Je dána čtvercová matice, ve které každá buňka představuje buď prázdné místo, nebo překážku. Můžeme umístit zrcadla na prázdnou pozici. Všechna zrcadla budou umístěna pod úhlem 45 stupňů, tj. mohou přenášet světlo zdola doprava, pokud jim v cestě nestojí žádná překážka.
V této otázce musíme spočítat, kolik takových zrcadel lze umístit do čtvercové matice, která může přenášet světlo zdola doprava.
Příklady:
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.
Tento problém můžeme vyřešit kontrolou polohy takových zrcadel v matici, zrcadlo, které může přenášet světlo zdola doprava, nebude mít v cestě žádnou překážku, tzn.
pokud je na indexu (i j) zrcadlo
na indexu (k j) nebude pro všechna k i žádná překážka< k <= N
na indexu (i k) nebude pro všechna k j žádná překážka< k <= N
Pokud vezmeme v úvahu výše uvedené dvě rovnice, můžeme najít nejspodnější překážku na každém řádku v jedné iteraci dané matice a nejspodnější překážku v každém sloupci v další iteraci dané matice. Po uložení těchto indexů do samostatného pole můžeme u každého indexu zkontrolovat, zda nesplňuje žádnou překážkovou podmínku nebo ne, a poté počet odpovídajícím způsobem zvýšit.
Níže je implementováno řešení na výše uvedené koncepci, které vyžaduje O(N^2) čas a O(N) prostor navíc.
C++// C++ program to find how many mirror can transfer // light from bottom to right #include
// 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 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# 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
<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>
Výstup
2
Časová složitost: O(n2).
Pomocný prostor: O(n)