题目链接：点击打开链接
Magnus decided to play a classic chess game. Though what he saw in his locker shocked him! His favourite chessboard got broken into 4pieces, 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 2nby 2n. You are allowed to move pieces but not allowed to rotate or flip them.
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.
Print one number — minimum number of squares Magnus should recolor to be able to obtain a valid chessboard.
1 0010
1
3 101 010 101101 000 101010 101 011010 101 010
2
题目大意：问最少染多少色，可以将四块拼成一个完整的国际象棋棋盘
思路：四块碎片是无序的，随便以什么顺序拼，只要能完成一块棋盘就行
定义模式一：棋盘碎片的第一格为1， 如样例二中的第一块碎片
定义模式二：棋盘碎片的第一格为0， 如样例二中的第四块碎片
显然一个完整的棋盘应该是由两块模式一、两块模式二组成的
于是我们只需要求出每块碎片变成模式一、变成模式二需要的染色数，然后暴力枚举哪两块碎片变成模式一（另外两块碎片变成模式二）总共需要的染色数最少，就行了
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<set>
#include<queue>
#include<cmath>
typedef long long ll;
using namespace std;
const ll maxn = 1e6 + 10;
char a[105][105];
int m1[4], m2[4];
int main(){int n;memset(m1, 0, sizeof(m1));memset(m2, 0, sizeof(m2));scanf("%d",&n);getchar();for(int i = 0 ; i < 4 ; i++){for(int j = 0; j < n ;j++){scanf("%s", a[j]);for(int k = 0; k < n; k++){if((j+k)%2==0){if(a[j][k]-'0' == 1) m2[i]++;else if(a[j][k] - '0' == 0) m1[i]++;} else if((j+k)%2==1){if(a[j][k]-'0' == 1)m1[i]++;else if(a[j][k] - '0' == 0) m2[i]++;}}}}int ans = n * n * 4;for(int i = 0; i < 3; i++){//第i和j个棋盘碎片作为模式一，其他两个模式二 for(int j = i+1; j < 4; j++){int sum = 0;for(int k = 0; k < 4; k++){if(k!=i && k!=j)sum+=m2[k];}sum+=(m1[i]+m1[j]);//cout << sum << endl;ans = min(ans, sum);}}cout << ans << endl;return 0;
}