[洛谷P4390][BOI2007]Mokia 摩基亚

题目大意： 维护一个W*W的矩阵，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。

题解：CDQ分治，把询问拆成四个小矩形

卡点：无

C++ Code：

#include <cstdio>
#include <algorithm>
#define int long long
#define lb(x) (x & -x)
using namespace std;
struct node {
	int op, id, x, y, num;
}q[200010], tmp[200010];
int S, W, op, num, cnt;
int T[2000010];
inline bool cmp(node a, node b) {return a.id < b.id;}
void add(int p, int num) {for (int i = p; i <= W; i += lb(i)) T[i] += num;}
int ask(int p) {int res = 0; for(int i = p; i; i -= lb(i)) res += T[i]; return res;}
void clear(int p) {for (int i = p; i <= W; i += lb(i)) T[i] = 0;}
void CDQ(int l, int r) {
	if (l >= r) return ;
	int mid = l + r >> 1;
	CDQ(l, mid); CDQ(mid + 1, r);
	int p1 = l, p2 = mid + 1, p3 = l;
	while (p1 <= mid && p2 <= r) {
		if (q[p1].x <= q[p2].x) {
			if (!q[p1].op) add(q[p1].y, q[p1].num);
			tmp[p3++] = q[p1++];
		} else {
			if (q[p2].op) q[p2].num += ask(q[p2].y);
			tmp[p3++] = q[p2++];
		}
	}
	while (p1 <= mid) {
		if (!q[p1].op) add(q[p1].y, q[p1].num);
		tmp[p3++] = q[p1++];
	}
	while (p2 <= r) {
		if (q[p2].op) q[p2].num += ask(q[p2].y);
		tmp[p3++] = q[p2++];
	}
	for (int i = l; i <= mid; i++) if (!q[i].op) clear(q[i].y);
	for (int i = l; i <= r; i++) q[i] = tmp[i];
}
signed main() {
	scanf("%lld%lld", &S, &W);
	while (scanf("%lld", &op)){
		if (op == 3) break;
		if (op == 1){
			int x, y, num;
			scanf("%lld%lld%lld", &x, &y, &num);cnt++;
			q[cnt] = (node){0, cnt, x, y, num};
		}else {
			int x1, y1, x2, y2;
			scanf("%lld%lld%lld%lld", &x1, &y1, &x2, &y2);
			cnt++; q[cnt] = (node) {1, cnt, x1 - 1, y1 - 1, 0};
			cnt++; q[cnt] = (node) {1, cnt, x1 - 1, y2, 0};
			cnt++; q[cnt] = (node) {1, cnt, x2, y1 - 1, 0};
			cnt++; q[cnt] = (node) {1, cnt, x2, y2, 0};
		}
	}
	CDQ(1, cnt);
	sort(q + 1, q + cnt + 1, cmp);
	for (int i = 1; i <= cnt; i++) if (q[i].op) {
		printf("%lld\n", q[i + 3].num - q[i + 1].num - q[i + 2].num + q[i].num + S * (q[i + 3].y - q[i].y) * (q[i + 3].x - q[i].x));
		i += 3;
	}
	return 0;
}