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]);
}