HDOJ 2870 Largest Submatrix

逐层的找最大面积。

Largest Submatrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 967    Accepted Submission(s): 467

Problem Description

Now here is a matrix with letter 'a','b','c','w','x','y','z' and you can change 'w' to 'a' or 'b', change 'x' to 'b' or 'c', change 'y' to 'a' or 'c', and change 'z' to 'a', 'b' or 'c'. After you changed it, what's the largest submatrix with the same letters you can make?

 

Input

The input contains multiple test cases. Each test case begins with m and n (1 ≤ m, n ≤ 1000) on line. Then come the elements of a matrix in row-major order on m lines each with n letters. The input ends once EOF is met.

 

Output

For each test case, output one line containing the number of elements of the largest submatrix of all same letters.

 

Sample Input

2 4
abcw
wxyz

 

Sample Output

3

 

Source

2009 Multi-University Training Contest 7 - Host by FZU

 

Recommend

gaojie

 

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

const int N=1005;

char a[N][N];

int high[N];

int C,R;

bool isH(char stc,char c)

{

    if(stc=='a')

        return c=='a'||c=='w'||c=='y'||c=='z';

    if(stc=='b')

        return c=='b'||c=='w'||c=='x'||c=='z';

    if(stc=='c')

        return c=='c'||c=='x'||c=='y'||c=='z';

    return false;

}

int main()

{

    while(scanf("%d%d",&R,&C)!=EOF)

    {

        for(int i=0;i<R;i++)

          scanf("%s",a);

    int ans=-1;

    for(char stc='a';stc<='c';stc++)

    {

        memset(high,0,sizeof(high));

        for(int i=0;i<R;i++)

        {

            for(int j=0;j<C;j++)

            {

                if(isH(stc,a[j]))

                    high[j]++;

                else

                    high[j]=0;

            }

            for(int j=0;j<C;j++)

            {

                int cnt=1;

                if(high[j]==0) continue;

                for(int k=j-1;k>=0&&high[k]>=high[j];k--)

                    cnt++;

                for(int k=j+1;k<C&&high[k]>=high[j];k++)

                    cnt++;

                ans=max(ans,high[j]*cnt);

            }

        }

    }

    printf("%d\n",ans);

    }

    return 0;

}