[Locked] Shortest Distance from All Buildings

Shortest Distance from All Buildings

You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values 0, 1 or 2, where:

• Each 0 marks an empty land which you can pass by freely.

• Each 1 marks a building which you cannot pass through.

• Each 2 marks an obstacle which you cannot pass through.

For example, given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2):

1 - 0 - 2 - 0 - 1
|   |   |   |   |
0 - 0 - 0 - 0 - 0
|   |   |   |   |
0 - 0 - 1 - 0 - 0

The point (1,2) is an ideal empty land to build a house, as the total travel distance of 3+3+1=7 is minimal. So return 7.

Note:
There will be at least one building. If it is not possible to build such house according to the above rules, return -1.

分析：

　　这题如果不考虑obstacle的存在的话，与另一道leetcode题目一样，分成x轴和y轴，根据值为1的点的坐标，直接算出最小距离；然而多了一个obstacle，这题又更像是gates and walls这题了，不同的是，对于每个为0的点，各个建筑物到它的最近的距离都要计算出来并累加，而不是算最近距离的最小值。K为building个数，M、N分别为长和宽，时间复杂度为O(KMN)，空间复杂度为O(MN)

代码：

//计算每个岛到坐标为(i, j)的building的最短距离
void dfs(int i, int j, int cur, vector<vector<int> > &dist, vector<vector<int> > &grids) {
    if(cur > dist[i][j])
        return;
    dist[i][j] = cur++;
    if(grids[i][j + ] == )
        dfs(i, j + , cur, dist, grids);
    if(grids[i][j - ] == )
       dfs(i, j - , cur, dist, grids);
    if(grids[i + ][j] == )
        dfs(i + , j, cur, dist, grids);
    if(grids[i - ][j] == )
        dfs(i - , j, cur, dist, grids);
    return;
}
//迭代计算总距离矩阵，并重置距离矩阵
void postProcess(vector<vector<int> > &dist, vector<vector<int> > &totaldist, vector<vector<int> > &grids) {
    for(int i = ; i < grids.size(); i++)
        for(int j = ; j < grids[].size(); j++) {
            if(grids[i][j] == )
                totaldist[i][j] += dist[i][j];
            dist[i][j] = INT_MAX;
        }
    return;
}
//主要功能函数
int shortestDist(vector<vector<int> > &grids) {
    //设立岗哨
    grids.insert(grids.begin(), vector<int> (grids[].size(), ));
    grids.push_back(vector<int> (grids[].size(), ));
    for(auto &v : grids) {
        v.insert(v.begin(), );
        v.push_back();
    }
    //声明并初始化距离矩阵
    vector<vector<int> > dist(grids);
    for(auto &v : dist)
        for(int &i : v)
            i = INT_MAX;
    //声明并初始化总距离矩阵
    vector<vector<int> > totaldist(grids.size(), vector<int> (grids[].size(), ));
    //对每个building进行扩展，计算其到周边岛屿的最小距离
    for(int i = ; i < grids.size(); i++) {
        for(int j = ; j < grids[].size(); j++) {
            if(grids[i][j] == ) {
                dfs(i, j, , dist, grids);
                postProcess(dist, totaldist, grids);
            }
        }
    }
    //在总距离矩阵中找到最小距离
    int sd = INT_MAX;
    for(int i = ; i < grids.size(); i++)
        for(int j = ; j < grids[].size(); j++)
            if(grids[i][j] == )
                sd = min(sd, totaldist[i][j]);
    //去除岗哨，还原输入矩阵
    grids.pop_back();
    grids.erase(grids.begin());
    for(auto &v : grids) {
        v.pop_back();
        v.erase(v.begin());
    }
    return sd;
}