Largest Submatrix

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/38481357

本文介绍了一种解决最大相同字母子矩阵问题的算法，通过三次计算最大空白子矩阵来求解，利用动态规划思想更新左边界和右边界，实现高效求解。

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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1488 Accepted Submission(s): 716

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

思路：其实就是算3次最大空白子矩阵问题，用n方算法，l[i][j]和r[i][j]表示在i行j点上方连续的空白列可以向左右扩展的最远距离。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[1010][1010];
int v[1010][1010],ans,n,m,l[1010][1010],r[1010][1010];
void solve()
{ int i,j,k;
  for(i=1;i<=n;i++)
  { v[i][0]=-1;
    v[i][m+1]=-1;
    for(j=1;j<=m;j++)
     while(v[i][j]<=v[i][l[i][j]-1])
      l[i][j]=l[i][l[i][j]-1];
    for(j=m;j>=1;j--)
     while(v[i][j]<=v[i][r[i][j]+1])
      r[i][j]=r[i][r[i][j]+1];
    for(j=1;j<=m;j++)
     ans=max(ans,v[i][j]*(r[i][j]-l[i][j]+1));
  }
}
int main()
{ int i,j,k;
  while(~scanf("%d%d",&n,&m))
  { ans=0;
    for(i=1;i<=n;i++)
     scanf("%s",s[i]+1);
    for(i=1;i<=n;i++)
     for(j=1;j<=m;j++)
     { l[i][j]=r[i][j]=j;
       if(s[i][j]=='a' || s[i][j]=='w' || s[i][j]=='y' || s[i][j]=='z')
        v[i][j]=v[i-1][j]+1;
       else
        v[i][j]=0;
     }
    solve();
    for(i=1;i<=n;i++)
     for(j=1;j<=m;j++)
     { l[i][j]=r[i][j]=j;
       if(s[i][j]=='b' || s[i][j]=='w' || s[i][j]=='x' || s[i][j]=='z')
        v[i][j]=v[i-1][j]+1;
       else
        v[i][j]=0;
     }
    solve();
    for(i=1;i<=n;i++)
     for(j=1;j<=m;j++)
     { l[i][j]=r[i][j]=j;
       if(s[i][j]=='c' || s[i][j]=='x' || s[i][j]=='y' || s[i][j]=='z')
        v[i][j]=v[i-1][j]+1;
       else
        v[i][j]=0;
     }
    solve();
    printf("%d\n",ans);
  }
}