Codeforces (869E || #439 Div.2 E || #439 Div.1 C)

Description

biu~
给一个方格图，支持以下操作：

1. 在一个子矩形外围套一圈栅栏。
2. 去掉一个子矩形外围的栅栏（保证存在）。
3. 询问从\((x1, y1)\)到\((x2, y2)\)是否可以不穿过栅栏

保证栅栏间无交，无重边，无共点，且和边界不交。
\(r, c \leqslant 2500\), \(q \leqslant 100000\)。

Solution

每一“层”haxh不同的值，二维BIT维护，判断两点的值是否相等即可。

#include<bits/stdc++.h>
using namespace std;

inline int read() {
	int x = 0, flag = 1; char ch = getchar();
	while (ch > '9' || ch < '0') { if (ch == '-') flag = -1; ch = getchar(); }
	while (ch <= '9' && ch >= '0') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * flag;
}

#define N 2550
#define ll long long
const ll hashValue = 817;

int n, m;

struct BITType {
	ll c[N][N];
	inline int lowbit(int k) { return (k & (-k)); }
	void add(int i, int j, ll delta) {
		for (int x = i; x <= n; x += lowbit(x)) for (int y = j; y <= m; y += lowbit(y))
			c[x][y] += delta;
	}
	ll query(int i, int j) {
		ll ans = 0;
		for (int x = i; x; x -= lowbit(x)) for (int y = j; y; y -= lowbit(y)) ans += c[x][y];
		return ans;
	}
	void update(int r1, int c1, int r2, int c2, ll x) {
		add(r1, c1, x),	add(r1, c2 + 1, -x),
		add(r2 + 1, c1, -x), add(r2 + 1, c2 + 1, x);
	}
}bit;

int main() {
	cin >> n >> m; int q = read();
	while (q--) {
		int op = read();
		int r1 = read(), c1 = read(), r2 = read(), c2 = read();
		if (op != 3) {
			ll x = r1; (x *= hashValue) += c1; (x *= hashValue) += r2; (x *= hashValue) += c2;
			if (op == 2) x *= -1;
			bit.update(r1, c1, r2, c2, x);
		}
		else puts(bit.query(r1, c1) == bit.query(r2, c2) ? "Yes" : "No");
	}
	return 0;
}