HDU2870 Largest Submatrix

Largest Submatrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3105    Accepted Submission(s): 1476

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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Statistic |  Submit |  Discuss | Note

有个字母矩阵，包含字母"a、b、c、w、x、y、z"，其中，w能变为"a、b"，x能变为"b、c"，y能变为"a、c"，z能变为"a、b、c"。问能构成的最大字母完全一样的子矩阵面积为多大？

分析

对三种字符分别考虑，每次尽量转换成一种字符。

问题就变成了最大01子矩阵。上单调栈即可。

时间复杂度\(O(mn)\)

代码

#include<iostream>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') w=-1;ch=getchar();}
    while(isdigit(ch)) data=data*10+ch-'0',ch=getchar();
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;
using namespace std;

co int N=1002;
int m,n,a[N][N],b[N][N],c[N][N],s[N],w[N];
char buf[N];
void Largest_Submatrix(){
    for(int i=1;i<=m;++i){
        scanf("%s",buf+1);
        for(int j=1;j<=n;++j){
            if(buf[j]=='a'||buf[j]=='w'||buf[j]=='y'||buf[j]=='z') a[i][j]=a[i-1][j]+1;
            else a[i][j]=0;
            if(buf[j]=='b'||buf[j]=='w'||buf[j]=='x'||buf[j]=='z') b[i][j]=b[i-1][j]+1;
            else b[i][j]=0;
            if(buf[j]=='c'||buf[j]=='x'||buf[j]=='y'||buf[j]=='z') c[i][j]=c[i-1][j]+1;
            else c[i][j]=0;
        }
    }
    int ans=0;
    for(int i=1;i<=m;++i){
// max of a
        int p=0;
        for(int j=1;j<=n+1;++j){
            if(a[i][j]>s[p]) s[++p]=a[i][j],w[p]=1;
            else{
                int width=0;
                while(s[p]>a[i][j]){
                    width+=w[p];
                    ans=max(ans,width*s[p]);
                    --p;
                }
                s[++p]=a[i][j],w[p]=width+1;
            }
        }
// max of b
        p=0;
        for(int j=1;j<=n+1;++j){
            if(b[i][j]>s[p]) s[++p]=b[i][j],w[p]=1;
            else{
                int width=0;
                while(s[p]>b[i][j]){
                    width+=w[p];
                    ans=max(ans,width*s[p]);
                    --p;
                }
                s[++p]=b[i][j],w[p]=width+1;
            }
        }
// max of c
        p=0;
        for(int j=1;j<=n+1;++j){
            if(c[i][j]>s[p]) s[++p]=c[i][j],w[p]=1;
            else{
                int width=0;
                while(s[p]>c[i][j]){
                    width+=w[p];
                    ans=max(ans,width*s[p]);
                    --p;
                }
                s[++p]=c[i][j],w[p]=width+1;
            }
        }
    }
    printf("%d\n",ans);
}
int main(){
    while(~scanf("%d %d",&m,&n)) Largest_Submatrix();
    return 0;
}