bzoj 1176 cdq分治套树状数组

题面：

维护一个W*W的矩阵，初始值均为S.每次操作可以增加某格子的权值,或询问某子矩阵的总权值.修改操作数M<=160000,询问数Q<=10000,W<=2000000.

Input

第一行两个整数,S,W;其中S为矩阵初始值;W为矩阵大小

接下来每行为一下三种输入之一(不包含引号):

"1 x y a"

"2 x1 y1 x2 y2"

"3"

输入1:你需要把(x,y)(第x行第y列)的格子权值增加a

输入2:你需要求出以左下角为(x1,y1),右上角为(x2,y2)的矩阵内所有格子的权值和,并输出

输入3:表示输入结束

Output

对于每个输入2,输出一行,即输入2的答案

思路：正常思路用树状数组套树状数组，但是w范围过大，开不下。考虑用cdq分治套树状数组。我们用cdq分治对x坐标排序，用树状数组维护y坐标，把询问用差分思想拆分成四个不同的询问即可。

代码：

#include <bits/stdc++.h>
#define LL long long
#define lowbit (x & (-x))
using namespace std;
const int maxn = 200010;
int tot, tot_ans, n;
struct query{
	int type, x, y, pos, flag;
};
query q[maxn], tmp[maxn];
int c[maxn * 10], ans[maxn];
void add(int x, int y) {
	for (; x <= n; x += lowbit)
		c[x] += y;
}
int ask(int x) {
	int ans = 0;
	for (; x; x-= lowbit) {
		ans += c[x];
	}
	return ans;
}
void clear(int x) {
	for (; x <= n; x += lowbit) {
		c[x] = 0;
	}
}
void cdq(int l, int r) {
	if(l == r) return;
	int mid = (l + r) >> 1;
	cdq(l, mid);
	cdq(mid + 1, r);
	int l1 = l, l2 = mid + 1, pos = l;
	while(l1 <= mid && l2 <= r) {
		if(q[l1].x <= q[l2].x) {
			if(q[l1].type == 1) {
				add(q[l1].y, q[l1].pos);
			}
			tmp[pos++] = q[l1++];
		} else {
			if(q[l2].type == 2) {
				ans[q[l2].pos] += q[l2].flag * ask(q[l2].y);
			}
			tmp[pos++] = q[l2++];
		}
	}
	while(l1 <= mid) tmp[pos++] = q[l1++];
	while(l2 <= r) {
		if(q[l2].type == 2) {
			ans[q[l2].pos] += q[l2].flag * ask(q[l2].y);
		}
		tmp[pos++] = q[l2++];
	}
	for (int i = l; i <= mid; i++) {
		if(q[i].type == 1) {
			clear(q[i].y);
		}
	}
	for (int i = l; i <= r; i++)
		q[i] = tmp[i];
}
int main() {
	int l1, r1, l2, r2, x, op, val;
	scanf("%d%d", &x, &n);
	while(~scanf("%d", &op) && op != 3) {
		if(op == 1) {
			scanf("%d%d%d", &l1, &r1, &val);
			q[++tot] = (query){1, l1, r1, val, 0};
		} else {
			scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
			tot_ans++;
			ans[tot_ans] = x * (l2 - l1 + 1) * (r2 - r1 + 1);
			q[++tot] = (query){2, l1 - 1, r1 - 1, tot_ans, 1};
			q[++tot] = (query){2, l1 - 1, r2, tot_ans, -1};
			q[++tot] = (query){2, l2, r1 - 1, tot_ans, -1};
			q[++tot] = (query){2, l2, r2, tot_ans, 1};
		}
	}
	cdq(1, tot);
	for (int i = 1; i <= tot_ans; i++) {
		printf("%d\n", ans[i]);
	}
}