HDU OJ Largest Submatrix

Largest Submatrix

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 11   Accepted Submission(s) : 6

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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

分析：将w,x,y,z全部转化为a,b,c,然后分别求关于a,b,c,的最大子矩阵，tempa,tempb,tempc,三者的最大值即为结果。

代码如下：

#include <iostream>
#include <cstdio>
using namespace std;

int numa[1005],numb[1005],numc[1005];
int Maxa,Maxb,Maxc;
int Mina,Minb,Minc;
int result;
char ch;
int m,n;
int bit[30];
int counta,countb,countc,tempa,tempb,tempc;

int main()
{
    int i,j,k;
    bit[0]=4;
    bit[1]=2;
    bit[2]=1;
    bit[22]=6;
    bit[23]=3;
    bit[24]=5;
    bit[25]=7;
    while(cin>>m>>n)
    {
        getchar();
        memset(numa,0,sizeof(numa));
        memset(numb,0,sizeof(numb));
        memset(numc,0,sizeof(numc));
        tempa=tempb=tempc=0;
        for(i=0;i<m;i++)
        {
            Maxa=Maxb=Maxc=-1;
            Mina=Minb=Minc=0x7fffffff;
            for(j=0;j<n;j++)
            {
                ch=getchar();
                if((bit[ch-'a']&4)!=0)
                {
                    numa[j]++;
                }
                else
                {
                    numa[j]=0;
                }
                Maxa=max(numa[j],Maxa);
                Mina=min(Mina,numa[j]);
                if((bit[ch-'a']&2)!=0)
                {
                    numb[j]++;
                }
                else
                {
                    numb[j]=0;
                }
                Maxb=max(numb[j],Maxb);
                Minb=min(Minb,numb[j]);
                if((bit[ch-'a']&1)!=0)
                {
                    numc[j]++;
                }
                else
                {
                    numc[j]=0;
                }
                Maxc=max(numc[j],Maxc);
                Minc=min(Minc,numc[j]);
            }
            getchar();
            for(j=Mina;j<=Maxa;j++)
            {
                counta=0;
                for(k=0;k<n;k++)
                {
                    if(j<=numa[k])
                    {
                        counta+=j;
                        tempa=max(counta,tempa);
                    }
                    else
                    {
                        counta=0;
                    }
                }
            }
            for(j=Minb;j<=Maxb;j++)
            {
                countb=0;
                for(k=0;k<n;k++)
                {
                    if(j<=numb[k])
                    {
                        countb+=j;
                        tempb=max(countb,tempb);
                    }
                    else
                    {
                        countb=0;
                    }
                }
            }
            for(j=Minc;j<=Maxc;j++)
            {
                countc=0;
                for(k=0;k<n;k++)
                {
                    if(j<=numc[k])
                    {
                        countc+=j;
                        tempc=max(countc,tempc);
                    }
                    else
                    {
                        countc=0;
                    }
                }
            }
        }
        result=max(tempa,max(tempb,tempc));
        cout<<result<<endl;
    }
    return 0;
}