Je uvedená štvorcová matica, v ktorej každá bunka predstavuje buď prázdne miesto, alebo prekážku. Zrkadlá môžeme umiestniť na prázdne miesto. Všetky zrkadlá budú umiestnené pod uhlom 45 stupňov, to znamená, že môžu prenášať svetlo zdola doprava, ak im v ceste nie je žiadna prekážka.
zoznam náboženstiev
V tejto otázke musíme spočítať, koľko takýchto zrkadiel je možné umiestniť do štvorcovej matice, ktorá dokáže prenášať svetlo zdola doprava.
Príklady:
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 môžeme vyriešiť tak, že skontrolujeme polohu takýchto zrkadiel v matici, zrkadlo, ktoré dokáže prenášať svetlo zdola doprava, nebude mať v ceste žiadnu prekážku, t.j.
ak je zrkadlo na indexe (i j), potom
nebude prekážka na indexe (k j) pre všetky k i< k <= N
nebude prekážka na indexe (i k) pre všetky k j< k <= N
Ak vezmeme do úvahy vyššie uvedené dve rovnice, môžeme nájsť prekážku úplne vpravo v každom riadku v jednej iterácii danej matice a môžeme nájsť najspodnejšiu prekážku v každom stĺpci v ďalšej iterácii danej matice. Po uložení týchto indexov do samostatného poľa môžeme pri každom indexe skontrolovať, či nevyhovuje žiadnej prekážkovej podmienke alebo nie, a následne zodpovedajúcim spôsobom zvýšiť počet.
Nižšie je implementované riešenie vyššie uvedeného konceptu, ktoré vyžaduje O(N^2) čas a O(N) priestor navyše.
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 zložitosť: O(n2).
Pomocný priestor: O(n)
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