C. Chessboard time limit per test1 second memory limit per test256 megabytes inputstandard input out_3. oleg and chess time limit per test 6.5 seconds -优快云博客

本文链接：https://blog.youkuaiyun.com/SunMoonVocano/article/details/79865816

本文介绍了一道关于修复损坏棋盘的算法题。任务是将四个大小为n×n的棋盘碎片组合成一个2n×2n的有效棋盘，其中n为奇数。目标是最少地更改方格颜色，确保每相邻方格颜色不同。

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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 ak, 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
inputCopy
1
0

0

1

0
output
1
inputCopy
3
101
010
101

101
000
101

010
101
011

010
101
010
output
2

这个题感觉蛮有意思，四块棋盘，每个格子或者是0或者是1，让你把四个长为奇数的棋盘合成一个棋盘，顺序不计，问你最少1修改几个格子，使得每个格子与相邻的格子数字不同。

我们先想想，正确的棋盘是什么样子那？无非是1个0,1个1交替出现，与坐标联系起来，i+j 为奇数则为某个状态，否则为另一个状态。其实一个完整的棋盘分成四个，再把它合起来，能产生两种棋盘，第一个为0 或者第一个为 1.

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long ll;

char s[4][150][150];
char mp[250][250];
int ii[] = {0,1,2,3};
int n;


int main()
{
	cin >> n;
	for(int i = 0; i < 4; i++)
	for(int j = 0; j < n; j++)
	for(int k = 0; k < n; k++)
	    cin >> s[i][j][k];
	int ans = 1e8;
	do{
		int tmp = 0;
		for(int i = 0; i < 2*n; i++)
		for(int j = 0; j < 2*n; j++)
		{
  	        if(i <= n && j <= n)
			{
				mp[i][j] = s[ii[0]][i][j];
			}
			if(i <= n && j > n)
			{
				mp[i][j] = s[ii[1]][i][j-n];
			}
			if(i > n && j <= n)
			{
				mp[i][j] = s[ii[2]][i-n][j];
			}
			if(i > n && j > n)
			{
				mp[i][j] = s[ii[3]][i-n][j-n];
			}
			char c = (i+j) % 2 ? '1' : '0';
			if(c != mp[i][j]) tmp++;
		}
		ans = min(ans,tmp);
	}while(next_permutation(ii,ii+4));
	cout << ans <<endl;
    return 0;
}