Educational Codeforces Round 41 (Rated for Div. 2) C.Chessboard (DP)

C. Chessboard

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Magnus decided to play a classic chess game. Though what he saw in his locker shocked him! His favourite chessboard got broken into 4 pieces, each of size n by n, n is always odd. And what's even worse, some squares were of wrong color. j-th square of the i-th row of k-th piece of the board has color a_k, i, j; 1 being black and 0 being white.

Now Magnus wants to change color of some squares in such a way that he recolors minimum number of squares and obtained pieces form a valid chessboard. Every square has its color different to each of the neightbouring by side squares in a valid board. Its size should be 2n by 2n. You are allowed to move pieces but not allowed to rotate or flip them.

Input

The first line contains odd integer n (1 ≤ n ≤ 100) — the size of all pieces of the board.

Then 4 segments follow, each describes one piece of the board. Each consists of n lines of n characters; j-th one of i-th line is equal to 1 if the square is black initially and 0 otherwise. Segments are separated by an empty line.

Output

Print one number — minimum number of squares Magnus should recolor to be able to obtain a valid chessboard.

Examples

Input

Copy

1
0

0

1

0

Output

Copy

1

Input

Copy

3
101
010
101

101
000
101

010
101
011

010
101
010

Output

Copy2

题意，现在有2n*2n的棋盘，被分成了n*n的四个部分，现在可以以任意的顺序拼回原来的棋盘
然后需要进行操作将棋盘变得相邻的方格的数字不同，每次操作可以将一个0变成1，或者0变成1
问最少需要多少此操作

分析：棋盘拼接的方式有24种，所以可以直接枚举所有的拼接方式。
对于每种拼接方式完成后，用dp[i][j][k] 表示拼接到i行j列,且该方格的数字为k时所消耗的操作数
则有转移方程如下（r为该方格最初的颜色）
dp[i][j][r]=dp[i-1][j][r^1]+dp[i][j-1][r^1]-dp[i-1][j-1][r];
dp[i][j][r^1]=dp[i-1][j][r]+dp[i][j-1][r]-dp[i-1][j-1][r^1]+1;


代码如下:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
#define INF 0x3f3f3f3f
int map1[220][220];
int dp[220][220][2];
int n;
int nn;
int minn=INF;
int str[5][220][220];
char ch;
int x[5]={0,1,2,3,4};
void trans(int a,int b,int c,int d) 
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        map1[i][j]=str[a][i][j];

     for(int i=1;i<=n;i++)
        for(int j=n+1;j<=nn;j++)
          map1[i][j]=str[b][i][j-n];

    for(int i=n+1;i<=nn;i++)
        for(int j=1;j<=n;j++)
        map1[i][j]=str[c][i-n][j];

     for(int i=n+1;i<=nn;i++)
          for(int j=n+1;j<=nn;j++)
          map1[i][j]=str[d][i-n][j-n];

}

void  solve()
{
  for(int i=0;i<=nn;i++)
  {
     for(int k=0;k<=1;k++)
     {
     dp[i][0][k]=0;
     dp[0][i][k]=0;
     }
  }
    dp[1][1][map1[1][1]]=0;
    dp[1][1][1^map1[1][1]]=1;

  for(int i=1;i<=nn;i++)
    for(int j=1;j<=nn;j++)
  {
     if(i==1&&j==1)continue;
     int r=map1[i][j];
     dp[i][j][r]=dp[i-1][j][r^1]+dp[i][j-1][r^1]-dp[i-1][j-1][r];
     dp[i][j][r^1]=dp[i-1][j][r]+dp[i][j-1][r]-dp[i-1][j-1][r^1]+1;
  }


  minn=min(dp[nn][nn][0],minn);
  minn=min(dp[nn][nn][1],minn);
}

int main()
{
   scanf("%d",&n);
   nn=2*n;

  for(int k=1;k<=4;k++)
   for(int i=1;i<=n;i++)
     for(int j=1;j<=n;j++)
     {
         scanf(" %c",&ch);
         str[k][i][j]=ch-'0';
     }

   int num=0;
    do
    {
      trans(x[1],x[2],x[3],x[4]);
      solve();
      num++;
    }while(next_permutation(x+1,x+5));
  printf("%d
",minn);

   return 0;
}