Snake and Ladder Problem

We know about the famous game Snake and Ladder. In this game, some rooms are present on the board, with the room number. Some rooms are connected with a ladder or with snakes. When we get a ladder, we can climb up to some rooms to reach near to the destination without moving sequentially. Similarly, when we get some snake, it sends us to a lower room to start the journey again from that room.

In this problem, we have to find the minimum number of the dice throw is required to reach start to destination.

Input and Output

Input:
The starting and ending location of the snake and ladders.
Snake: From 26 to 0, From 20 to 8, From 16 to 3, From 18 to 6
Ladder From 2 to 21, From 4 to 7, From 10 to 25, from 19 to 28
Output:
Min Dice throws required is 3

Algorithm

minDiceThrow(move, cell)

Input: jump location for snake or ladder, and the total number of cells.
Output: Minimum number of dice throw required to reach to the final cell.

Begin
   initially mark all cell as unvisited
   define queue q
   mark the staring vertex as visited

   for starting vertex the vertex number := 0 and distance := 0
   add starting vertex s into q
   while q is not empty, do
      qVert := front element of the queue
      v := vertex number of qVert
      if v = cell -1, then //when it is last vertex
         break the loop
      delete one item from queue
      for j := v + 1, to v + 6 and j 
Example
#include
#include 
using namespace std;

struct vertex {
   int vert;
   int dist;       // Distance of this vertex from source
};

int minDiceThrow(int move[], int cell) {
   bool visited[cell];
   for (int i = 0; i  q;

   visited[0] = true;       //initially starting from 0
   vertex s = {0, 0};
   q.push(s);             // Enqueue 0'th vertex

   vertex qVert;
   while (!q.empty()) {
      qVert = q.front();
      int v = qVert.vert;

      if (v == cell-1)    //when v is the destination vertex
         break;

      q.pop();
      for (int j=v+1; j
Output
Min Dice throws required is 3