ty++;
	}
	for (int i = l; i < tx; i++) if (a[i].op == 1) modify(a[i].y, -a[i].v);
	cdq(l, mid), cdq(mid + 1, r);
}
main() {
	read(n);
	int len = 0, cntq = 0, op; read(op);
	while (op != 3) {
		if (op == 1) {
			int x, y, v;
			read(x), read(y), read(v);
			a[++len] = (Node) {op, len, x, y, v};
		} else {
			int x1, y1, x2, y2;
			read(x1), read(y1), read(x2), read(y2);
			a[++len] = (Node) {2, len, x2, y2, ++cntq};
			a[++len] = (Node) {3, len, x1 - 1, y2, cntq};
			a[++len] = (Node) {3, len, x2, y1 - 1, cntq};
			a[++len] = (Node) {2, len, x1 - 1, y1 - 1, cntq};
		}
		read(op);
	}
	sort(a + 1, a + len + 1);
	cdq(1, len);
	for (int i = 1; i <= cntq; i++) cout << ans[i] << "\n";
	return 0;
}