正题

题目链接:https://www.luogu.org/problemnew/show/P1169

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题目大意

一个矩阵中求一个最大的子矩阵和子正方形使得它们其中都是01交错。

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解题思路

$lef{t}_{i,j}$表示$(i,j)$往左扩展多远，$righ{t}_{i,j}$表示$(i,j)$往右扩展多远，$u{p}_{i,j}$表示$(i,j)$向上扩展多少个。

然后对于每个点作为最下边的点时答案为
leni,j=upi,j+max{leftk,j,rightk,j}(j−k&lt;upi,j)len_{i,j}=up_{i,j}+max\{left_{k,j},right_{k,j}\}(j-k&lt;up_{i,j})leni,j​=upi,j​+max{leftk,j​,rightk,j​}(j−k<upi,j​)
$$ans1=u{p}_{i,j}\ast le{n}_{i,j}$$
ans2=min{upi,j,leni,j}2ans2=min\{up_{i,j},len_{i,j}\}^2ans2=min{upi,j​,leni,j​}2

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$code$

// luogu-judger-enable-o2
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=2100;
int n,m,a[N][N],right[N][N],left[N][N],up[N][N],ans1,ans2;
int main()
{scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)scanf("%d",&a[i][j]);for(int i=1;i<=n;i++)for(int j=m;j>=1;j--){right[i][j]=(a[i][j]!=a[i][j+1]?right[i][j+1]:j);if(j==m) right[i][j]=m;}for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){left[i][j]=(a[i][j]!=a[i][j-1]?left[i][j-1]:j);if(j==1) left[i][j]=1;}for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){up[i][j]=1;if(i>1&&a[i][j]!=a[i-1][j]){left[i][j]=max(left[i][j],left[i-1][j]);right[i][j]=min(right[i][j],right[i-1][j]);up[i][j]=up[i-1][j]+1;}int l=right[i][j]-left[i][j]+1;int w=min(up[i][j],l); ans1=max(l*up[i][j],ans1);ans2=max(w*w,ans2);}printf("%d\n%d",ans2,ans1);
}