HDU OJ Largest Submatrix-优快云博客

本文介绍了一种解决矩阵中字母转换后形成相同字母最大子矩阵的问题，通过将特定字母转换为'a'、'b'、'c'，并分别求解关于每个字母的最大子矩阵，最终找出最大值作为结果。

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

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;
}