Максимално огледала која могу да преносе светлост одоздо на десно
Дата је квадратна матрица у којој свака ћелија представља или празно или препреку. Можемо поставити огледала на празно место. Сва огледала ће бити смештена под углом од 45 степени, односно могу да преносе светлост одоздо на десно ако на њиховом путу нема препрека.
У овом питању треба да избројимо колико се таквих огледала може поставити у квадратну матрицу која може да преноси светлост одоздо на десно.
Примери:
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.
Овај проблем можемо решити провером положаја таквих огледала у матрици, огледало које може да преноси светлост одоздо на десно неће имати никакву препреку на свом путу тј.
ако се огледало налази на индексу (и ј) онда
неће бити препрека на индексу (к ј) за све к и < k <= N
неће бити препрека на индексу (и к) за све к ј < k <= N
Имајући на уму горње две једначине, можемо пронаћи крајњу десну препреку у сваком реду у једној итерацији дате матрице и можемо пронаћи најдоњу препреку у свакој колони у другој итерацији дате матрице. Након чувања ових индекса у засебном низу, можемо да проверимо на сваком индексу да ли не задовољава услов препреке или не, а затим да повећамо број у складу са тим.
Испод је имплементирано решење горе наведеног концепта које захтева О(Н^2) времена и О(Н) додатног простора.
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>
Излаз
2
Временска сложеност: О(н 2 ).
Помоћни простор: О(н)
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