to detect path in bfs - c++

Question

i am doing a standard problem to calculate min moves to reach target by a knight but i also want to keep track of path but its showing error.it dispalys
prog.cpp: In function
'int minStepToReachTarget(int*, int*, int)':
prog.cpp:76:42: error: no match for 'operator[]' (operand types are
'std::vector<cell>' and 'cell')
{q.push(cell(x, y, t.dis + 1));parent[cell(x, y, t.dis + 1)]=t;}
I have commented down the line 76 in my code.
struct cell {
int x, y;
int dis;
cell() {}
cell(int x, int y, int dis): x(x), y(y), dis(dis) {}
};
//cell parent[10000];
typedef cell c;
vector<c> parent(10000);
bool isInside(int x, int y, int N) {
if (x >= 1 && x <= N && y >= 1 && y <= N)
return true;
return false;
}
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[knightPos[0]][knightPos[1]] = true;
// parent[cell(knightPos[0], knightPos[1], 0)]=t;
// loop untill we have one element in queue
while (!q.empty()) {
t = q.front();
//parent[t]=t;
q.pop();
visit[t.x][t.y] = true;
// if current cell is equal to target cell,
// return its distance
if (t.x == targetPos[0] && t.y == targetPos[1])
return t.dis;
// loop for all reahable states
for (int i = 0; i < 8; i++) {
x = t.x + dx[i];
y = t.y + dy[i];
// If rechable state is not yet visited and
// inside board, push that state into queue
if (isInside(x, y, N) && !visit[x][y]) {
q.push(cell(x, y, t.dis + 1));
//76 ERROR: parent[cell(x, y, t.dis + 1)]=t;}
}
}
}
int main() {
// size of square board
int N = 6;
int knightPos[] = {4, 5};
int targetPos[] = {1, 1};
int m= minStepToReachTarget(knightPos, targetPos, N);
cout<<m<<endl;
return 0;
}

Answer 1

#Code
int dx[8] = {-1,-2, 1 ,2 ,-1, -2, 1, 2};
int dy[8] = {-2,-1, -2,-1, 2, 1, 2, 1};
q.push(ppi(pi(kx, ky), 0));
while(!q.empty()){
pi cur = q.front().first; int d = q.front().second; q.pop();
int x = cur.first;
int y = cur.second;
// printf("%d %d %d\n", x,y, visited[x][y]);
if(visited[x][y]) continue;
visited[x][y] = true;
if (x == tx && y == ty){
printf("%d", d);
return 0;
}
for(int i = 0; i<8; i++){
int nx = x+dx[i];
int ny = y+dy[i];
if(nx <= 0 || nx > n || ny <= 0 || ny > n) continue;
q.push(ppi(pi(nx, ny), d+1));
}
}
printf("-1");
Explanation
Here, this is the bfs which I implemented. I am storing the grid as a boolean value, I can keep track of which squares I have traveled too. If I am unable to reach the square, I output -1. I would also recommend not using function calls unless really necessary, as this is a simple BFS question.
Extension
Just in case you want to find out more, I took the following chunk from my code for a harder problem, where there exists a grid with forbidden squares where the knight cannot travel. In that case, forbidden squares are initially marked as 'true' instead of 'false' to signify that is traveled so I will not go on it.
I hope my above solution helps.