Day3-T3

原题目

　　Describe：又是这种最大子矩阵捆绑一堆条件的题

　　code：

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define jxcjulao 1111
using namespace std;
long long lft[jxcjulao][jxcjulao],up[jxcjulao][jxcjulao],a[jxcjulao][jxcjulao];
long long n,m,minn,ans,f,l;
inline long long read(){
    long long ret=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9') {if (ch=='-') f=-f;ch=getchar();}
    while (ch>='0'&&ch<='9') ret=ret*10+ch-'0',ch=getchar();
    return ret*f;
}
inline double read2(){
    double X=0,Y=1.0;int w=0;char ch=0;
    while(!isdigit(ch)){w|=ch=='-';ch=getchar();}
    while(isdigit(ch))X=X*10+(ch^48),ch=getchar();
    ch=getchar();
    while(isdigit(ch)) X+=(Y/=10)*(ch^48),ch=getchar();
    return w?-X:X;
}
inline void write(int x){
    if(x<0){putchar('-');write(-x);return;}
    if(x/10) write(x/10);
    putchar(x%10+'0');
}
int main(){
//    freopen("matrix.in","r",stdin);
//    freopen("matrix.out","w",stdout);
    n=read(),m=read();
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    a[i][j]=read();
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    {
        up[i][j]=lft[i][j]=1;
        for(int k=i;k>1;k--)                         //向上至多延伸 
        if(a[k][j]>a[k-1][j])up[i][j]++;
        else break;
        for(int k=j;k>1;k--)                         //向左至多延伸
        if(a[i][k]>a[i][k-1])lft[i][j]++;
        else break;
    }
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    for(int k=up[i][j];k>=1;k--)
    {
        if(k*lft[i][j]<ans)break;                    //剪枝 
        minn=lft[i][j];                              //Lft至多到哪里 
        for(l=i;l>=i-k+1;l--)
        minn=min(minn,lft[l][j]);                    //(i,j)->(i+up[i][j],j)中最小的 lft[x][y] 
        if(k*minn<ans)continue;                      //剪枝 
        for(l=j;l>=j-minn+1;l--)                     //check
        if(up[i][l]<k)break;
        minn=min(minn,j-l);                          //取min 
        minn*=k;ans=max(ans,minn);
    }
    cout<<ans;
}
/*
5 5
3 1 2 5 7
1 2 3 4 6
2 4 5 6 9
1 5 7 9 7
3 6 8 10 1
*/