S obzirom na 2D mrežu veličine n*n gdje svaka ćelija predstavlja trošak prolaska kroz tu ćeliju zadatak je pronaći minimalni trošak premjestiti se iz gore lijevo ćelija za dolje desno ćelija. Iz zadane ćelije možemo se useliti 4 smjera : lijevo desno gore dolje.
Bilješka: Pretpostavlja se da negativni troškovni ciklusi ne postoje u ulaznoj matrici.
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Primjer:
Ulazni: mreža = {{9 4 9 9}
{6 7 6 4}
{8 3 3 7}
{7 4 9 10}}
Izlaz: 43
Obrazloženje: Minimalni trošak je 9 + 4 + 7 + 3 + 3 + 7 + 10.
Pristup:
Ideja je koristiti Dijkstrin algoritam pronaći put minimalne cijene kroz mrežu. Ovaj pristup rešetku tretira kao grafikon gdje je svaka ćelija čvor, a algoritam dinamički istražuje najisplativiju stazu do donje desne ćelije uvijek prvo proširujući najjeftinije staze.
Pristup korak po korak:
- Upotrijebite min-heap kako biste uvijek prvo obradili najjeftiniji put i gurnuli gornju lijevu ćeliju u njega.
- Inicijalizirajte matricu troškova s maksimalnim vrijednostima postavljajući trošak početne ćelije na njezinu vrijednost mreže.
- Za svaku ćeliju označite sve 4 susjedne ćelije
- Ako se pronađe staza niže cijene, ažurirajte cijenu ćelije i gurnite je u hrpu.
- Vratite minimalnu cijenu da dođete do donje desne ćelije.
U nastavku je implementacija gornjeg pristupa:
C++// C++ program to find minimum Cost Path with // Left Right Bottom and Up moves allowed #include using namespace std; // Function to check if cell is valid. bool isValidCell(int i int j int n) { return i>=0 && i<n && j>=0 && j<n; } int minimumCostPath(vector<vector<int>> &grid) { int n = grid.size(); // Min heap to implement dijkstra priority_queue<vector<int> vector<vector<int>> greater<vector<int>>> pq; // 2d grid to store minimum cost // to reach every cell. vector<vector<int>> cost(n vector<int>(n INT_MAX)); cost[0][0] = grid[0][0]; // Direction vector to move in 4 directions vector<vector<int>> dir = {{-10} {10} {0-1} {01}}; pq.push({grid[0][0] 0 0}); while (!pq.empty()) { vector<int> top = pq.top(); pq.pop(); int c = top[0] i = top[1] j = top[2]; // Check for all 4 neighbouring cells. for (auto d: dir) { int x = i + d[0]; int y = j + d[1]; // If cell is valid and cost to reach this cell // from current cell is less if (isValidCell(x y n) && cost[i][j]+grid[x][y]<cost[x][y]) { // Update cost to reach this cell. cost[x][y] = cost[i][j]+grid[x][y]; // Push the cell into heap. pq.push({cost[x][y] x y}); } } } // Return minimum cost to // reach bottom right cell. return cost[n-1][n-1]; } int main() { vector<vector<int>> grid = {{9499}{6764}{8337}{74910}}; cout << minimumCostPath(grid) << endl; return 0; }
Java // Java program to find minimum Cost Path with // Left Right Bottom and Up moves allowed import java.util.PriorityQueue; import java.util.Arrays; class GfG { // Function to check if cell is valid. static boolean isValidCell(int i int j int n) { return i >= 0 && i < n && j >= 0 && j < n; } static int minimumCostPath(int[][] grid) { int n = grid.length; // Min heap to implement Dijkstra PriorityQueue<int[]> pq = new PriorityQueue<>((a b) -> Integer.compare(a[0] b[0])); // 2D grid to store minimum cost // to reach every cell. int[][] cost = new int[n][n]; for (int[] row : cost) { Arrays.fill(row Integer.MAX_VALUE); } cost[0][0] = grid[0][0]; // Direction vector to move in 4 directions int[][] dir = {{-1 0} {1 0} {0 -1} {0 1}}; pq.offer(new int[]{grid[0][0] 0 0}); while (!pq.isEmpty()) { int[] top = pq.poll(); int c = top[0] i = top[1] j = top[2]; // Check for all 4 neighbouring cells. for (int[] d : dir) { int x = i + d[0]; int y = j + d[1]; // If cell is valid and cost to reach this cell // from current cell is less if (isValidCell(x y n) && cost[i][j] + grid[x][y] < cost[x][y]) { // Update cost to reach this cell. cost[x][y] = cost[i][j] + grid[x][y]; // Push the cell into heap. pq.offer(new int[]{cost[x][y] x y}); } } } // Return minimum cost to // reach bottom right cell. return cost[n - 1][n - 1]; } public static void main(String[] args) { int[][] grid = { {9 4 9 9} {6 7 6 4} {8 3 3 7} {7 4 9 10} }; System.out.println(minimumCostPath(grid)); } }
Python # Python program to find minimum Cost Path with # Left Right Bottom and Up moves allowed import heapq # Function to check if cell is valid. def isValidCell(i j n): return i >= 0 and i < n and j >= 0 and j < n def minimumCostPath(grid): n = len(grid) # Min heap to implement Dijkstra pq = [] # 2D grid to store minimum cost # to reach every cell. cost = [[float('inf')] * n for _ in range(n)] cost[0][0] = grid[0][0] # Direction vector to move in 4 directions dir = [[-1 0] [1 0] [0 -1] [0 1]] heapq.heappush(pq [grid[0][0] 0 0]) while pq: c i j = heapq.heappop(pq) # Check for all 4 neighbouring cells. for d in dir: x y = i + d[0] j + d[1] # If cell is valid and cost to reach this cell # from current cell is less if isValidCell(x y n) and cost[i][j] + grid[x][y] < cost[x][y]: # Update cost to reach this cell. cost[x][y] = cost[i][j] + grid[x][y] # Push the cell into heap. heapq.heappush(pq [cost[x][y] x y]) # Return minimum cost to # reach bottom right cell. return cost[n - 1][n - 1] if __name__ == '__main__': grid = [ [9 4 9 9] [6 7 6 4] [8 3 3 7] [7 4 9 10] ] print(minimumCostPath(grid))
C# // C# program to find minimum Cost Path with // Left Right Bottom and Up moves allowed using System; using System.Collections.Generic; class GfG { // Function to check if cell is valid. static bool isValidCell(int i int j int n) { return i >= 0 && i < n && j >= 0 && j < n; } static int minimumCostPath(int[][] grid) { int n = grid.Length; // Min heap to implement Dijkstra var pq = new SortedSet<(int cost int x int y)>(); // 2D grid to store minimum cost // to reach every cell. int[][] cost = new int[n][]; for (int i = 0; i < n; i++) { cost[i] = new int[n]; Array.Fill(cost[i] int.MaxValue); } cost[0][0] = grid[0][0]; // Direction vector to move in 4 directions int[][] dir = { new int[] {-1 0} new int[] {1 0} new int[] {0 -1} new int[] {0 1} }; pq.Add((grid[0][0] 0 0)); while (pq.Count > 0) { var top = pq.Min; pq.Remove(top); int i = top.x j = top.y; // Check for all 4 neighbouring cells. foreach (var d in dir) { int x = i + d[0]; int y = j + d[1]; // If cell is valid and cost to reach this cell // from current cell is less if (isValidCell(x y n) && cost[i][j] + grid[x][y] < cost[x][y]) { // Update cost to reach this cell. cost[x][y] = cost[i][j] + grid[x][y]; // Push the cell into heap. pq.Add((cost[x][y] x y)); } } } // Return minimum cost to // reach bottom right cell. return cost[n - 1][n - 1]; } static void Main(string[] args) { int[][] grid = new int[][] { new int[] {9 4 9 9} new int[] {6 7 6 4} new int[] {8 3 3 7} new int[] {7 4 9 10} }; Console.WriteLine(minimumCostPath(grid)); } }
JavaScript // JavaScript program to find minimum Cost Path with // Left Right Bottom and Up moves allowed function comparator(a b) { if (a[0] > b[0]) return -1; if (a[0] < b[0]) return 1; return 0; } class PriorityQueue { constructor(compare) { this.heap = []; this.compare = compare; } enqueue(value) { this.heap.push(value); this.bubbleUp(); } bubbleUp() { let index = this.heap.length - 1; while (index > 0) { let element = this.heap[index] parentIndex = Math.floor((index - 1) / 2) parent = this.heap[parentIndex]; if (this.compare(element parent) < 0) break; this.heap[index] = parent; this.heap[parentIndex] = element; index = parentIndex; } } dequeue() { let max = this.heap[0]; let end = this.heap.pop(); if (this.heap.length > 0) { this.heap[0] = end; this.sinkDown(0); } return max; } sinkDown(index) { let left = 2 * index + 1 right = 2 * index + 2 largest = index; if ( left < this.heap.length && this.compare(this.heap[left] this.heap[largest]) > 0 ) { largest = left; } if ( right < this.heap.length && this.compare(this.heap[right] this.heap[largest]) > 0 ) { largest = right; } if (largest !== index) { [this.heap[largest] this.heap[index]] = [ this.heap[index] this.heap[largest] ]; this.sinkDown(largest); } } isEmpty() { return this.heap.length === 0; } } // Function to check if cell is valid. function isValidCell(i j n) { return i >= 0 && i < n && j >= 0 && j < n; } function minimumCostPath(grid) { let n = grid.length; // Min heap to implement Dijkstra const pq = new PriorityQueue(comparator) // 2D grid to store minimum cost // to reach every cell. let cost = Array.from({ length: n } () => Array(n).fill(Infinity)); cost[0][0] = grid[0][0]; // Direction vector to move in 4 directions let dir = [[-1 0] [1 0] [0 -1] [0 1]]; pq.enqueue([grid[0][0] 0 0]); while (!pq.isEmpty()) { let [c i j] = pq.dequeue(); // Check for all 4 neighbouring cells. for (let d of dir) { let x = i + d[0]; let y = j + d[1]; // If cell is valid and cost to reach this cell // from current cell is less if (isValidCell(x y n) && cost[i][j] + grid[x][y] < cost[x][y]) { // Update cost to reach this cell. cost[x][y] = cost[i][j] + grid[x][y]; // Push the cell into heap. pq.enqueue([cost[x][y] x y]); } } } // Return minimum cost to // reach bottom right cell. return cost[n - 1][n - 1]; } let grid = [ [9 4 9 9] [6 7 6 4] [8 3 3 7] [7 4 9 10] ]; console.log(minimumCostPath(grid));
Izlaz
43
Vremenska složenost: O(n^2 log(n^2))
Pomoćni prostor: O(n^2 log(n^2))
Zašto se ne može koristiti dinamičko programiranje?
array vs arraylist
Dinamičko programiranje ovdje ne uspijeva jer dopuštanje kretanja u sva četiri smjera stvara cikluse u kojima se ćelije mogu ponovno posjetiti kršeći pretpostavku optimalne podstrukture. To znači da cijena dolaska do ćelije iz dane ćelije nije fiksna, već ovisi o cijelom putu.
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