[Locked] Best Meeting Point

Best Meeting Point

A group of two or more people wants to meet and minimize the total travel distance. You are given a 2D grid of values 0 or 1, where each 1 marks the home of someone in the group. The distance is calculated using Manhattan Distance, where distance(p1, p2) = |p2.x - p1.x| + |p2.y - p1.y|.

For example, given three people living at (0,0), (0,4), and(2,2):

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

The point (0,2) is an ideal meeting point, as the total travel distance of 2+2+2=6 is minimal. So return 6.

分析：

　　第一反应就是求二维平面上的距离最优值；由于是曼哈顿距离，所以x维和y维可以分开求最小，最终结果也会最小，这样转化成了一维轴上绝对值之和最小，中学学过嘛，画图可知。

解法：

　　找到x轴上所有为1的点，然后从外到内依次两两index求差，这些差的总和为x轴上的最小值；y轴上也做同样的操作得到y轴最小值。两轴上的最小值之和为最小曼哈顿距离。时间复杂度为m*n，空间复杂度为1的个数。

代码：

int minDist(vector<vector<int> > v) {
    deque<int> inum, jnum;
    int dist = 0;
    for(int i = 0; i < v.size(); i++) {
        for(int j = 0; j < v[0].size(); j++) {
            if(v[i][j])
                inum.push_back(i);
        }
    }
    while(inum.size() >= 2) {
        dist += inum.back() - inum.front();
        inum.pop_front();
        inum.pop_back();
    }
    for(int j = 0; j < v[0].size(); j++) {
        for(int i = 0; i < v.size(); i++) {
            if(v[i][j])
                jnum.push_back(j);
        }
    }
    while(jnum.size() >= 2) {
        dist += jnum.back() - jnum.front();
        jnum.pop_front();
        jnum.pop_back();
    }
    return dist;
}