本文是博主原创文章，未经允许不得转载。
我在博客园也同步发布了此文，链接 http://www.cnblogs.com/umbrellalalalala/p/8824234.html
题目来源 http://codeforces.com/problemset/problem/961/C
【题目】（看不懂可以往下翻，有翻译）
Magnus decided to play a classic chess game. Though what he saw in his locker shocked him! His favourite chessboard got broken into4 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 integern (1 ≤ n ≤ 100) — the size of all pieces of the board.
Then 4 segments follow, each describes one piece of the board. Each consists ofn 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
1
0
0
1
0Output
1Input
3
101
010
101
101
000
101
010
101
011
010
101
010Output
2
2【分析】
大概意思就是我的棋盘碎成了四块一样大小的n×n正方形碎片，其中n为奇数，而且棋盘碎片上的方格颜色也不太对。现在我要将四个碎片拼成一个完整的棋盘，按照常理，棋盘的颜色是黑白相间的，但是由于碎片的颜色不太对，所以我需要改变一些碎片上的方格的颜色。问：我需要至少改变多少个方格的颜色，才能将碎片拼成一个完整的棋盘。样例输入就是四片碎片的形态，其中0代表白色方格，1代表个黑色方格。在拼接的时候只能平移碎片，不能旋转、反转。
【示例代码】（注释包含思路，注意理解其中0型棋盘和1型棋盘的含义，概念是我自定义的。我们最终只需要以最小的颜色改变数将四片棋盘变成两个0型棋盘和两个1型棋盘即可）
#include<stdio.h>
#include<stdlib.h>
#define MAX_N 105
//分别用于输入四片碎片
char chessboad1[MAX_N][MAX_N];
char chessboad2[MAX_N][MAX_N];
char chessboad3[MAX_N][MAX_N];
char chessboad4[MAX_N][MAX_N];
int main() {
int n;
int type[4][2];//0型棋盘是0比1多一位，1型棋盘则相反。type[i][0]存放若将第i片改变成0型需要改变几片的颜色，type[i][1]同理。最终我们需要将四片棋盘变成两个0型棋盘和两个1型棋盘
//The input model
scanf("%d", &n);
getchar();
for(int i=0;i<4;i++)
for (int j = 0; j < 2; j++) {
type[i][j] = 0;
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%c", &chessboad1[i][j]);
if ((i + j) % 2 == 0) {
if (chessboad1[i][j] == '1')type[0][0]++;
else type[0][1]++;
}
else {
if (chessboad1[i][j] == '1')type[0][1]++;
else type[0][0]++;
}
}
getchar();
}
getchar();
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%c", &chessboad2[i][j]);
if ((i + j) % 2 == 0) {
if (chessboad2[i][j] == '1')type[1][0]++;
else type[1][1]++;
}
else {
if (chessboad2[i][j] == '1')type[1][1]++;
else type[1][0]++;
}
}
getchar();
}
getchar();
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%c", &chessboad3[i][j]);
if ((i + j) % 2 == 0) {
if (chessboad3[i][j] == '1')type[2][0]++;
else type[2][1]++;
}
else {
if (chessboad3[i][j] == '1')type[2][1]++;
else type[2][0]++;
}
}
getchar();
}
getchar();
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
scanf("%c", &chessboad4[i][j]);
if ((i + j) % 2 == 0) {
if (chessboad4[i][j] == '1')type[3][0]++;
else type[3][1]++;
}
else {
if (chessboad4[i][j] == '1')type[3][1]++;
else type[3][0]++;
}
}
getchar();
}
//The input model end
int temp;
for(int i=0;i<4-1;i++)
for (int j = 0; j < 4 - i - 1; j++) {
if (type[j][0] > type[j + 1][0]) {
temp = type[j][0];
type[j][0] = type[j + 1][0];
type[j + 1][0] = temp;
temp = type[j][1];
type[j][1] = type[j + 1][1];
type[j + 1][1] = temp;
}
}
printf("%d\n", type[0][0] + type[1][0] + type[2][1] + type[3][1]);
system("pause");
return 0;
}