C++ Knight moves path - c++

Question

I am working on the knight moves problem and managed to print the number of moves but still need to print the path e.g. "3 moves: path is: [3,3] [4,5] [2,4] [4,3]". I tried to print the Queue but got instead all visited paths. I tried also to work backward to the previous points (function minStepToReachTarget) but i think my novice skill did not help.
i already got the number of moves but is there a function or piece of code that help me print the path?
Best,
James.
#include <bits/stdc++.h>
#include <iostream>
#include <queue>
using namespace std;
// structure for storing a cell's data
class cell {
public:
int x, y;
int dis;
cell() {}
cell(int x, int y, int dis) : x(x), y(y), dis(dis) { }
};
// Utility method returns true if (x, y) lies
// inside Board
bool isInside(int x, int y, int N)
{
if (x >= 1 && x <= N && y >= 1 && y <= N)
return true;
return false;
}
// Method returns minimum step
// to reach target position
int minStepToReachTarget( int knightPos[], int targetPos[], int N)
{
// x and y direction, where a knight can move
int dx[] = { -2, -1, 1, 2, -2, -1, 1, 2 };
int dy[] = { -1, -2, -2, -1, 1, 2, 2, 1 };
// queue for storing states of knight in board
queue<cell> q;
// push starting position of knight with 0 distance
q.push(cell(knightPos[0], knightPos[1], 0));
cell t;
int x, y;
bool visit[N + 1][N + 1];
// make all cell unvisited
for (int i = 1; i <= N; i++)
for (int j = 1; j <= N; j++)
visit[i][j] = false;
// visit starting state
visit[knightPos[0]][knightPos[1]] = true;
// loop until we have one element in queue
while (!q.empty()) {
t = q.front();
q.pop();
// cout << "[" << t.x << " "<< t.y<<"]\n";
// if current cell is equal to target cell,
// return its distance
if (t.x == targetPos[0] && t.y == targetPos[1]) {
// cout << "[" << targetPos[0] << " " << targetPos[1] << "]\n";
return t.dis;
}
// loop for all reachable states
for (int i = 0; i < 8; i++) {
x = t.x + dx[i];
y = t.y + dy[i];
// If reachable state is not yet visited and
// inside board, push that state into queue
if (isInside(x, y, N) && !visit[x][y]) {
visit[x][y] = true;
//** cout << "[" << x << " " << y << "]\n";
q.push(cell(x, y, t.dis + 1));
}
}
}
return t.dis;
}
int main(){
int N = 8, knightPos[2], targetPos[2];
cout <<"=> Enter the current Knight’s location: ";
for (int i = 0; i < 2; i++) std::cin>> knightPos[i];
cout <<"=> Enter the destination location: ";
for (int i = 0; i < 2; i++) std::cin>> targetPos[i];
cout <<"=> You made it in " << minStepToReachTarget(knightPos, targetPos, N) <<
" moves from [" << knightPos[0] <<"," << knightPos[1] << "] "
"to [" << targetPos[0] <<"," << targetPos[1] <<"]! Here is your path: ";
return 0;
/*
=> Enter the current Knight’s location: 3 3
=> Enter the destination location: 4 3
=> You made it in 3 moves from [3,3] to [4,3]! Here is your path:
[3,3] [4,5] [2,4] [4,3]
*/
}

Answer 1

You're doing a breadth-first search on your board. Once your reach your destination, the function minStepToReachTarget has no idea how it got there. What you can do is, for each cell in the queue, keep track of its 'parent', i.e. from which cell it was visited first. Once you've found the target, instead of returning t.dis, you could trace the parents of t and return an array representing the path. t.dis would then be equal to the length of the array minus one, since you include both the starting point and target point in the path.