今天看了几个中位数的题,感觉的确应该好好总结下。
带权中位数的概念:http://baike.baidu.com/view/1209446.htm
中位数要解决的问题:一维平面就是一条直线上汇集到一点的距离的最小值,求这一点。
二维的就是平面上所有点汇集到一点的最小距离,同求这一点。
带权中位数就是这个距离变成了权值。
其实这点满足这个最重要的公式就OK了:
sigma( i=1 to k-1) Wi <= W/2
sigma( i=k+1 to n) Wi <= W/2
但正是这个公式让人头痛,因为公式后面由着大量的证明。
我通俗的在这里说下,有两群人,第一群人共有A个,第二群人共有B个,要求把这两群人合并,使所有人走的最少,求这个步数。如果A>B的话,应该让B走完所有的路径去A,反之一样(可以举例子试试)。这里就是有两群人归纳之N堆时的情况,找到这个临界值。
下面列举一个变种的题。
士兵站队
问题:
在一个划分成网格的操场上,n个士兵散乱地站在网格点上。网格点由整数坐标(x,y)表示。士兵们可以沿网格边上、下、左、右移动一步,但在同一时刻任一网格点上只能有一名士兵。按照军官的命令,士兵们要整齐地列成一个水平队列,即排列成(x,y),(x+1,y),…,(x+n-1,y)。如何选择x和y的值才能使士兵们以最少的总移动步数排成一列。
算法:
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代码源于网络。
这里有个nth_element(a,n,b)的函数,第一次见到,发现又是个很好的东东。
STL中的nth_element()方法的使用 通过调用nth_element(start, start+n, end) 方法可以使第n大元素处于第n位置(从0开始,其位置是下标为 n的元素),并且比这个元素小的元素都排在这个元素之前,比这个元素大的元素都排在这个元素之后,但不能保证他们是有序的。
通俗的说就是去到第几大的元素,而且能够分半放。