codeforces B. Ohana Cleans Up
2015-06-26 00:05
330 查看
B. Ohana Cleans Up
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.
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
4 0101 1000 1111 0101
output
2
input
3 111 111 111
output
3
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; }
相关文章推荐
- nagios邮件报警脚本
- 设计模式学习--建造者模式
- 企业版证书($299)In-House方式发布指南
- windows编程学习笔记(三)ListBox的使用方法
- 得到view坐标的各种方法
- CSS定位规则之BFC 你竟然一直不知道的东西!!!!!
- spring的Ioc详解
- 如何成为一名优秀的iOS开发工程师
- Leetcode#1 Two Sum
- oracle 索引是什么
- 2015062509 - 思考
- 不借助 Wine 和云服务:新项目能让 Linux 完整运行微软 Office 套件
- WPS Office:Linux 上的 Microsoft Office 的免费替代品
- 在Python中使用zlib模块进行数据压缩的教程
- python读取TXT到数组及列表去重后按原来顺序排序的方法
- Thinkphp关闭缓存的方法
- 【吾日三省吾身】2015.6.25-涅槃行动第三十八天
- PHP多态代码实例
- PHP批量去除BOM头代码分享
- 以文件形式缓存php变量的方法