[ABC264F] Monochromatic Path

Problem Statement

We have a grid with $H$ rows and $W$ columns. Each square is painted either white or black.
For each integer pair $(i, j)$ such that $1 \leq i \leq H$ and $1 \leq j \leq W$,
the color of the square at the $i$-th row from the top and $j$-th column from the left (we simply denote this square by Square $(i, j)$) is represented by $A_{i, j}$.
Square $(i, j)$ is white if $A_{i, j} = 0$, and black if $A_{i, j} = 1$.

You may perform the following operations any number of (possibly $0$) times in any order:

• Choose an integer $i$ such that $1 \leq i \leq H$, pay $R_i$ yen (the currency in Japan), and invert the color of each square in the $i$-th row from the top in the grid. (White squares are painted black, and black squares are painted white.)
• Choose an integer $j$ such that $1 \leq j \leq W$, pay $C_j$ yen, and invert the color of each square in the $j$-th column from the left in the grid.

Print the minimum total cost to satisfy the following condition:

• There exists a path from Square $(1, 1)$ to Square $(H, W)$
  that can be obtained by repeatedly moving down or to the right, such that all squares in the path (including Square $(1, 1)$ and Square $(H, W)$) have the same color.

We can prove that it is always possible to satisfy the condition in a finite number of operations under the Constraints of this problem.

Constraints

• $2 \leq H, W \leq 2000$
• $1 \leq R_i \leq 10^9$
• $1 \leq C_j \leq 10^9$
• $A_{i, j} \in \lbrace 0, 1\rbrace$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$H$ $W$
$R_1$ $R_2$ $\ldots$ $R_H$
$C_1$ $C_2$ $\ldots$ $C_W$
$A_{1, 1}A_{1, 2}\ldots A_{1, W}$
$A_{2, 1}A_{2, 2}\ldots A_{2, W}$
$\vdots$
$A_{H, 1}A_{H, 2}\ldots A_{H, W}$

Output

Print the answer.

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Sample Input 1

3 4
4 3 5
2 6 7 4
0100
1011
1010

Sample Output 1

9

We denote a white square by 0 and a black square by 1.
On the initial grid, you can pay $R_2 = 3$ yen to invert the color of each square in the $2$-nd row from the top to make the grid:

0100
0100
1010

Then, you can pay $C_2 = 6$ yen to invert the color of each square in the $2$-nd row from the left to make the grid:

0000
0000
1110

Now, there exists a path from Square $(1, 1)$ to Square $(3, 4)$ such that all squares in the path have the same color (such as the path $(1, 1) \rightarrow (2, 1) \rightarrow (2, 2) \rightarrow (2, 3) \rightarrow (2, 4) \rightarrow (3, 4)$).
The total cost paid is $3+6 = 9$ yen, which is the minimum possible.

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Sample Input 2

15 20
29 27 79 27 30 4 93 89 44 88 70 75 96 3 78
39 97 12 53 62 32 38 84 49 93 53 26 13 25 2 76 32 42 34 18
01011100110000001111
10101111100010011000
11011000011010001010
00010100011111010100
11111001101010001011
01111001100101011100
10010000001110101110
01001011100100101000
11001000100101011000
01110000111011100101
00111110111110011111
10101111111011101101
11000011000111111001
00011101011110001101
01010000000001000000

Sample Output 2

125

思考如何走出这条路径。

如果我现在是往右走的，那么如果现在颜色不对，我可以把这一列给修改了。只要我不转成往下走，修改这一列是不会有影响别的格子的。往下走时同理。所以题目说了一定有解，而我们构造出了一个可行解。

但是我们还可以往右走时修改这一行，那么现在有两个影响因素。定义 \(dp_{i,j,k,l}\) 表示走到了点 \((i,j)\),原来方向为 \(k\)， \(l\) 表示在这个方向上是否经过反转。

那么现在有可能把整个棋盘变成 \(0\) 或者 \(1\)，这不妨分开考虑。观察这一位是否需要经过整行/整列反转，计算代价。同时看要不要转弯，转弯后是否经过反转。

注意一件事，走在 \((1,1)\) 时是可以决定行反转还是列反转的，都要试一下。

感觉写成记忆化搜索直接点

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=2005;
int h,w,r[N],c[N],a[N][N];
LL dp[N][N][2][2],ans;//0表示下，1表示右
LL dfs(int x,int y,int i,int j)
{
	if(dp[x][y][i][j]!=-1)
		return dp[x][y][i][j];
	LL ret=1e18;
	if(i)
	{
		int c=a[x][y]^j? ::c[y]:0;
		if(x==h&&y==w)
			return c;
		if(y<w)
			ret=min(ret,dfs(x,y+1,i,j)+c);
		if(x<h)
		{
			if(a[x][y]^j)
				ret=min(ret,dfs(x+1,y,0,1)+c);
			else
				ret=min(ret,dfs(x+1,y,0,0));
		}
	}
	else
	{
		int c=a[x][y]^j? r[x]:0;
		if(x==h&&y==w)
			return c;
		if(x<h)
			ret=min(ret,dfs(x+1,y,i,j)+c);
		if(y<w)
		{
			if(a[x][y]^j)
				ret=min(ret,dfs(x,y+1,1,1)+c);
			else
				ret=min(ret,dfs(x,y+1,1,0));
		}
	}
//	printf("%d %d %d %d %lld\n",x,y,i,j,ret);
	return dp[x][y][i][j]=ret;
}
int main()
{
	scanf("%d%d",&h,&w);
	for(int i=1;i<=h;i++)
		scanf("%d",r+i);
	for(int i=1;i<=w;i++)
		scanf("%d",c+i);
	for(int i=1;i<=h;i++)
	{
		char ch;
		for(int j=1;j<=w;j++)
			scanf(" %c",&ch),a[i][j]=ch-'0';
	}
	memset(dp,-1,sizeof(dp));
	ans=min(min(dfs(1,1,0,0),dfs(1,1,1,0)),min(dfs(1,1,0,1)+c[1],dfs(1,1,1,1)+r[1]));
	for(int i=1;i<=h;i++)
		for(int j=1;j<=w;j++)
			a[i][j]^=1;
	memset(dp,-1,sizeof(dp));
	ans=min(ans,min(min(dfs(1,1,0,0),dfs(1,1,1,0)),min(dfs(1,1,0,1)+c[1],dfs(1,1,1,1)+r[1])));
	printf("%lld",ans);
}