Codeforces 364E 分治

题意：给你一个01矩阵，问此矩阵有多少个和恰好为k的子矩形。

思路：分治，对于当前矩形，用一条中线把矩形分成两半，分治之后计算跨过中线的矩形个数。更具体的来说（假设划了一条水平中线），我们枚举矩形左右边界，然后用指针维护一下到中线的连续和为k的边界。之后通过差分就可以计算出对应的左右边界的矩形的贡献数目。对于一个n * m的矩阵，计算贡献的时间复杂度是O(n * (m * k + n))的，带有n * n项，所以计算的时候需要用交替画水平线和竖直线，不然就超时了。总复杂度O（n * m * k * ( log(n) + log(m) ) );

代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2510;
int sum[maxn][maxn];
char s[maxn][maxn];
int n, m, K;
long long ans;
int p1[10], p2[10];
int cal(int x1, int y1, int x2, int y2) {
	return sum[x2][y2] - sum[x1 - 1][y2] - sum[x2][y1 - 1] + sum[x1 - 1][y1 - 1];
}
void div(int x1, int y1, int x2, int y2, bool flag) {
	if(x1 > x2 || y1 > y2) return;
	if(x1 == x2 && y1 == y2) {
		ans += (cal(x1, y1, x2, y2) == K);
		return;
	}
	if(flag) {
		int mid = (x1 + x2) >> 1;
		div(x1, y1, mid, y2, !flag);
		div(mid + 1, y1, x2, y2, !flag);
		for (int i = y1; i <= y2; i++) {
			for (int k = 0; k <= K; k++) {
				p1[k] = mid, p2[k] = mid + 1;
			}
			for (int j = y2; j >= i; j--) {
				for (int k = 0; k <= K; k++) {
					while(p1[k] >= x1 && cal(p1[k], i, mid, j) <= k) p1[k]--;
					while(p2[k] <= x2 && cal(mid + 1, i, p2[k], j) <= k) p2[k]++;
				}
				for (int k = 1; k < K; k++) {
					ans += (p1[k - 1] - p1[k]) * (p2[K - k] - p2[K - k - 1]);
				}
				if(K > 0) {
					ans += (mid - p1[0]) * (p2[K] - p2[K - 1]);
					ans += (p2[0] - mid - 1) * (p1[K - 1] - p1[K]);
				} else if(K == 0) {
					ans += (mid - p1[0]) * (p2[0] - mid - 1);
				}
			}
		}		
	}
	else {
		int mid = (y1 + y2) >> 1;
		div(x1, y1, x2, mid, !flag);
		div(x1, mid + 1, x2, y2, !flag);
		for (int i = x1; i <= x2; i++) {
			for (int k = 0; k <= K; k++) {
				p1[k] = mid, p2[k] = mid + 1;
			}
			for (int j = x2; j >= i; j--) {
				for (int k = 0; k <= K; k++) {
					while(p1[k] >= y1 && cal(i, p1[k], j, mid) <= k) p1[k]--;
					while(p2[k] <= y2 && cal(i, mid + 1, j, p2[k]) <= k) p2[k]++;
				}
				for (int k = 1; k < K; k++) {
					ans += (p1[k - 1] - p1[k]) * (p2[K - k] - p2[K - k - 1]);
				}
				if(K > 0) {
					ans += (mid - p1[0]) * (p2[K] - p2[K - 1]);
					ans += (p2[0] - mid - 1) * (p1[K - 1] - p1[K]);
				} else if(K == 0) {
					ans += (mid - p1[0]) * (p2[0] - mid - 1);
				}
				//printf("%d %d %d %d %d %d
", x1, y1, x2, y2, 2, ans);
			}
		}
	}
}
int main() {
	//freopen("out.txt", "r", stdin);
	scanf("%d%d%d", &n, &m, &K);
	for (int i = 1; i <= n; i++) {
		scanf("%s", s[i] + 1); 
	}
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++) {
			sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + (s[i][j] == '1');
		}
	div(1, 1, n, m, 0);
	printf("%lld
", ans);
}