[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.

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.

Sample Input 1

Sample Output 1

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:

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:

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.

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

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);

}