Largest Submatrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 694    Accepted Submission(s): 331


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
 
代码：

#include<stdio.h>
#include<string.h>
#define N 1005
#define MX(a,b) (a>b?a:b)
int num[N];
char s[N][N];
int n,m;

bool check(char c,char ss)
{
   if(c=='a')
   return ss=='a'||ss=='w'||ss=='y'||ss=='z';
   if(c=='b')
   return ss=='b'||ss=='w'||ss=='x'||ss=='z';
   if(c=='c')
   return ss=='c'||ss=='x'||ss=='y'||ss=='z';  
}

int main()
{
   while(scanf("%d%d",&n,&m)!=-1)
   {
      for(int i=0;i<n;i++)
   scanf("%s",s[i]);
   int max=0;
   for(char c='a';c<='c';c++)
   {
      memset(num,0,sizeof(num));
   for(int i=0;i<n;i++)
   {
      for(int j=0;j<m;j++)
   {
      if(check(c,s[i][j]))
      num[j]++;
      else
      num[j]=0;   
      }   
      for(int j=0;j<m;j++)
      {
        int cnt=1;
        if(!num[j])
        continue;

      //枚举每个高度，然后以它做基准，向左向右扩张，扩张条件是h[i-k]>=h[k]、h[i+k]>=h[k]。记录扩张的个数即可。
        for(int k=1;j-k>=0&&num[j-k]>=num[j];k++)
        cnt++;
        for(int k=1;j+k<m&&num[j+k]>=num[j];k++ )
        cnt++;
        max=MX(max,cnt*num[j]);
       }
         }     
   }       
   printf("%d\n",max);
   }  
   return 0;
}

链接：http://acm.hdu.edu.cn/showproblem.php?pid=2870