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: Print the minimum total cost to satisfy the following condition: We can prove that it is always possible to satisfy the condition in a finite number of operations under the Constraints of this problem.Problem Statement
Constraints
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
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.
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);
}