POI2006 TET-Tetris 3D

题目传送门

第一次写二维线段树

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二维线段树模板题
感觉二维线段树很迷代码倒是挺好记
还得再理解理解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
#define ls p << 1
#define rs p << 1 | 1
using namespace std;
LL read() {
	LL k = 0, f = 1; char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9')
		k = k * 10 + c - 48, c = getchar();
	return k * f;
}
int n, m, t;
struct zzz {
	int maxn[901 << 2], tag[901 << 2];
	void update(int nl, int nr, int k, int l = 1, int r = m, int p = 1) {
		maxn[p] = max(maxn[p], k);
		if(l >= nl && r <= nr) {
			tag[p] = max(tag[p], k); return ;
		}
		int mid = (l+r) >> 1;
		if(nl <= mid) update(nl, nr, k, l, mid, ls);
		if(nr > mid) update(nl, nr, k, mid+1, r, rs);
	}
	int query(int nl, int nr, int l = 1, int r = m, int p = 1) {
		if(l >= nl && r <= nr) return max(maxn[p], tag[p]);
		int mid = (l+r) >> 1, ans = tag[p];
		if(nl <= mid) ans = max(ans, query(nl, nr, l, mid, ls));
		if(nr > mid) ans = max(ans, query(nl, nr, mid+1, r, rs));
		return ans;
	}
};
struct hhh {
	zzz maxn[901 << 2], tag[901 << 2];
	void update(int nl, int nr, int kl, int kr, int k, int l = 1, int r = n, int p = 1) {
		maxn[p].update(kl, kr, k);
		if(l >= nl && r <= nr) {
			tag[p].update(kl, kr, k); return ;
		}
		int mid = (l+r) >> 1;
		if(nl <= mid) update(nl, nr, kl, kr, k, l, mid, ls);
		if(nr > mid) update(nl, nr, kl, kr, k, mid+1, r, rs);
	}
	int query(int nl, int nr, int kl, int kr, int l = 1, int r = n, int p = 1) {
		if(l >= nl && r <= nr) return maxn[p].query(kl, kr);
		int mid = (l+r) >> 1, ans = tag[p].query(kl, kr);
		if(nl <= mid) ans = max(ans, query(nl, nr, kl, kr, l, mid, ls));
		if(nr > mid) ans = max(ans, query(nl, nr, kl, kr, mid+1, r, rs));
		return ans;
	}
}tree;
int main() {
	n = read(), m = read(), t = read();
	while(t--) {
		int d = read(), s = read(), h = read(), x = read()+1, y = read()+1;
		h += tree.query(x, x+d-1, y, y+s-1);
		tree.update(x, x+d-1, y, y+s-1, h);
	}
	cout << tree.query(1, n, 1, m);
	return 0;
}