TLS 9.2C

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这个题目我觉得我做的\(50\)分做法比\(100\)分的\(SBDP\)更具有价值.

因为这个\(DP\)真的很简单.

令\(f_{i,j}\)表示以\((i,j)\)为右下角的最大正方形的边长.则有转移方程:

\[f_{i,j} = min ( f_{i-1,j-1} , f_{i-1,j} , f_{i,j-1} ) + 1\]

最后的答案就在所有的\(f_{i,j}\)中取一个\(max\)就行了.

\(Code:\)

#include 
    
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         #include 
         
          #include 
          
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                #define MEM(x,y) memset ( x , y , sizeof ( x ) )#define rep(i,a,b) for (int i = a ; i <= b ; ++ i)#define per(i,a,b) for (int i = a ; i >= b ; -- i)#define pii pair < int , int >#define X first#define Y second#define rint read
                
                 #define int long long#define pb push_backusing std::set ;using std::pair ;using std::max ;using std::min ;using std::priority_queue ;using std::vector ;template < class T > inline T read () { T x = 0 , f = 1 ; char ch = getchar () ; while ( ch < '0' || ch > '9' ) { if ( ch == '-' ) f = - 1 ; ch = getchar () ; } while ( ch >= '0' && ch <= '9' ) { x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ; ch = getchar () ; } return f * x ;}const int N = 3e3 + 100 ;int n , m , f[N][N] , ans ;bool e[N][N] ;signed main (int argc , char * argv[] ) { freopen ("photo.in" , "r" , stdin) ; freopen ("photo.out" , "w" , stdout) ; n = rint () ; m = rint () ; rep ( i , 1 , m ) e[rint()][rint()] = true ; rep ( i , 1 , n ) rep ( j , 1 , n ) if ( ! e[i][j] ) f[i][j] = min ( f[i-1][j-1] , min ( f[i-1][j] , f[i][j-1] ) ) + 1 ; else f[i][j] = 0 ; rep ( i , 1 , n ) rep ( j , 1 , n ) ans = max ( ans , f[i][j] ) ; printf ("%lld\n" , ans ) ; return 0 ;}

这是\(O(n^2)\)的\(DP\)做法.

我接下里要说的这个\(O(n^3)\)的暴力做法其实是非常优美的.

它其实是悬线法求极大正方形的基础,这里我只维护了一个向右的最远扩展距离.

这个数组可以\(O(n^3)\)预处理,但其实可以做到\(O(n^2)\),从最右边向左递推就可以做到\(O(n^2)\).

这样借助这个数组,我们就可以枚举这个极大正方形的左右边界,然后从上向下去枚举,保证必须要连续一段都大于等于这个枚举出来的边长(枚举左右边界就知道了边长),每次遇到合法的就取\(max\)就可以了.

\(Code:\)

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              #include 
               
                #define MEM(x,y) memset ( x , y , sizeof ( x ) )#define rep(i,a,b) for (int i = a ; i <= b ; ++ i)#define per(i,a,b) for (int i = a ; i >= b ; -- i)#define pii pair < int , int >#define X first#define Y second#define rint read
                
                 #define int long long#define pb push_backusing std::set ;using std::pair ;using std::max ;using std::min ;using std::priority_queue ;using std::vector ;template < class T > inline T read () { T x = 0 , f = 1 ; char ch = getchar () ; while ( ch < '0' || ch > '9' ) { if ( ch == '-' ) f = - 1 ; ch = getchar () ; } while ( ch >= '0' && ch <= '9' ) { x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ; ch = getchar () ; } return f * x ;}const int N = 3e3 + 100 ;int n , m , r[N][N] , ans ;bool e[N][N] ;signed main (int argc , char * argv[] ) { freopen ("photo.in" , "r" , stdin) ; freopen ("photo.out" , "w" , stdout) ; n = rint () ; m = rint () ; rep ( i , 1 , m ) e[rint()][rint()] = true ; rep ( i , 1 , n ) rep ( j , 1 , n ) rep ( k , j , n ) if ( ! e[i][k] ) ++ r[i][j] ; else break ; rep ( i , 1 , n ) rep ( j , i , n ) { int last = 0 , len = j - i + 1 ; rep ( k , 1 , n ) { if ( ! last ) { if ( r[k][i] >= len ) last = k ; else last = 0 ; } else { if ( r[k][i] >= len ) { if ( k - last + 1 == len ) ans = max ( ans , len ) ; } else last = 0 ; } } } printf ("%lld\n" , ans ) ; system ("pause") ; return 0 ;}

转载于:https://www.cnblogs.com/Equinox-Flower/p/11448050.html

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