Luo Gu P1169 [ZJOI2007] board production line suspension method

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First look at the hanging line method: "On the use of great ideas to solve the biggest problem of sub-matrices"

For this question, barrier point is the current color to the same point, the other did as a template to do just fine.

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;
int n,m,bd[2010][2010],l[2010][2010],r[2010][2010],ht[2010][2010];

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        for( Int J = . 1 ; J <= m; J ++ ) 
        { 
            Scanf ( " % D " , & BD [I] [J]); 
            L [I] [J] = J; R & lt [I] [J] = J ; HT [I] [J] = . 1 ; // to suspension wires ij initial value 
        }
     for ( int I = . 1 ; I <= n-; I ++ )
         for ( int J = 2 ; J <= m; J ++ )
             IF (! BD [I] [J] BD = [I] [J- . 1 ]) 
                L [I] [J] = L [I] [J- . 1 ];   // suspension wires may extend leftward ij 
    for(int i=1;i<=n;i++)
        for(int j=m-1;j>=1;j--)
            if(bd[i][j]!=bd[i][j+1])
                r[i][j]=r[i][j+1];  //悬线ij向右延伸 
    for(int i=2;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(bd[i][j]!=bd[i-1][j])
            {
                ht[i][j]HT = [I- . 1 ] [J] + . 1 ;   // suspension wires extending upwardly ij 
                L [I] [J] = max (L [I] [J], L [I- . 1 ] [J]);   / / suspended line interval ij shrinkage left 
                R & lt [I] [J] = min (R & lt [I] [J], R & lt [I- . 1 ] [J]);   // shrink the right section 
            } 
            
    int ANS1 = 0 , ANS2 = 0 ;
     for ( int I = . 1 ; I <= n-; I ++ )
         for ( int J = . 1 ; J <= m; J ++ ) 
        { 
            int A = R & lt [I] [J] -l [I] [J] + . 1 , B = HT [I] [J];   //Ij corresponding to the width of the rectangle of the suspension wires extend around the suspension wires ij distance, high-ij extends upwardly a distance of 
            ANS1 = max (ANS1, min (A, B) * min (A, B));   // the corresponding area of the square 
            ans2 = max (ANS2, A * B);   // rectangular area 
        } 
    the printf ( " % D \ D n-% \ n- " , ANS1, ANS2);
     return  0 ; 
}