[ABC233G] Strongest Takahashi

Problem Statement

There is a $N \times N$ grid, with blocks on some squares.
The grid is described by $N$ strings $S_1,S_2,\dots,S_N$, as follows.

• If the $j$-th character of $S_i$ is #, there is a block on the square at the $i$-th row from the top and $j$-th column from the left.
• If the $j$-th character of $S_i$ is ., there is not a block on the square at the $i$-th row from the top and $j$-th column from the left.

Takahashi can do the operation below zero or more times.

• First, choose an integer $D$ between $1$ and $N$ (inclusive), and a $D \times D$ subsquare within the grid.
• Then, consume $D$ stamina points to destroy all blocks within the subsquare.

Find the minimum number of stamina points needed to destroy all the blocks.

Constraints

• $N$ is an integer.
• $1 \le N \le 50$
• $S_i$ consists of # and ..
• $|S_i|=N$

---

Input

Input is given from Standard Input in the following format:

$N$
$S_1$
$S_2$
$\vdots$
$S_N$

Output

Print the answer as an integer.

---

Sample Input 1

5
##...
.##..
#.#..
.....
....#

Sample Output 1

4

By choosing the subsquares below, Takahashi will consume $4$ stamina points, which is optimal.

• The $3 \times 3$ subsquare whose top-left square is at the $1$-st row from the top and $1$-st column from the left.
• The $1 \times 1$ subsquare whose top-left square is at the $5$-th row from the top and $5$-th column from the left.

---

Sample Input 2

3
...
...
...

Sample Output 2

0

There may be no block on the grid.

---

Sample Input 3

21
.....................
.....................
...#.#...............
....#.............#..
...#.#...........#.#.
..................#..
.....................
.....................
.....................
..........#.....#....
......#..###.........
........#####..#.....
.......#######.......
.....#..#####........
.......#######.......
......#########......
.......#######..#....
......#########......
..#..###########.....
.........###.........
.........###.........

Sample Output 3

19

50的数据范围，基本上高次dp或者折半搜索的。这题不太像折半搜索，那就试一下高维dp。

定义 $dp_{x1,y1,x2,y2}$ 为解决以 $(x1,y1)$ 为左上角，以 $(x2,y2)$ 为右下角矩形 所需的最小代价.

首先肯定有一种 $\max(y2-y1+1,x2-x1+1)$ 的方案，那就是把他填满。

有一个引理，在我们填的矩阵中，一定不可能接壤，不然的话就可以选一个更大的矩阵，在代价一样的情况下框住的面积更大了。

所以如果不是全选的话，一定可以沿着某一行或者某一列把大矩阵割成两个矩阵解决。枚举对应的这一行或一列，递归下去就行了

这个dp的顺序应该按照区间dp的顺序。

#include<bits/stdc++.h>
using namespace std;
const int N=55;
int n,dp[N][N][N][N],c1[N][N],c2[N][N],r1,r2;
char s[N][N];
void tomax(int&a,int b) 
{
	a=min(a,b);
}
int main() 
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%s",s[i]+1);
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			c1[i][j]=c1[i][j-1]+(s[i][j]=='#');
			c2[i][j]=c2[i-1][j]+(s[i][j]=='#');
		}
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			for(int l1=1;l1+i-1<=n;l1++)
			{
				for(int l2=1;l2+j-1<=n;l2++)
				{
					r1=l1+i-1,r2=l2+j-1;
					dp[l1][l2][r1][r2]=max(i,j);
					for(int k=l1;k<=r1;k++)
						if(c1[k][r2]==c1[k][l2-1])
							tomax(dp[l1][l2][r1][r2],dp[l1][l2][k-1][r2]+dp[k+1][l2][r1][r2]);
					for(int k=l2;k<=r2;k++)
						if(c2[l1-1][k]==c2[r1][k])
							tomax(dp[l1][l2][r1][r2],dp[l1][l2][r1][k-1]+dp[l1][k+1][r1][r2]);
//					printf("%d %d %d %d %d\n",l1,l2,r1,r2,dp[l1][l2][r1][r2]);
				}
			}
		}
	}
	printf("%d",dp[1][1][n][n]);
}