HDU 2870 Largest Submatrix（最大子矩阵面积）

Largest Submatrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2570    Accepted Submission(s): 1262


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

 

Source

2009 Multi-University Training Contest 7 - Host
by FZU 

 

POINT：
用的单调栈，还有一种方法是求上界和下界。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <stack>
using namespace std;
#define  LL long long
const int N = 1000;
char mp[N+3][N+3];
int a[N+3][N+3];
int n,m;
int ans;
void doit()
{
for(int j=1;j<=m;j++)
{
stack<int>q;
int pos[N+3];
for(int i=1;i<=n;i++)
{
pos[i]=i;
while(!q.empty()&&a[i][j]<=a[q.top()][j])
{
ans=max(ans,a[q.top()][j]*(i-pos[q.top()]));
pos[i]=pos[q.top()];
q.pop();
}
q.push(i);
}
while(!q.empty())
{
ans=max(ans,a[q.top()][j]*(n-pos[q.top()]+1));
q.pop();
}
}

}
int main()
{
while(~scanf("%d %d",&n,&m))
{
ans=0;
for(int i=1;i<=n;i++)
{
scanf("%s",mp[i]+1);
}
for(int i=1;i<=n;i++)
{
int now=0;
for(int j=1;j<=m;j++)
{
if(mp[i][j]=='a'||mp[i][j]=='w'||mp[i][j]=='y'||mp[i][j]=='z')
now++;
else now=0;
a[i][j]=now;
}
}
//        for(int i=1;i<=n;i++)
//        {
//            for(int j=1;j<=m;j++)
//            {
//                printf("%d ",a[i][j]);
//            }
//            printf("\n");
//        }
doit();
for(int i=1;i<=n;i++)
{
int now=0;
for(int j=1;j<=m;j++)
{
if(mp[i][j]=='b'||mp[i][j]=='w'||mp[i][j]=='x'||mp[i][j]=='z')
now++;
else now=0;
a[i][j]=now;
}
}
doit();
for(int i=1;i<=n;i++)
{
int now=0;
for(int j=1;j<=m;j++)
{
if(mp[i][j]=='c'||mp[i][j]=='x'||mp[i][j]=='y'||mp[i][j]=='z')
now++;
else now=0;
a[i][j]=now;
}
}
doit();
printf("%d\n",ans);
}

}