P1169 [ZJOI2007] Chessboard making (suspended line method dp)

https://www.luogu.com.cn/problem/P1169

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Purpose:
 Solve the largest sub-matrix that meets the conditions in a given matrix.
Practice:
Use a line (both horizontal and vertical) to move left and right until the constraint conditions are not met or the boundary
is reached. Define a few things:
    left[i][j]: represents from ( i,j) the leftmost position that can be reached
    right[i][j]: represents the rightmost position that can be reached from (i,j)
    up[i][j]: represents the longest extension from (i,j) Length.
Here is the recurrence formula:
   left[i][j]=max(left[i][j],left[i-1][j])
   right[i][j]=min(right[i ][j],right[i-1][j])

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=2010;
typedef int LL;
LL ma[maxn][maxn],qleft[maxn][maxn],qright[maxn][maxn],up[maxn][maxn];
LL n,m;
int main(void)
{
  cin.tie(0);std::ios::sync_with_stdio(false);
  cin>>n>>m;
  for(LL i=1;i<=n;i++){
    for(LL j=1;j<=m;j++){
        cin>>ma[i][j];
        qleft[i][j]=qright[i][j]=j;
        up[i][j]=1;
    }
  }
  for(LL i=1;i<=n;i++){
    for(LL j=2;j<=m;j++){
        if(ma[i][j]!=ma[i][j-1]){
            qleft[i][j]=qleft[i][j-1];
        }
    }
  }
  for(LL i=1;i<=n;i++){
    for(LL j=m-1;j>=1;j--){
        if(ma[i][j]!=ma[i][j+1]){
            qright[i][j]=qright[i][j+1];
        }
    }
  }
  LL area1=0;LL area2=0;
  for(LL i=1;i<=n;i++){
    for(LL j=1;j<=m;j++){
        if(i>1&&ma[i][j]!=ma[i-1][j]){
            qleft[i][j]=max(qleft[i][j],qleft[i-1][j]);
            qright[i][j]=min(qright[i][j],qright[i-1][j]);
            up[i][j]=up[i-1][j]+1;
        }
        LL a=qright[i][j]-qleft[i][j]+1;
        LL b=min(a,up[i][j]);
        area1=max(area1,b*b);
        area2=max(area2,a*up[i][j]);
    }
  }
  cout<<area1<<endl<<area2<<endl;
  return 0;
}