Minimum Cost Path with Left, Right, Bottom and Up moves allowed - GeeksforGeeks


Minimum Cost Path with Left, Right, Bottom and Up moves allowed - GeeksforGeeks
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.


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];
}

http://www.geeksforgeeks.org/dynamic-programming-set-6-min-cost-path/
    private static int minCost(int cost[][], int m, int n)
    {
        int i, j;
        int tc[][]=new int[m+1][n+1];
 
        tc[0][0] = cost[0][0];
 
        /* Initialize first column of total cost(tc) array */
        for (i = 1; i <= m; i++)
            tc[i][0] = tc[i-1][0] + cost[i][0];
 
        /* Initialize first row of tc array */
        for (j = 1; j <= n; j++)
            tc[0][j] = tc[0][j-1] + cost[0][j];
 
        /* Construct rest of the tc array */
        for (i = 1; i <= m; i++)
            for (j = 1; j <= n; j++)
                tc[i][j] = min(tc[i-1][j-1],
                               tc[i-1][j],
                               tc[i][j-1]) + cost[i][j];
 
        return tc[m][n];
    }


int minCost(int cost[R][C], int m, int n)
{
   if (n < 0 || m < 0)
      return INT_MAX;
   else if (m == 0 && n == 0)
      return cost[m][n];
   else
      return cost[m][n] + min( minCost(cost, m-1, n-1),
                               minCost(cost, m-1, n),
                               minCost(cost, m, n-1) );
}
Read full article from Minimum Cost Path with Left, Right, Bottom and Up moves allowed - GeeksforGeeks

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