LA3029

题解:

一个类似尺取法的算法

代码:

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=1005;
int T,n,m,up[N][N],left[N][N],right[N][N];
char s[N][N];
int main()
{
    scanf("%d",&T);
    while (T--)
     {
         scanf("%d%d",&m,&n);
         for (int i=0;i<m;i++)
          for (int j=0;j<n;j++)
           for (s[i][j]=getchar();s[i][j]!='F'&&s[i][j]!='R';s[i][j]=getchar());
         int ans=0;
         for (int i=0;i<m;i++)
          { 
             int l=-1,r=n;
              for (int j=0;j<n;j++)
               {
                   if (s[i][j]=='R')
                    {
                        up[i][j]=left[i][j]=0;
                        l=j;
                    }
                   else
                 {
                     up[i][j]=i==0?1:up[i-1][j]+1;
                     left[i][j]=i==0?l+1:max(left[i-1][j],l+1);
                 } 
               }
            for (int j=n-1;j>=0;j--)
             if (s[i][j]=='R')
              {
                right[i][j]=n;
                r=j;
              } 
             else
              {
                right[i][j]=i==0?r-1:min(right[i-1][j],r-1);
                ans=max(ans,up[i][j]*(right[i][j]-left[i][j]+1));
              }                     
          }
      printf("%d
",ans*3);     
     }
}
原文地址:https://www.cnblogs.com/xuanyiming/p/8207128.html