best meeting point

最新推荐文章于 2020-05-03 15:39:48 发布

转载最新推荐文章于 2020-05-03 15:39:48 发布 · 764 阅读

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本文讨论了如何在无限网格上为N个人找到一个最优的会议地点，使得所有人到达该地点的总行程时间最小。通过将点映射到x轴和y轴，并分别找到中位数位置，最终确定会议地点。

There is an infinite integer grid at which N people have their houses on. They decide to unite at a common meeting place, which is someone's house. From any given cell, all 8 adjacent cells are reachable in 1 unit of time. eg: (x,y) can be reached from (x-1,y+1) in a single unit of time. Find a common meeting place which minimizes the sum of the travel times of all the persons.

这个题目可以参考之前做的加油站问题，只是一维简化版本

有n个加油站，选择中间的那个加油站，作为总加油站，其他加油站到它的距离最短

可以先把所有的点映射到x轴，找到一个点位于所有点的中位数位置

再把所有的点映射到Y轴，找到一个点位于所有点的中位数位置

这两条线的交点就是meeting place

struct POS 
{
        int x;
        int y;
};

bool CompX(POS a, POS b) { return a.x < b.x; }
bool CompY(POS a, POS b) { return a.y < b.y; }

POS FindMeetingPlace(POS a[], int n)
{
        assert(a && n>0);

        POS pos;
        sort(a, a+n, CompX);
        pos.x = a[n/2].x;

        sort(a, a+n, CompY);
        pos.y = a[n/2].y;

        return pos;
}