Codeforces Round #309 (Div. 2) B 暴力

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B. Ohana Cleans Up

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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.

Examples
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：

给出n*n的矩阵，0表示脏，1表示干净。每次只能扫一列，扫完之后原来干净的会变脏，脏的会变干净

问最佳的扫法使得干净的行最多。输出最大的干净行

思路：

枚举当前行如果作为干净行的所有干净行数，暴力更新取值

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include <ctime>#include<queue>#include<set>#include<map>#include<stack>#include<iomanip>#include<cmath>#define mst(ss,b) memset((ss),(b),sizeof(ss))#define maxn 0x3f3f3f3f#define MAX 1000100///#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef unsigned long long ull;#define INF (1ll<<60)-1using namespace std;int a[110][110],b[110][110];int n;int main(){    string s;    scanf("%d",&n);    for(int i=1;i<=n;i++){        cin>>s;        for(int j=0;j<s.size();j++){            a[i][j+1]=s[j]-'0';        }    }    int ans=0;    for(int i=1;i<=n;i++){        for(int j=1;j<=n;j++){            for(int k=1;k<=n;k++){                b[j][k]=a[j][k];            }        }        for(int j=1;j<=n;j++){            if(b[i][j]==0){                for(int k=1;k<=n;k++){                    b[k][j]^=1;                }            }        }        int num=0;        for(int j=1;j<=n;j++){            int t=0;            for(int k=1;k<=n;k++){                t+=b[j][k];            }            if(t==n) num++;        }        ans=max(ans,num);    }    cout<<ans<<endl;    return 0;}