hdu 2870 Largest Submatrix

Largest Submatrix

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

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

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

 

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指定字母可以全部变为某指定字母

问相同字母组成的最大子矩阵

hdu 1505 加强版

#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 1001
using namespace std;
int n,m,k,minn,ans;
char s[N+2][N+2],t[N+2][N+2];
int l[N],r[N];
int a[N],b[N];
int q[N],tmp[N],head,tail;
void change(char w,char x,char y,char z)
{
    for(int i=1;i<=n;i++)
     for(int j=1;j<=m;j++)
      if(s[i][j]==x || s[i][j]==y || s[i][j]==z) t[i][j]=w;
      else t[i][j]=s[i][j]; 
}
void monotonous(int *c,int *d)
{
    int h=1;
    while(h<=m && !c[h]) h++;
    if(h>m) return;
    q[0]=c[h]; tmp[0]=h;
    head=0; tail=1;
    for(int i=h+1;i<=m;i++)
    {
        if(!c[i]) while(head<tail) d[tmp[head++]]=i-1;
        else if(head==tail) q[tail]=c[i],tmp[tail++]=i;
        else
        {
            while(head<tail && c[i]<q[tail-1]) d[tmp[--tail]]=i-1;
            q[tail]=c[i];
            tmp[tail++]=i;
        }
    }
    while(head<tail) d[tmp[head++]]=tmp[tail-1];
}
void solve(char x)
{
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(t[i][j]!=x) a[j]=b[m-j+1]=0;
            else
            {
                if(t[i][j]==t[i-1][j]) a[j]++,b[m-j+1]++;
                 else a[j]=b[m-j+1]=1;
            }
        }
        monotonous(a,r);
        monotonous(b,l);
        for(int j=1;j<=m;j++) tmp[j]=l[j];
        for(int j=1;j<=m;j++) l[m-j+1]=m-tmp[j]+1;
        for(int j=1;j<=m;j++) ans=max(ans,(r[j]-l[j]+1)*a[j]);
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        ans=0;
        for(int i=1;i<=n;i++) scanf("%s",s[i]+1);
        change('a','w','y','z'); solve('a');
        change('b','w','x','z'); solve('b');
        change('c','x','y','z'); solve('c');
        printf("%d
",ans);
    }
}