Maksymalne lustra, które mogą przenosić światło z dołu do prawej
Dana jest macierz kwadratowa, w której każda komórka reprezentuje półfabrykat lub przeszkodę. Lustra możemy umieścić w pozycji pustej. Wszystkie zwierciadła będą ustawione pod kątem 45 stopni, co oznacza, że będą mogły przenosić światło z dołu do prawej, jeśli na ich drodze nie będzie żadnej przeszkody.
W tym pytaniu musimy policzyć, ile takich luster można umieścić w kwadratowej matrycy, która może przenosić światło od dołu do prawej.
Przykłady:
Output for above example is 2. In above diagram mirror at (3 1) and (5 5) are able to send light from bottom to right so total possible mirror count is 2.
Problem ten możemy rozwiązać sprawdzając położenie takich zwierciadeł w matrycy, że zwierciadło przekazujące światło z dołu do prawej nie będzie miało na swojej drodze żadnej przeszkody tj.
jeśli lustro znajduje się w indeksie (i j), to
nie będzie żadnych przeszkód w indeksie (k j) dla wszystkich k i < k <= N
nie będzie żadnej przeszkody w indeksie (i k) dla wszystkich k j < k <= N
Mając na uwadze powyższe dwa równania, możemy znaleźć najbardziej wysuniętą na prawo przeszkodę w każdym rzędzie w jednej iteracji danej macierzy i możemy znaleźć najniższą przeszkodę w każdej kolumnie w innej iteracji danej macierzy. Po zapisaniu tych indeksów w osobnej tablicy możemy sprawdzić przy każdym indeksie, czy spełnia on warunek braku przeszkód, a następnie odpowiednio zwiększyć liczbę.
Poniżej zaimplementowano rozwiązanie oparte na powyższej koncepcji, które wymaga O(N^2) czasu i O(N) dodatkowej przestrzeni.
C++ // C++ program to find how many mirror can transfer // light from bottom to right #include using namespace std ; // method returns number of mirror which can transfer // light from bottom to right int maximumMirrorInMatrix ( string mat [] int N ) { // To store first obstacles horizontally (from right) // and vertically (from bottom) int horizontal [ N ] vertical [ N ]; // initialize both array as -1 signifying no obstacle memset ( horizontal -1 sizeof ( horizontal )); memset ( vertical -1 sizeof ( vertical )); // looping matrix to mark column for obstacles for ( int i = 0 ; i < N ; i ++ ) { for ( int j = N -1 ; j >= 0 ; j -- ) { if ( mat [ i ][ j ] == 'B' ) continue ; // mark rightmost column with obstacle horizontal [ i ] = j ; break ; } } // looping matrix to mark rows for obstacles for ( int j = 0 ; j < N ; j ++ ) { for ( int i = N -1 ; i >= 0 ; i -- ) { if ( mat [ i ][ j ] == 'B' ) continue ; // mark leftmost row with obstacle vertical [ j ] = i ; break ; } } int res = 0 ; // Initialize result // if there is not obstacle on right or below // then mirror can be placed to transfer light for ( int i = 0 ; i < N ; i ++ ) { for ( int j = 0 ; j < N ; j ++ ) { /* if i > vertical[j] then light can from bottom if j > horizontal[i] then light can go to right */ if ( i > vertical [ j ] && j > horizontal [ i ]) { /* uncomment this code to print actual mirror position also cout < < i < < ' ' < < j < < endl; */ res ++ ; } } } return res ; } // Driver code to test above method int main () { int N = 5 ; // B - Blank O - Obstacle string mat [ N ] = { 'BBOBB' 'BBBBO' 'BBBBB' 'BOOBO' 'BBBOB' }; cout < < maximumMirrorInMatrix ( mat N ) < < endl ; return 0 ; }
Java // Java program to find how many mirror can transfer // light from bottom to right import java.util.* ; class GFG { // method returns number of mirror which can transfer // light from bottom to right static int maximumMirrorInMatrix ( String mat [] int N ) { // To store first obstacles horizontally (from right) // and vertically (from bottom) int [] horizontal = new int [ N ] ; int [] vertical = new int [ N ] ; // initialize both array as -1 signifying no obstacle Arrays . fill ( horizontal - 1 ); Arrays . fill ( vertical - 1 ); // looping matrix to mark column for obstacles for ( int i = 0 ; i < N ; i ++ ) { for ( int j = N - 1 ; j >= 0 ; j -- ) { if ( mat [ i ] . charAt ( j ) == 'B' ) { continue ; } // mark rightmost column with obstacle horizontal [ i ] = j ; break ; } } // looping matrix to mark rows for obstacles for ( int j = 0 ; j < N ; j ++ ) { for ( int i = N - 1 ; i >= 0 ; i -- ) { if ( mat [ i ] . charAt ( j ) == 'B' ) { continue ; } // mark leftmost row with obstacle vertical [ j ] = i ; break ; } } int res = 0 ; // Initialize result // if there is not obstacle on right or below // then mirror can be placed to transfer light for ( int i = 0 ; i < N ; i ++ ) { for ( int j = 0 ; j < N ; j ++ ) { /* if i > vertical[j] then light can from bottom if j > horizontal[i] then light can go to right */ if ( i > vertical [ j ] && j > horizontal [ i ] ) { /* uncomment this code to print actual mirror position also cout < < i < < ' ' < < j < < endl; */ res ++ ; } } } return res ; } // Driver code public static void main ( String [] args ) { int N = 5 ; // B - Blank O - Obstacle String mat [] = { 'BBOBB' 'BBBBO' 'BBBBB' 'BOOBO' 'BBBOB' }; System . out . println ( maximumMirrorInMatrix ( mat N )); } } /* This code is contributed by PrinciRaj1992 */
Python3 # Python3 program to find how many mirror can transfer # light from bottom to right # method returns number of mirror which can transfer # light from bottom to right def maximumMirrorInMatrix ( mat N ): # To store first obstacles horizontally (from right) # and vertically (from bottom) horizontal = [ - 1 for i in range ( N )] vertical = [ - 1 for i in range ( N )]; # looping matrix to mark column for obstacles for i in range ( N ): for j in range ( N - 1 - 1 - 1 ): if ( mat [ i ][ j ] == 'B' ): continue ; # mark rightmost column with obstacle horizontal [ i ] = j ; break ; # looping matrix to mark rows for obstacles for j in range ( N ): for i in range ( N - 1 - 1 - 1 ): if ( mat [ i ][ j ] == 'B' ): continue ; # mark leftmost row with obstacle vertical [ j ] = i ; break ; res = 0 ; # Initialize result # if there is not obstacle on right or below # then mirror can be placed to transfer light for i in range ( N ): for j in range ( N ): ''' if i > vertical[j] then light can from bottom if j > horizontal[i] then light can go to right ''' if ( i > vertical [ j ] and j > horizontal [ i ]): ''' uncomment this code to print actual mirror position also''' res += 1 ; return res ; # Driver code to test above method N = 5 ; # B - Blank O - Obstacle mat = [ 'BBOBB' 'BBBBO' 'BBBBB' 'BOOBO' 'BBBOB' ]; print ( maximumMirrorInMatrix ( mat N )); # This code is contributed by rutvik_56.
C# // C# program to find how many mirror can transfer // light from bottom to right using System ; class GFG { // method returns number of mirror which can transfer // light from bottom to right static int maximumMirrorInMatrix ( String [] mat int N ) { // To store first obstacles horizontally (from right) // and vertically (from bottom) int [] horizontal = new int [ N ]; int [] vertical = new int [ N ]; // initialize both array as -1 signifying no obstacle for ( int i = 0 ; i < N ; i ++ ) { horizontal [ i ] =- 1 ; vertical [ i ] =- 1 ; } // looping matrix to mark column for obstacles for ( int i = 0 ; i < N ; i ++ ) { for ( int j = N - 1 ; j >= 0 ; j -- ) { if ( mat [ i ][ j ] == 'B' ) { continue ; } // mark rightmost column with obstacle horizontal [ i ] = j ; break ; } } // looping matrix to mark rows for obstacles for ( int j = 0 ; j < N ; j ++ ) { for ( int i = N - 1 ; i >= 0 ; i -- ) { if ( mat [ i ][ j ] == 'B' ) { continue ; } // mark leftmost row with obstacle vertical [ j ] = i ; break ; } } int res = 0 ; // Initialize result // if there is not obstacle on right or below // then mirror can be placed to transfer light for ( int i = 0 ; i < N ; i ++ ) { for ( int j = 0 ; j < N ; j ++ ) { /* if i > vertical[j] then light can from bottom if j > horizontal[i] then light can go to right */ if ( i > vertical [ j ] && j > horizontal [ i ]) { /* uncomment this code to print actual mirror position also cout < < i < < ' ' < < j < < endl; */ res ++ ; } } } return res ; } // Driver code public static void Main ( String [] args ) { int N = 5 ; // B - Blank O - Obstacle String [] mat = { 'BBOBB' 'BBBBO' 'BBBBB' 'BOOBO' 'BBBOB' }; Console . WriteLine ( maximumMirrorInMatrix ( mat N )); } } // This code is contributed by Princi Singh
JavaScript < script > // JavaScript program to find how many mirror can transfer // light from bottom to right // method returns number of mirror which can transfer // light from bottom to right function maximumMirrorInMatrix ( mat N ) { // To store first obstacles horizontally (from right) // and vertically (from bottom) var horizontal = Array ( N ). fill ( - 1 ); var vertical = Array ( N ). fill ( - 1 ); // looping matrix to mark column for obstacles for ( var i = 0 ; i < N ; i ++ ) { for ( var j = N - 1 ; j >= 0 ; j -- ) { if ( mat [ i ][ j ] == 'B' ) { continue ; } // mark rightmost column with obstacle horizontal [ i ] = j ; break ; } } // looping matrix to mark rows for obstacles for ( var j = 0 ; j < N ; j ++ ) { for ( var i = N - 1 ; i >= 0 ; i -- ) { if ( mat [ i ][ j ] == 'B' ) { continue ; } // mark leftmost row with obstacle vertical [ j ] = i ; break ; } } var res = 0 ; // Initialize result // if there is not obstacle on right or below // then mirror can be placed to transfer light for ( var i = 0 ; i < N ; i ++ ) { for ( var j = 0 ; j < N ; j ++ ) { /* if i > vertical[j] then light can from bottom if j > horizontal[i] then light can go to right */ if ( i > vertical [ j ] && j > horizontal [ i ]) { /* uncomment this code to print actual mirror position also cout < < i < < ' ' < < j < < endl; */ res ++ ; } } } return res ; } // Driver code var N = 5 ; // B - Blank O - Obstacle var mat = [ 'BBOBB' 'BBBBO' 'BBBBB' 'BOOBO' 'BBBOB' ]; document . write ( maximumMirrorInMatrix ( mat N )); < /script>
Wyjście
2
Złożoność czasowa: O(n 2 ).
Przestrzeń pomocnicza: O(n)
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