hdu1506 的二维情况,对于每一个点,已他所在行为矩形的最上边。预处理该点下面有多少个连续的一就是h[i] [j],然后求出l[i][j]和r[i][j].
#include<bits/stdc++.h>
using namespace std;
const int N=1024;
int g[N][N],r[N][N],h[N][N],l[N][N];
int main()
{
int _,n,m,i,j,k,ans;
char s[4000];
scanf("%d",&_);
while(_--)
{
scanf("%d%d",&n,&m);
//getchar();
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
scanf("%s",s);
if(s[0]=='R') g[i][j]=0;
else g[i][j]=1;
r[i][j]=l[i][j]=j;
}
}
memset(h,0,sizeof(h));
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
if(g[i][j]==0) h[i][j]=0;
else h[i][j]=h[i-1][j]+1;
}
}
ans=0;
//printf("11111\n");
for(i=1;i<=n;i++)
{
//printf("%d",i);
h[i][0]=h[i][m+1]=-1;
for(j=m;j>=1;j--)
{
while(h[i][r[i][j]+1]>=h[i][r[i][j]])
r[i][j]=r[i][r[i][j]+1];
}
for(j=1;j<=m;j++)
{
while(h[i][l[i][j]-1]>=h[i][l[i][j]])
l[i][j]=l[i][l[i][j]-1];
}
for(j=1;j<=m;j++)
{
//printf("%d %d %d %d\n",(l[i][j]-r[i][j]+1)*h[i][j],r[i][j],l[i][j],h[i][j]);
ans=max(ans,(r[i][j]-l[i][j]+1)*h[i][j]);
}
}
printf("%d\n",3*ans);
}
return 0;
}