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Chessboard
阅读量:4710 次
发布时间:2019-06-10

本文共 2842 字,大约阅读时间需要 9 分钟。

Chessboard

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 nn 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, j1 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 nlines 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
10010
Output
1
Input
3101010101101000101010101011010101010
Output
2
题解:
#include 
#include
#include
#include
using namespace std;int n;char map[4][150][150];int move[4][2] = {1,0,0,1,-1,0,0,-1};int dfs_1(int q){ int count = 0; if (map[q][0][0] == '0') { } else { count++; } int p = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n ; j++) { if (i == 0 && j == 0) { continue; } else { if (map[q][i][j]-'0' == !p) { p = map[q][i][j] - '0'; } else { p = !p; count++; } } } } return count;}int dfs_2(int q){ int count = 0; if (map[q][0][0] == '1') { } else { count++; } int p = 1; for (int i = 0; i < n; i++) { for (int j = 0; j < n ; j++) { if (i == 0 && j == 0) { continue; } else { if (map[q][i][j]-'0' == !p) { p = map[q][i][j] - '0'; } else { p = !p; count++; } } } } return count;}int main(){ scanf("%d", &n); for (int i = 0; i < 4; i++) { for (int j = 0; j < n; j++) { scanf("%s", map[i][j]); } } int answer[4][2]; for (int i = 0; i < 4 ;i++) { answer[i][0] = dfs_1(i); answer[i][1] = dfs_2(i); } int m = 100000000; m = min(answer[0][0]+ answer[1][0]+answer[2][1]+answer[3][1], m); m = min(answer[0][0]+ answer[1][1]+answer[2][0]+answer[3][1], m); m = min(answer[0][0]+ answer[1][1]+answer[2][1]+answer[3][0], m); m = min(answer[0][1]+ answer[1][1]+answer[2][0]+answer[3][0], m); m = min(answer[0][1]+ answer[1][0]+answer[2][1]+answer[3][0], m); m = min(answer[0][1]+ answer[1][0]+answer[2][0]+answer[3][1], m); printf("%d\n", m); return 0;}

 

转载于:https://www.cnblogs.com/focus5679/p/9286183.html

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