Trova la sequenza di serpenti di lunghezza massima
Dato una griglia di numeri trova la sequenza di serpenti di lunghezza massima e stampala. Se esistono più sequenze di serpenti con la massima lunghezza stampare uno di essi.
Una sequenza di serpenti è costituita da numeri adiacenti nella griglia in modo tale che per ogni numero il numero a destra o il numero sotto è +1 o -1 il suo valore. Ad esempio, se si è in posizione (x y) nella griglia, puoi spostarsi a destra, ovvero (x y+1) se quel numero è ± 1 o spostarsi verso il basso, ovvero (x+1 y) se quel numero è ± 1.
For example 9 6 5 2 8 7 6 5 7 3 1 6 1 1 1 7 In above grid the longest snake sequence is: (9 8 7 6 5 6 7)
Sotto la figura mostra tutti i percorsi possibili:
Ti consigliamo vivamente di ridurre al minimo il browser e provarlo da solo.
L'idea è di utilizzare la programmazione dinamica. Per ogni cella della matrice manteniamo la massima lunghezza di un serpente che termina nella cella corrente. La sequenza di serpenti di lunghezza massima avrà un valore massimo. La cella di valore massimo corrisponderà alla coda del serpente. Per stampare il serpente, dobbiamo tornare indietro dalla coda fino alla testa di Snake.
Let T[i][i] represent maximum length of a snake which ends at cell (i j) then for given matrix M the DP relation is defined as T[0][0] = 0 T[i][j] = max(T[i][j] T[i][j - 1] + 1) if M[i][j] = M[i][j - 1] ± 1 T[i][j] = max(T[i][j] T[i - 1][j] + 1) if M[i][j] = M[i - 1][j] ± 1
Di seguito è riportata l'implementazione dell'idea
C++ // C++ program to find maximum length // Snake sequence and print it #include using namespace std ; #define M 4 #define N 4 struct Point { int x y ; }; // Function to find maximum length Snake sequence path // (i j) corresponds to tail of the snake list < Point > findPath ( int grid [ M ][ N ] int mat [ M ][ N ] int i int j ) { list < Point > path ; Point pt = { i j }; path . push_front ( pt ); while ( grid [ i ][ j ] != 0 ) { if ( i > 0 && grid [ i ][ j ] - 1 == grid [ i - 1 ][ j ]) { pt = { i - 1 j }; path . push_front ( pt ); i -- ; } else if ( j > 0 && grid [ i ][ j ] - 1 == grid [ i ][ j - 1 ]) { pt = { i j - 1 }; path . push_front ( pt ); j -- ; } } return path ; } // Function to find maximum length Snake sequence void findSnakeSequence ( int mat [ M ][ N ]) { // table to store results of subproblems int lookup [ M ][ N ]; // initialize by 0 memset ( lookup 0 sizeof lookup ); // stores maximum length of Snake sequence int max_len = 0 ; // store coordinates to snake's tail int max_row = 0 ; int max_col = 0 ; // fill the table in bottom-up fashion for ( int i = 0 ; i < M ; i ++ ) { for ( int j = 0 ; j < N ; j ++ ) { // do except for (0 0) cell if ( i || j ) { // look above if ( i > 0 && abs ( mat [ i - 1 ][ j ] - mat [ i ][ j ]) == 1 ) { lookup [ i ][ j ] = max ( lookup [ i ][ j ] lookup [ i - 1 ][ j ] + 1 ); if ( max_len < lookup [ i ][ j ]) { max_len = lookup [ i ][ j ]; max_row = i max_col = j ; } } // look left if ( j > 0 && abs ( mat [ i ][ j - 1 ] - mat [ i ][ j ]) == 1 ) { lookup [ i ][ j ] = max ( lookup [ i ][ j ] lookup [ i ][ j - 1 ] + 1 ); if ( max_len < lookup [ i ][ j ]) { max_len = lookup [ i ][ j ]; max_row = i max_col = j ; } } } } } cout < < 'Maximum length of Snake sequence is: ' < < max_len < < endl ; // find maximum length Snake sequence path list < Point > path = findPath ( lookup mat max_row max_col ); cout < < 'Snake sequence is:' ; for ( auto it = path . begin (); it != path . end (); it ++ ) cout < < endl < < mat [ it -> x ][ it -> y ] < < ' (' < < it -> x < < ' ' < < it -> y < < ')' ; } // Driver code int main () { int mat [ M ][ N ] = { { 9 6 5 2 } { 8 7 6 5 } { 7 3 1 6 } { 1 1 1 7 } }; findSnakeSequence ( mat ); return 0 ; }
Java // Java program to find maximum length // Snake sequence and print it import java.util.* ; class GFG { static int M = 4 ; static int N = 4 ; static class Point { int x y ; public Point ( int x int y ) { this . x = x ; this . y = y ; } }; // Function to find maximum length Snake sequence path // (i j) corresponds to tail of the snake static List < Point > findPath ( int grid [][] int mat [][] int i int j ) { List < Point > path = new LinkedList <> (); Point pt = new Point ( i j ); path . add ( 0 pt ); while ( grid [ i ][ j ] != 0 ) { if ( i > 0 && grid [ i ][ j ] - 1 == grid [ i - 1 ][ j ] ) { pt = new Point ( i - 1 j ); path . add ( 0 pt ); i -- ; } else if ( j > 0 && grid [ i ][ j ] - 1 == grid [ i ][ j - 1 ] ) { pt = new Point ( i j - 1 ); path . add ( 0 pt ); j -- ; } } return path ; } // Function to find maximum length Snake sequence static void findSnakeSequence ( int mat [][] ) { // table to store results of subproblems int [][] lookup = new int [ M ][ N ] ; // initialize by 0 // stores maximum length of Snake sequence int max_len = 0 ; // store coordinates to snake's tail int max_row = 0 ; int max_col = 0 ; // fill the table in bottom-up fashion for ( int i = 0 ; i < M ; i ++ ) { for ( int j = 0 ; j < N ; j ++ ) { // do except for (0 0) cell if ( i != 0 || j != 0 ) { // look above if ( i > 0 && Math . abs ( mat [ i - 1 ][ j ] - mat [ i ][ j ] ) == 1 ) { lookup [ i ][ j ] = Math . max ( lookup [ i ][ j ] lookup [ i - 1 ][ j ] + 1 ); if ( max_len < lookup [ i ][ j ] ) { max_len = lookup [ i ][ j ] ; max_row = i ; max_col = j ; } } // look left if ( j > 0 && Math . abs ( mat [ i ][ j - 1 ] - mat [ i ][ j ] ) == 1 ) { lookup [ i ][ j ] = Math . max ( lookup [ i ][ j ] lookup [ i ][ j - 1 ] + 1 ); if ( max_len < lookup [ i ][ j ] ) { max_len = lookup [ i ][ j ] ; max_row = i ; max_col = j ; } } } } } System . out . print ( 'Maximum length of Snake ' + 'sequence is: ' + max_len + 'n' ); // find maximum length Snake sequence path List < Point > path = findPath ( lookup mat max_row max_col ); System . out . print ( 'Snake sequence is:' ); for ( Point it : path ) System . out . print ( 'n' + mat [ it . x ][ it . y ] + ' (' + it . x + ' ' + it . y + ')' ); } // Driver code public static void main ( String [] args ) { int mat [][] = {{ 9 6 5 2 } { 8 7 6 5 } { 7 3 1 6 } { 1 1 1 7 }}; findSnakeSequence ( mat ); } } // This code is contributed by 29AjayKumar
C# // C# program to find maximum length // Snake sequence and print it using System ; using System.Collections.Generic ; class GFG { static int M = 4 ; static int N = 4 ; public class Point { public int x y ; public Point ( int x int y ) { this . x = x ; this . y = y ; } }; // Function to find maximum length Snake sequence path // (i j) corresponds to tail of the snake static List < Point > findPath ( int [ ] grid int [ ] mat int i int j ) { List < Point > path = new List < Point > (); Point pt = new Point ( i j ); path . Insert ( 0 pt ); while ( grid [ i j ] != 0 ) { if ( i > 0 && grid [ i j ] - 1 == grid [ i - 1 j ]) { pt = new Point ( i - 1 j ); path . Insert ( 0 pt ); i -- ; } else if ( j > 0 && grid [ i j ] - 1 == grid [ i j - 1 ]) { pt = new Point ( i j - 1 ); path . Insert ( 0 pt ); j -- ; } } return path ; } // Function to find maximum length Snake sequence static void findSnakeSequence ( int [ ] mat ) { // table to store results of subproblems int [ ] lookup = new int [ M N ]; // initialize by 0 // stores maximum length of Snake sequence int max_len = 0 ; // store coordinates to snake's tail int max_row = 0 ; int max_col = 0 ; // fill the table in bottom-up fashion for ( int i = 0 ; i < M ; i ++ ) { for ( int j = 0 ; j < N ; j ++ ) { // do except for (0 0) cell if ( i != 0 || j != 0 ) { // look above if ( i > 0 && Math . Abs ( mat [ i - 1 j ] - mat [ i j ]) == 1 ) { lookup [ i j ] = Math . Max ( lookup [ i j ] lookup [ i - 1 j ] + 1 ); if ( max_len < lookup [ i j ]) { max_len = lookup [ i j ]; max_row = i ; max_col = j ; } } // look left if ( j > 0 && Math . Abs ( mat [ i j - 1 ] - mat [ i j ]) == 1 ) { lookup [ i j ] = Math . Max ( lookup [ i j ] lookup [ i j - 1 ] + 1 ); if ( max_len < lookup [ i j ]) { max_len = lookup [ i j ]; max_row = i ; max_col = j ; } } } } } Console . Write ( 'Maximum length of Snake ' + 'sequence is: ' + max_len + 'n' ); // find maximum length Snake sequence path List < Point > path = findPath ( lookup mat max_row max_col ); Console . Write ( 'Snake sequence is:' ); foreach ( Point it in path ) Console . Write ( 'n' + mat [ it . x it . y ] + ' (' + it . x + ' ' + it . y + ')' ); } // Driver code public static void Main ( String [] args ) { int [ ] mat = { { 9 6 5 2 } { 8 7 6 5 } { 7 3 1 6 } { 1 1 1 7 } }; findSnakeSequence ( mat ); } } // This code is contributed by Princi Singh
Python3 def snakesequence ( S m n ): sequence = {} DP = [[ 1 for x in range ( m + 1 )] for x in range ( n + 1 )] a b maximum = 0 0 0 position = [ 0 0 ] for i in range ( 0 n + 1 ): for j in range ( 0 m + 1 ): a b = 0 0 p = 'initial' if ( i > 0 and abs ( S [ i ][ j ] - S [ i - 1 ][ j ]) == 1 ): a = DP [ i - 1 ][ j ] if ( j > 0 and abs ( S [ i ][ j ] - S [ i ][ j - 1 ]) == 1 ): b = DP [ i ][ j - 1 ] if a != 0 and a >= b : p = str ( i - 1 ) + ' ' + str ( j ) elif b != 0 : p = str ( i ) + ' ' + str ( j - 1 ) q = str ( i ) + ' ' + str ( j ) sequence [ q ] = p DP [ i ][ j ] = DP [ i ][ j ] + max ( a b ) if DP [ i ][ j ] >= maximum : maximum = DP [ i ][ j ] position [ 0 ] = i position [ 1 ] = j snakeValues = [] snakePositions = [] snakeValues . append ( S [ position [ 0 ]][ position [ 1 ]]) check = 'found' str_next = str ( position [ 0 ]) + ' ' + str ( position [ 1 ]) findingIndices = sequence [ str_next ] . split () while ( check == 'found' ): if sequence [ str_next ] == 'initial' : snakePositions . insert ( 0 str_next ) check = 'end' continue findingIndices = sequence [ str_next ] . split () g = int ( findingIndices [ 0 ]) h = int ( findingIndices [ 1 ]) snakeValues . insert ( 0 S [ g ][ h ]) snake_position = str ( g ) + ' ' + str ( h ) snakePositions . insert ( 0 str_next ) str_next = sequence [ str_next ] return [ snakeValues snakePositions ] S = [[ 9 6 5 2 ] [ 8 7 6 5 ] [ 7 3 1 6 ] [ 1 1 10 7 ]] m = 3 n = 3 seq = snakesequence ( S m n ) for i in range ( len ( seq [ 0 ])): print ( seq [ 0 ][ i ] '' seq [ 1 ][ i ] . split ())
JavaScript function snakesequence ( S m n ) { let sequence = {} let DP = new Array ( n + 1 ) for ( var i = 0 ; i <= n ; i ++ ) DP [ i ] = new Array ( m + 1 ). fill ( 1 ) let a = 0 b = 0 maximum = 0 let position = [ 0 0 ] for ( var i = 0 ; i <= n ; i ++ ) { for ( var j = 0 ; j <= m ; j ++ ) { a = 0 b = 0 let p = 'initial' if ( i > 0 && Math . abs ( S [ i ][ j ] - S [ i - 1 ][ j ]) == 1 ) a = DP [ i - 1 ][ j ] if ( j > 0 && Math . abs ( S [ i ][ j ] - S [ i ][ j - 1 ]) == 1 ) b = DP [ i ][ j - 1 ] if ( a != 0 && a >= b ) p = String ( i - 1 ) + ' ' + String ( j ) else if ( b != 0 ) p = String ( i ) + ' ' + String ( j - 1 ) let q = String ( i ) + ' ' + String ( j ) sequence [ q ] = p DP [ i ][ j ] = DP [ i ][ j ] + Math . max ( a b ) if ( DP [ i ][ j ] >= maximum ) { maximum = DP [ i ][ j ] position [ 0 ] = i position [ 1 ] = j } } } let snakeValues = [] let snakePositions = [] snakeValues . push ( S [ position [ 0 ]][ position [ 1 ]]) let check = 'found' let String_next = String ( position [ 0 ]) + ' ' + String ( position [ 1 ]) let findingIndices = sequence [ String_next ]. split ( ' ' ) while ( check == 'found' ) { if ( sequence [ String_next ] == 'initial' ) { snakePositions . unshift ( String_next ) check = 'end' continue } findingIndices = sequence [ String_next ]. split ( ' ' ) let g = parseInt ( findingIndices [ 0 ]) let h = parseInt ( findingIndices [ 1 ]) snakeValues . unshift ( S [ g ][ h ]) let snake_position = String ( g ) + ' ' + String ( h ) snakePositions . unshift ( String_next ) String_next = sequence [ String_next ] } return [ snakeValues snakePositions ] } // Driver Code let S = [[ 9 6 5 2 ] [ 8 7 6 5 ] [ 7 3 1 6 ] [ 1 1 10 7 ]] let m = 3 let n = 3 let seq = snakesequence ( S m n ) for ( var i = 0 ; i < seq [ 0 ]. length ; i ++ ) console . log ( seq [ 0 ][ i ] + '' seq [ 1 ][ i ]. split ( ' ' ))
Produzione
Maximum length of Snake sequence is: 6 Snake sequence is: 9 (0 0) 8 (1 0) 7 (1 1) 6 (1 2) 5 (1 3) 6 (2 3) 7 (3 3)
La complessità temporale della soluzione sopra è O (m*n). Lo spazio ausiliario utilizzato dalla soluzione sopra è O (m*n). Se non siamo tenuti a stampare, lo spazio del serpente può essere ulteriormente ridotto a O (N) poiché usiamo solo il risultato dall'ultima riga.
