[USACO11MAR] Brownie Slicing G题解(二分+二维前缀和+矩阵分割）

[USACO11MAR] Brownie Slicing G

题目地址

P3017 [USACO11MAR] Brownie Slicing G

思路

二分最大化最小值
切割思路：

一行一行进行切割，如果这一行可以切割出b块大于等于mid的块，就开始切割下一行
如果无法切割出b块，就把正在切割的行与下一行拼起来一起切割
最后通过能切割出b块的水平块块够不够a条来判断m是否合适

代码

#include <iostream>using namespace std;int a[1010][1010], s[1010][1010];
int r, c, x, y;bool check(int m) {int lrow = 0;int rows = 0;for (int i = 1; i <= r; i ++) {int num = 0, sum = 0;for (int j = 1; j <= c; j ++) {if (sum + (s[i][j]-s[i][j-1])-(s[lrow][j]-s[lrow][j-1]) < m)sum += (s[i][j]-s[i][j-1])-(s[lrow][j]-s[lrow][j-1]);else {sum = 0;num ++;}}if (num >= y) {lrow = i;++ rows;}}return rows >= x;
}int main() {cin >> r >> c >> x >> y;for (int i = 1; i <= r; i ++)for (int j = 1; j <= c; j ++) {cin >> a[i][j];s[i][j] = s[i-1][j]+s[i][j-1]-s[i-1][j-1]+a[i][j];}int left = 0, right = s[r][c];//m 越小越容易成功while (left < right) {int m = left + right + 1 >> 1;if (check(m))left = m;elseright = m - 1;}cout << left;return 0;
}