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最新推荐文章于 2024-05-04 16:55:46 发布

siyutt

最新推荐文章于 2024-05-04 16:55:46 发布

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分类专栏：思维题目笔记

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本文介绍了一个关于修复2*nX2*n大小的混乱棋盘的问题。一个完整的棋盘被分为四个n*n大小的部分，颜色分布混乱。文章探讨了如何重新排列这些部分并更改最少数量的方格颜色以恢复棋盘原有的黑白相间模式。

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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 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.

题意:有一个2n X 2n大小的棋盘，棋盘上的颜色有两种它们分别用1和0表示，它们的位置相互交错。现在有一个棋盘分裂为4个n*n大小的块，并且上面的颜色已经乱了，要求你把这些块拼在一起，要求求出使棋盘颜色符合要求所需改颜色的次数最少是多少。

#include <iostream>
#include <fstream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <cmath>
#include <algorithm>
#include <functional>
#define inf 1000000000
using namespace std;
typedef long long ll;
const int MAXN=1e9+10;
const int MAX=1e5+10;
const double eps=1e-6;

int n;
char s[5][110][110];

int main(){
    #ifdef ONLINE_JUDGE
    #else
    freopen("in.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    #endif

    cin>>n;
    for(int k=1;k<=4;k++)
        for(int i=0;i<n;i++)
                scanf("%s",&s[k][i]);
    int ans=100000;
    for(int k=1;k<=4;k++){
        for(int h=k+1;h<=4;h++){
            int y,x;
            x=y=0;
            for(int i=1;i<=4;i++)
                if(i!=h&&i!=k){
                    if(!x)  x=i;
                    else    y=i;
                }
            int temp=0;
            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    if((i%2==1&&j%2==1)||(i%2==0&&j%2==0)){
                        if(s[k][i][j]!='1')   temp++;
                        if(s[h][i][j]!='1')   temp++;
                        if(s[x][i][j]!='0')   temp++;
                        if(s[y][i][j]!='0')   temp++;

                    }
                    else{
                        if(s[k][i][j]!='0')   temp++;
                        if(s[h][i][j]!='0')   temp++;
                        if(s[x][i][j]!='1')   temp++;
                        if(s[y][i][j]!='1')   temp++;
                    }
                }
            }
            ans=min(ans,temp);
        }
    }
    cout<<ans<<endl;

    return 0;
}