HDU -- 2830 -- matrix of any swap Column

What we do in the past two days is related to the matrix =-= ....

There are so many flowers in the small matrix...

This topic features that we can swap any two columns any times can be arbitrarily exchanged between column X and column Y at any time.

At first, I was worried that my method would not be able to see Ms... I always think there will be a more efficient method, but I did not expect it

My idea is to assume that the height of column Y in row X is 3 4 5 2 3 6

If the height is greater than or equal to 2, 6 to 0 columns in total. 0 can be regarded as (CNT [0] + CNT [1]). for the sum, you can assume that we have also placed 0 columns with a height of 0 and 1.

Similarly, the height> = 3 is 6-1...

So I am using this method for computation. It is quite simple to have some details, just pay attention to the processing. For example, the CNT [] array should be reset to 0 in each row because the H [] array in each row all are changing.

-------------- Actually not that nice =-= I have hidden the code first. If I haven't done it now, I can check it out ~

 1 #include <iostream> 2 using namespace std; 3  4 const int size = 1010; 5 char str[size]; 6 int h[size]; 7 int cnt[size]; 8  9 int main()10 {11     cin.sync_with_stdio(false);12     int n , m , ans , sum , maxH , area;13     while( cin >> n >> m )14     {15         memset( cnt , 0 , sizeof(cnt) );16         memset( h , 0 , sizeof(h) );17         ans = 0;18         for( int i = 1 ; i<=n ; i++ )19         {20             maxH = sum = 0;21             cin >> str;22             for( int j = 0 ; str[j]!=‘\0‘ ; j++ )23             {24                 if( str[j]==‘1‘ )25                     h[j+1] ++;26                 else27                     h[j+1] = 0;28                 cnt[ h[j+1] ] ++;29                 if( maxH < h[j+1] )30                     maxH = h[j+1];            31             }32             for( int k = 0 ; k<=maxH ; k++ )33             {34                 if( cnt[k]!=0 )35                 {36                     area = k * (m-sum);37                     sum += cnt[k];38                     if( area>ans )39                         ans = area;40                 }41                 cnt[k] = 0;42             }43         }44         cout << ans << endl;45     }46     return 0;47 }

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