Minimum Cost Path with Left, Right, Bottom and Up moves allowed in C++

Suppose we have a 2D array. Where each cell of which consists number cost which represents a cost to visit through that cell, we have to find a path from top-left cell to bottom-right cell by which total cost consumed is minimum.

So, if the input is like

then the output will be 340 as (32 + 11 + 14 + 48 + 66 + 13 + 19 + 7 + 34 + 12 + 21 + 42 + 21) = 340

To solve this, we will follow these steps −

• Define cell with x, y coordinate and distance parameter.
• Define an array matrix of size: row x col.
• fill the matrix with inf
• Define an array dx of size: 4 := { - 1, 0, 1, 0}
• Define an array dy of size: 4 := {0, 1, 0, - 1}
• Define one set of cell called st
• insert cell(0, 0, 0) into st
• matrix[0, 0] := grid[0, 0]
• while (not st is empty), do −
  • k := first element of st
  • delete first element of st from st
  • for initialize i := 0, when i < 4, update (increase i by 1), do −
    • x := k.x + dx[i]
    • y := k.y + dy[i]
    • if not isOk(x, y), then −
      • Ignore following part, skip to the next iteration
    • if matrix[x, y] > matrix[k.x, k.y] + grid[x, y], then −
      • if matrix[x, y] is not equal to inf, then −
        • find and delete cell(x, y, matrix[x, y]) from st
      • matrix[x, y] := matrix[k.x, k.y] + grid[x, y]
      • insert cell(x, y, matrix[x, y]) into st
• return matrix[row - 1, col - 1]

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
#define ROW 5
#define COL 5
class cell {
   public:
   int x, y;
   int distance;
   cell(int x, int y, int distance) :
   x(x), y(y), distance(distance) {}
};
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);
}
bool isOk(int i, int j) {
   return (i >= 0 && i < COL && j >= 0 && j < ROW);
}
int solve(int grid[ROW][COL], int row, int col) {
   int matrix[row][col];
   for (int i = 0; i < row; i++)
   for (int j = 0; j < col; j++)
   matrix[i][j] = INT_MAX;
   int dx[] = {-1, 0, 1, 0};
   int dy[] = {0, 1, 0, -1};
   set<cell> st;
   st.insert(cell(0, 0, 0));
   matrix[0][0] = grid[0][0];
   while (!st.empty()) {
      cell k = *st.begin();
      st.erase(st.begin());
      for (int i = 0; i < 4; i++) {
         int x = k.x + dx[i];
         int y = k.y + dy[i];  
         if (!isOk(x, y))
            continue;
         if (matrix[x][y] > matrix[k.x][k.y] + grid[x][y]){
            if (matrix[x][y] != INT_MAX)
               st.erase(st.find(cell(x, y, matrix[x][y])));
               matrix[x][y] = matrix[k.x][k.y] + grid[x][y];
               st.insert(cell(x, y, matrix[x][y]));
         }
      }
   }
   return matrix[row - 1][col - 1];
}
int main() {
   int grid[ROW][COL] = {
      32, 101, 66, 13, 19,
      11, 14, 48, 158, 7,
      101, 114, 175, 12, 34,
      89, 126, 42, 21, 141,
      100, 33, 112, 42, 21
   };
   cout << solve(grid, ROW, COL);
}

Input

{32, 101, 66, 13, 19,
11, 14, 48, 158, 7,
101, 114, 175, 12, 34,
89, 126, 42, 21, 141,
100, 33, 112, 42, 21
};

Output:

340

Arnab Chakraborty

e

Updated on: 27-Aug-2020

332 Views

Kickstart Your Career

Get certified by completing the course

Get Started