hdu 2870 Largest Submatrix （DP)

Largest Submatrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1534    Accepted Submission(s): 735

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
   
   
   
   

 

又是一个悲伤的故事，做了一上午，其实就是和1505和1506一样，这题枚举一下a，b，c，就行了，不知为什么酒一直wa，看别人的题解，没有用我这种方法的，都是记一下最左边大于等于它的位置，和右边大于等于它的位置，和我的方法在原理上是一样的，我是记左边大于等于它的位置，右边大于等于它的位置，debug了半天才发现把m打成了n。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=1000+100;
char s[maxn][maxn];
int a[maxn][maxn];
int dpl[maxn],dpr[maxn];
int n,m;
int ans;
void solve(char c,char w,char x,char y)
{
    for(int i=1;i<=n;i++)
    {
        a[i][0]=a[i][m+1]=-1;
        for(int j=1;j<=m;j++)
        {
            if(s[i][j]==c||s[i][j]==w||s[i][j]==x||s[i][j]==y)
            {
                if(i>1)
                a[i][j]=a[i-1][j]+1;
                else
                {
                    a[i][j]=1;
                }
            }
            else
            {
                a[i][j]=0;
            }
        }
       int temp;
       memset(dpl,0,sizeof(dpl));
       memset(dpr,0,sizeof(dpr));
       dpl[1]=1;
       dpr[m]=1;
       for(int j=2;j<=m;j++)
       {
           int k;
           dpl[j]=1;
           if(a[i][j])
           {
               k=j;
               while(a[i][j]<=a[i][k-1])
               {
                   dpl[j]+=dpl[k-1];
                   k-=dpl[k-1];
               }
               // while(a[i][dpl[j]-1]>=a[i][j])
               //  dpl[j]=dpl[dpl[j]-1];
           }
       }
       for(int j=m-1;j>=1;j--)
       {
           int k;
           dpr[j]=1;
           if(a[i][j])
           {
               k=j;
               while(a[i][j]<=a[i][k+1])
               {
                   dpr[j]+=dpr[k+1];
                   k=k+dpr[k+1];
               }
               // while(a[i][dpr[j]+1]>=a[i][j])
               //   dpr[j]=dpr[dpr[j]+1];
           }
       }
       for(int j=1;j<=m;j++)
       {
           if(a[i][j])
           {
           temp=a[i][j]*(dpr[j]+dpl[j]-1);
          // temp=a[i][j]*(dpr[j]-dpl[j]+1);
           if(ans<temp)
           ans=temp;
           }
       }
    }
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%s",s[i]+1);
        }
        ans=0;
        solve('a','w','y','z');
        solve('b','w','x','z');
        solve('c','x','y','z');
        printf("%d\n",ans);
    }
    return 0;
}