逐层的找最大面积。
Total Submission(s): 967 Accepted Submission(s): 467
Largest Submatrix
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 967 Accepted Submission(s): 467
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
abcw
wxyz
Sample Output
3
Source
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gaojie
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=1005;
char a[N][N];
int high[N];
int C,R;
bool isH(char stc,char c)
{
if(stc=='a')
return c=='a'||c=='w'||c=='y'||c=='z';
if(stc=='b')
return c=='b'||c=='w'||c=='x'||c=='z';
if(stc=='c')
return c=='c'||c=='x'||c=='y'||c=='z';
return false;
}
int main()
{
while(scanf("%d%d",&R,&C)!=EOF)
{
for(int i=0;i<R;i++)
scanf("%s",a);
int ans=-1;
for(char stc='a';stc<='c';stc++)
{
memset(high,0,sizeof(high));
for(int i=0;i<R;i++)
{
for(int j=0;j<C;j++)
{
if(isH(stc,a[j]))
high[j]++;
else
high[j]=0;
}
for(int j=0;j<C;j++)
{
int cnt=1;
if(high[j]==0) continue;
for(int k=j-1;k>=0&&high[k]>=high[j];k--)
cnt++;
for(int k=j+1;k<C&&high[k]>=high[j];k++)
cnt++;
ans=max(ans,high[j]*cnt);
}
}
}
printf("%d\n",ans);
}
return 0;
}