[Locked] Walls and Gates

Walls and Gates

You are given a m x n 2D grid initialized with these three possible values.

1. -1 - A wall or an obstacle.
2. 0 - A gate.
3. INF - Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.

Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.

For example, given the 2D grid:

INF  -1  0  INF
INF INF INF  -1
INF  -1 INF  -1
  0  -1 INF INF

After running your function, the 2D grid should be:

  3  -1   0   1
  2   2   1  -1
  1  -1   2  -1
  0  -1   3   4

分析：

　　深搜DFS即可，注意剪枝，注意overflow

代码：

//深搜，三种情况return: 房间编号已经比当前距离小的，墙和门，访问过的，而这三种情况可以用一个式子表示grids[i][j] < dist
void dfs(vector<vector<int> > &grids, int dist, int i, int j) {
    if(grids[i][j] < dist)
        return;
    grids[i][j] = dist;
    dfs(grids, dist + , i, j + );
    dfs(grids, dist + , i + , j);
    dfs(grids, dist + , i, j - );
    dfs(grids, dist + , i - , j);
    return;
}
void distanceFromGate(vector<vector<int> > &grids) {
    if(grids.empty())
        return;
    //设立边界岗哨
    grids.insert(grids.begin(), vector<int> (grids[].size(), -));
    grids.push_back(vector<int> (grids[].size(), -));
    for(auto &row : grids) {
        row.insert(row.begin(), -);
        row.push_back(-);
    }
    //从每个0开始进行递归
    for(int i = ; i < grids.size(); i++)
        for(int j = ; j < grids[].size(); j++)
            if(grids[i][j] == )
               dfs(grids, , i, j);
    //除去边界岗哨
    grids.erase(grids.begin());
    grids.pop_back();
    for(auto &row : grids) {
        row.erase(row.begin());
        row.pop_back();
    }
    return;
}