【Codeforces】-554B-Ohana Cleans Up(水)
来源:互联网 发布:c语言简单计算器程序 编辑:程序博客网 时间:2024/06/05 08:13
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.
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.
The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean.
40101100011110101
2
3111111111
3
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.
题解:一样的行出现的越多就就可以擦干净的越多。
不然都不一样只可以擦干净一行。
#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;#define CLR(a,b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3fint main(){int n;char a[110][110];bool vis[110];while(~scanf("%d",&n)){CLR(a,0);CLR(vis,false);for(int i=1;i<=n;i++)scanf("%s",a[i]+1);int num,ans=1;for(int i=1;i<n;i++){num=1;for(int j=i+1;j<=n;j++){int flag=1;if(vis[j]==true)continue;for(int x=1;x<=n;x++){if(a[i][x]!=a[j][x]){flag=0;break;}}if(flag){vis[j]=true;num++;}ans=max(ans,num);}}printf("%d\n",ans);}return 0;}
- CodeForces - 554B Ohana Cleans Up (模拟)水
- 【Codeforces】-554B-Ohana Cleans Up(水)
- codeforces 554B B. Ohana Cleans Up
- Codeforces 554B Ohana Cleans Up
- codeforces 554B Ohana Cleans Up
- CodeForces-554B Ohana Cleans Up
- Codeforces 554B Ohana Cleans Up 【水题】
- CodeForces 554B. Ohana Cleans Up【思维】
- 【codeforces 554B Ohana Cleans Up】
- CodeForces 554B Ohana Cleans Up
- CodeForces 554B Ohana Cleans Up
- CodeForces 554B Ohana Cleans Up
- 【59.49%】【codeforces 554B】Ohana Cleans Up
- 【CodeForces】554B - Ohana Cleans Up(思维)
- Codeforces 554B:Ohana Cleans Up(思维+水题)
- CodeForces 554B Ohana Cleans Up(思维)
- Codeforces 554B Ohana Cleans Up 扫房间
- CodeForces 554B Ohana Cleans Up 【思维+map】
- RocketMQ-- 建议去 slave 拉取消息
- zabbix安装过程
- 【POJ 1487】Single-Player Games(建方程高消)
- Java多线程系列--“JUC锁”03之 公平锁(一) (r)
- Light OJ 1031 区间dp
- 【Codeforces】-554B-Ohana Cleans Up(水)
- 搭建caffe问题总结
- 一个使用了 Bootstrap 的基本的 HTML 模板
- js做的稍微复杂地计算器
- 【codeforces】Dima and Guards
- ASP.NET MVC随想录——锋利的KATANA
- OpenStack云数据的护花使者Smaug
- Codeforces Round #354 (Div. 2) C. Vasya and String (尺取法)
- Java多线程系列--“JUC锁”04之 公平锁(二) (r)