Maksimiraj vsoto N X N zgornje leve podmatrike iz dane matrike 2N X 2N
Glede na a 2N x 2N matrika celih števil. Katero koli vrstico ali stolpec lahko obračate poljubno število krat in v poljubnem vrstnem redu. Naloga je izračunati največjo vsoto zgornjega levega N X N podmatriko, tj. vsoto elementov podmatrike od (0 0) do (N - 1 N - 1).
Primeri:
Vnos: z [][] = {
112 42 83 119
56 125 56 49
15 78 101 43
62 98 114 108
}
Izhod: 414
Dana matrika je velikosti 4 X 4, ki jo moramo maksimirati
vsota zgornje leve matrike 2 X 2, tj
vsota mat[0][0] + mat[0][1] + mat[1][0] + mat[1][1].
Naslednje operacije povečajo vsoto:
1. Obrnite stolpec 2
112 42 114 119
56 125 101 49
15 78 56 43
62 98 83 108
2. Obrnite vrstico 0
119 114 42 112
56 125 101 49
15 78 56 43
62 98 83 108
Vsota zgornje leve matrike = 119 + 114 + 56 + 125 = 414.
Če želite povečati vsoto zgornje leve podmatrike, opazujte, da za vsako celico zgornje leve podmatrike obstajajo štirje kandidati, ki pomenijo ustrezne celice v zgornji levi, zgornji desni, spodnji levi in spodnji desni podmatriki, s katerimi je mogoče zamenjati.
Zdaj opazujte vsako celico, kjer koli je, jo lahko zamenjamo z ustrezno vrednostjo kandidata v zgornji levi podmatriki, ne da bi spremenili vrstni red drugih celic v zgornji levi podmatriki. Diagram prikazuje primer, ko je največja vrednost 4 kandidatov v zgornji desni podmatriki. Če je v spodnji levi ali spodnji desni podmatriki, lahko najprej obrnemo vrstico ali stolpec, da ga postavimo v zgornjo desno podmatriko, nato pa sledimo istemu zaporedju operacij, kot je prikazano v diagramu.
V tej matriki recimo a 26 je največ 4 kandidati in a 23 je treba zamenjati z a 26 brez spreminjanja vrstnega reda celic v zgornji levi podmatriki.
Obrnite vrstico 2
Obratni stolpec 2
Obrnite vrstico 7
Obratni stolpec 6
Obrnite vrstico 2
Spodaj je izvedba tega pristopa:
C++ // C++ program to find maximum value of top N/2 x N/2 // matrix using row and column reverse operations #include #define R 4 #define C 4 using namespace std ; int maxSum ( int mat [ R ][ C ]) { int sum = 0 ; for ( int i = 0 ; i < R / 2 ; i ++ ) for ( int j = 0 ; j < C / 2 ; j ++ ) { int r1 = i ; int r2 = R - i - 1 ; int c1 = j ; int c2 = C - j - 1 ; // We can replace current cell [i j] // with 4 cells without changing affecting // other elements. sum += max ( max ( mat [ r1 ][ c1 ] mat [ r1 ][ c2 ]) max ( mat [ r2 ][ c1 ] mat [ r2 ][ c2 ])); } return sum ; } // Driven Program int main () { int mat [ R ][ C ] = { 112 42 83 119 56 125 56 49 15 78 101 43 62 98 114 108 }; cout < < maxSum ( mat ) < < endl ; return 0 ; }
Java // Java program to find maximum value of top N/2 x N/2 // matrix using row and column reverse operations class GFG { static int maxSum ( int mat [][] ) { int sum = 0 ; int maxI = mat . length ; int maxIPossible = maxI - 1 ; int maxJ = mat [ 0 ] . length ; int maxJPossible = maxJ - 1 ; for ( int i = 0 ; i < maxI / 2 ; i ++ ) { for ( int j = 0 ; j < maxJ / 2 ; j ++ ) { // We can replace current cell [i j] // with 4 cells without changing affecting // other elements. sum += Math . max ( Math . max ( mat [ i ][ j ] mat [ maxIPossible - i ][ j ] ) Math . max ( mat [ maxIPossible - i ] [ maxJPossible - j ] mat [ i ][ maxJPossible - j ] )); } } return sum ; } // Driven Program public static void main ( String [] args ) { int mat [][] = { { 112 42 83 119 } { 56 125 56 49 } { 15 78 101 43 } { 62 98 114 108 } }; System . out . println ( maxSum ( mat )); } } /* This Java code is contributed by Rajput-Ji*/
Python3 # Python3 program to find the maximum value # of top N/2 x N/2 matrix using row and # column reverse operations def maxSum ( mat ): Sum = 0 for i in range ( 0 R // 2 ): for j in range ( 0 C // 2 ): r1 r2 = i R - i - 1 c1 c2 = j C - j - 1 # We can replace current cell [i j] # with 4 cells without changing/affecting # other elements. Sum += max ( max ( mat [ r1 ][ c1 ] mat [ r1 ][ c2 ]) max ( mat [ r2 ][ c1 ] mat [ r2 ][ c2 ])) return Sum # Driver Code if __name__ == '__main__' : R = C = 4 mat = [[ 112 42 83 119 ] [ 56 125 56 49 ] [ 15 78 101 43 ] [ 62 98 114 108 ]] print ( maxSum ( mat )) # This code is contributed # by Rituraj Jain
C# // C# program to find maximum value // of top N/2 x N/2 matrix using row // and column reverse operations using System ; class GFG { static int R = 4 ; static int C = 4 ; static int maxSum ( int [ ] mat ) { int sum = 0 ; for ( int i = 0 ; i < R / 2 ; i ++ ) { for ( int j = 0 ; j < C / 2 ; j ++ ) { int r1 = i ; int r2 = R - i - 1 ; int c1 = j ; int c2 = C - j - 1 ; // We can replace current cell [i j] // with 4 cells without changing affecting // other elements. sum += Math . Max ( Math . Max ( mat [ r1 c1 ] mat [ r1 c2 ]) Math . Max ( mat [ r2 c1 ] mat [ r2 c2 ])); } } return sum ; } // Driven Code public static void Main () { int [ ] mat = { { 112 42 83 119 } { 56 125 56 49 } { 15 78 101 43 } { 62 98 114 108 } }; Console . Write ( maxSum ( mat )); } } // This code is contributed // by ChitraNayal
PHP // PHP program to find maximum value // of top N/2 x N/2 matrix using row // and column reverse operations function maxSum ( $mat ) { $R = 4 ; $C = 4 ; $sum = 0 ; for ( $i = 0 ; $i < $R / 2 ; $i ++ ) for ( $j = 0 ; $j < $C / 2 ; $j ++ ) { $r1 = $i ; $r2 = $R - $i - 1 ; $c1 = $j ; $c2 = $C - $j - 1 ; // We can replace current cell [i j] // with 4 cells without changing // affecting other elements. $sum += max ( max ( $mat [ $r1 ][ $c1 ] $mat [ $r1 ][ $c2 ]) max ( $mat [ $r2 ][ $c1 ] $mat [ $r2 ][ $c2 ])); } return $sum ; } // Driver Code $mat = array ( array ( 112 42 83 119 ) array ( 56 125 56 49 ) array ( 15 78 101 43 ) array ( 62 98 114 108 )); echo maxSum ( $mat ) . ' n ' ; // This code is contributed // by Mukul Singh ?>
JavaScript < script > // Javascript program to find maximum value of top N/2 x N/2 // matrix using row and column reverse operations let R = 4 ; let C = 4 ; function maxSum ( mat ) { let sum = 0 ; for ( let i = 0 ; i < R / 2 ; i ++ ) { for ( let j = 0 ; j < C / 2 ; j ++ ) { let r1 = i ; let r2 = R - i - 1 ; let c1 = j ; let c2 = C - j - 1 ; // We can replace current cell [i j] // with 4 cells without changing affecting // other elements. sum += Math . max ( Math . max ( mat [ r1 ][ c1 ] mat [ r1 ][ c2 ]) Math . max ( mat [ r2 ][ c1 ] mat [ r2 ][ c2 ])); } } return sum ; } // Driven Program let mat = [[ 112 42 83 119 ] [ 56 125 56 49 ] [ 15 78 101 43 ] [ 62 98 114 108 ]]; document . write ( maxSum ( mat )); // This code is contributed by avanitrachhadiya2155 < /script>
Izhod
414
Časovna zahtevnost: O(N 2 ).
Pomožni prostor: O(1) saj uporablja konstanten prostor za spremenljivke
Ustvari kviz
