B. Ohana Cleans Uptime limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Examplesinput
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.
这题题意倒是很快就懂了，，但却没有什么思路直接跳下一题了。。错失一血（心痛3s）
题意：n*n的房间，n行n列个格子。每个格子初始为1代表干净，反之是脏的。小明可以每次清理一整列，次数不限，但此列原先干净的变成脏的，脏的变成干净的。问最多有多少行可以全部是干净的。（结合样例理解吧）
思路：求最多的行数，很容易想到最终符合条件的这些行的初始状态肯定是一样的，所以直接判断有多少行相同即可，直接用map存string计数。（也是看到有人一血后才瞬间知道题意做法。。）
const int N=1e5+10;
string a;
int main()
{
int n;
while(~scanf("%d",&n))
{
map<string,int>q;
q.clear();
int num=0;
for(int i=0;i<n;i++)
{
cin>>a;
q[a]++;
num=max(num,q[a]);
}
printf("%d\n",num);
}
return 0;
}
再次错失一血，，心痛10000s+
