# What's wrong with my standard bfs code?

• I used standard BFS. Mark every position we reach 2 to avoid repeated search. The code fails at test case 70, which is a huge one. Could anyone see mistakes in my code?

``````class Solution {
public:
int shortestDistance(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
int N=maze.size(), M = maze[0].size();
int minlen=INT_MAX;
int dx[4] ={ 1, 0,-1, 0};
int dy[4] ={ 0, 1, 0,-1};
queue<vector<int>> Q;
Q.push({start[0],start[1],-1,0});
maze[start[0]][start[1]] = 2;
while( !Q.empty() ){
vector<int> F = Q.front();
Q.pop();
for(int i=0;i<4;i++){
int x=F[0],y=F[1],last=F[2],dist=F[3];
if( (last==1 || last==3) && (i==1 || i==3) ) continue;
if( (last==0 || last==2) && (i==0 || i==2) ) continue;
int d=0;
while( x+dx[i]>=0 && x+dx[i]<N && y+dy[i]>=0 && y+dy[i]<M && maze[x+dx[i]][y+dy[i]]!=1 ){
x += dx[i]; y += dy[i]; d++;
}
if( x==destination[0] && y==destination[1] && dist+d<minlen) minlen=dist+d;
if( maze[x][y]==2 ) continue;
maze[x][y] = 2;
Q.push({x,y,i,dist+d});
}
}
return minlen==INT_MAX?-1:minlen;
}
};
``````

• you might need to re process the position again if a smaller distance is found.

• I see. Searching from a searched position might yield shorter distance.

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