Given a two dimensional grid, each cell of which contains integer cost which represents a cost to traverse through that cell, we need to find a path from top left cell to bottom right cell by which total cost incurred is minimum.

**Note : **It is assumed that negative cost cycles do not exist in input matrix.

This problem is extension of below problem.

Min Cost Path with right and bottom moves allowed.

In previous problem only going right and bottom was allowed but in this problem we are allowed to go bottom, up, right and left i.e. in all 4 direction.

Examples:

A cost grid is given in below diagram, minimum cost to reach bottom right from top left is 327 (= 31 + 10 + 13 + 47 + 65 + 12 + 18 + 6 + 33 + 11 + 20 + 41 + 20) The chosen least cost path is shown in green.

It is not possible to solve this problem using dynamic programming similar to previous problem because here current state depends not only on right and bottom cells but also on left and upper cells. We solve this problem using dijkstra’s algorithm. Each cell of grid represents a vertex and neighbor cells adjacent vertices. We do not make an explicit graph from these cells instead we will use matrix as it is in our dijkstra’s algorithm.

In below code Dijkstra’ algorithm’s implementation using C++ STL is used. The code implemented below is changed to cope with matrix represented implicit graph. Please also see use of dx and dy arrays in below code, these arrays are taken for simplifying the process of visiting neighbor vertices of each cell.

// C++ program to get least cost path in a grid from // top-left to bottom-right #include <bits/stdc++.h> using namespace std; #define ROW 5 #define COL 5 // structure for information of each cell struct cell { int x, y; int distance; cell(int x, int y, int distance) : x(x), y(y), distance(distance) {} }; // Utility method for comparing two cells bool operator<(const cell& a, const cell& b) { if (a.distance == b.distance) { if (a.x != b.x) return (a.x < b.x); else return (a.y < b.y); } return (a.distance < b.distance); } // Utility method to check whether a point is // inside the grid or not bool isInsideGrid(int i, int j) { return (i >= 0 && i < COL && j >= 0 && j < ROW); } // Method returns minimum cost to reach bottom // right from top left int shortest(int grid[ROW][COL], int row, int col) { int dis[row][col]; // initializing distance array by INT_MAX for (int i = 0; i < row; i++) for (int j = 0; j < col; j++) dis[i][j] = INT_MAX; // direction arrays for simplification of getting // neighbour int dx[] = {-1, 0, 1, 0}; int dy[] = {0, 1, 0, -1}; set<cell> st; // insert (0, 0) cell with 0 distance st.insert(cell(0, 0, 0)); // initialize distance of (0, 0) with its grid value dis[0][0] = grid[0][0]; // loop for standard dijkstra's algorithm while (!st.empty()) { // get the cell with minimum distance and delete // it from the set cell k = *st.begin(); st.erase(st.begin()); // looping through all neighbours for (int i = 0; i < 4; i++) { int x = k.x + dx[i]; int y = k.y + dy[i]; // if not inside boundry, ignore them if (!isInsideGrid(x, y)) continue; // If distance from current cell is smaller, then // update distance of neighbour cell if (dis[x][y] > dis[k.x][k.y] + grid[x][y]) { // If cell is already there in set, then // remove its previous entry if (dis[x][y] != INT_MAX) st.erase(st.find(cell(x, y, dis[x][y]))); // update the distance and insert new updated // cell in set dis[x][y] = dis[k.x][k.y] + grid[x][y]; st.insert(cell(x, y, dis[x][y])); } } } // uncomment below code to print distance // of each cell from (0, 0) /* for (int i = 0; i < row; i++, cout << endl) for (int j = 0; j < col; j++) cout << dis[i][j] << " "; */ // dis[row - 1][col - 1] will represent final // distance of bottom right cell from top left cell return dis[row - 1][col - 1]; } // Driver code to test above methods int main() { int grid[ROW][COL] = { 31, 100, 65, 12, 18, 10, 13, 47, 157, 6, 100, 113, 174, 11, 33, 88, 124, 41, 20, 140, 99, 32, 111, 41, 20 }; cout << shortest(grid, ROW, COL) << endl; return 0; }

Output:

327

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