codeforces B. Ohana Cleans Up

最新推荐文章于 2021-08-17 22:19:40 发布

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原文链接：https://yq.aliyun.com/articles/567265

介绍了一个算法问题，关于如何通过翻转某些列的状态使二维网格中的尽可能多的行变得完全干净。该算法通过选定某些列并翻转其状态来最大化完全干净的行数。

　　Ohana Matsumae is trying to clean a room, which is divided up into an n by n grid of squares. Each square is initially either clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column.

Return the maximum number of rows that she can make completely clean.

Input

The first line of input will be a single integer n (1 ≤ n ≤ 100).

The next n lines will describe the state of the room. The i-th line will contain a binary string with n characters denoting the state of the i-th row of the room. The j-th character on this line is '1' if the j-th square in the i-th row is clean, and '0' if it is dirty.

Output

The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean.

Sample test(s)

input

output

input

output

Note

In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean.

In the second sample, everything is already clean, so Ohana doesn't need to do anything.

/*
    题意：选中某几列， 然后将这些列中为0的变为1， 为1的变为0，问最多能有多少行全为1 
    
    思路：假设最终答案包括第i行，那么如果a[i][j] 之前为0，则对应的这一列 j 一定是被选中的！ 
    对于每一行，将这一行某一列为0的列作为选中的列，然后再遍历一遍数组，计算全1的行的个数。 
*/ 
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include<set> 
using namespace std;
char a[105][105], aa[105][105];
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1; i<=n; ++i){
        scanf("%s", a[i]+1);
        for(int j=1; j<=n; ++j)
            aa[i][j] = a[i][j];
    }
    int ans = 0;
    for(int k=1; k<=n; ++k){
        for(int i=1; i<=n; ++i){
            if(a[k][i]=='0'){
                for(int j=1; j<=n; ++j)
                    if(a[j][i]=='0')
                        a[j][i]='1';
                    else a[j][i] = '0';
            }
        }
        int ss = 0;
        for(int i=1; i<=n; ++i)
            for(int j=1; j<=n; ++j)
                if(a[i][j] == '0')
                    break;
                else if(j==n)
                    ++ss;
        if(ans < ss)    ans = ss;
        for(int i=1; i<=n; ++i)
            for(int j=1; j<=n; ++j)
                a[i][j] = aa[i][j];
    }
    printf("%d\n", ans);
    return 0;
}