Atsižvelgiant į skaičių tinklelį, raskite maksimalaus ilgio gyvatės seką ir atsispausdinkite. Jei egzistuoja kelios gyvatės sekos su maksimaliu ilgiu, atspausdintą bet kurį iš jų.
Gyvatės seka yra sudaryta iš gretimų skaičių tinklelyje taip, kad kiekvienam skaičiui skaičius dešinėje arba žemiau esančio skaičius yra +1 arba -1 jo vertė. Pvz., Jei esate vietoje (x y) tinklelyje, galite judėti į dešinę, t. Y. (x y+1), jei šis skaičius yra ± 1, arba judėti žemyn, t. Y. (x+1 y), jei šis skaičius yra ± 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)
Žemiau esančiame paveikslėlyje pavaizduoti visi galimi keliai:
Mes labai rekomenduojame sumažinti savo naršyklę ir pirmiausia išbandyti tai patys.
Idėja yra naudoti dinaminį programavimą. Kiekvienai matricos ląstelei mes išlaikome maksimalų gyvatės ilgį, kuris baigiasi esamomis ląstelėmis. Maksimalaus ilgio gyvatės seka turės maksimalią vertę. Maksimali vertės ląstelė atitiks gyvatės uodegą. Norėdami atspausdinti gyvatę, turime atsitraukti nuo uodegos iki galo atgal į gyvatės galvą.
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
Žemiau yra idėjos įgyvendinimas
C++// C++ program to find maximum length // Snake sequence and print it #include
// 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# 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
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())
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(' '))
Išvestis
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)
Aukščiau pateikto tirpalo laiko sudėtingumas yra O (M*n). Pagalbinė erdvė, naudojama aukščiau esančiame tirpale, yra O (m*n). Jei neprivalome spausdinti gyvatės erdvės, galima dar labiau sumažinti iki O (n), nes mes naudojame tik paskutinės eilės rezultatą.