Monday, August 29, 2016

[TopCoder]SRM 645 Div2 Connecting Cars | 书影博客


[TopCoder]SRM 645 Div2 Connecting Cars | 书影博客
Janusz works in roller coaster maintenance. The station of the roller coaster is a long straight segment of railroad tracks. There are some cars on those tracks. The cars are currently not attached to each other, and there may be gaps between some of them. Janusz has to push them all together and connect them into a train.
You are given the tuple (integer)s positions and lengths. For each valid i, there is a car that is lengths[i] meters long and starts positions[i] meters from the beginning of the station. (In other words, the coordinates currently occupied by this car are in the interval from positions[i] to positions[i]+lengths[i].)
Moving a single car one meter in either direction costs Janusz one unit of energy. Compute the smallest total amount of energy sufficient to push all cars together. In the final configuration the cars must be located one after another with no gaps between them.
(Note that there is no restriction on the movement of cars or on the final position of the train. Janusz may push the cars in any order, and he may even push some cars by a non-integer number of meters if he wants to.)
Notes
-    You may assume that the optimal answer is always an integer that fits into a signed 64-bit integer data type.
Constraints
-    lengths and positions will have the same number of elements.
-    lengths will have between 2 and 50 elements, inclusive.
-    Each element of lengths and positions will be between 1 and 10^9, inclusive.
-    The segments occupied by the cars may touch but they will not overlap.

题目大意:

雅努什负责维护过山车。过山车的车站是一段长直铁轨。铁轨上停放着一些车厢,车厢现在互不相连,它们之间可能存在一定的间隔。雅努什需要把这些车厢首尾相连,组合成一列完整的过山车。
给定车厢的位置和距离的元组。对于每一个i,有一节长度为lengths[i]米的车厢停靠在距离起点positions[i]米处。(换言之,这节车厢的起止坐标为positions[i], positions[i]+lengths[i])
朝任意方向将一节车厢移动一米的距离需要花费雅努什一个单位的能量值。计算足以将所有车厢首尾无缝相接的最小总花费。
注意事项和限制条件如上所示。

解题思路:

题目可以转化为求各个线段的中位数,然后计算其余线段到中位数的距离和。
给定一个数列,中位数有这样的性质:所有数与中位数之间的距离之和最小。
The geometric median of a discrete set of sample points in a Euclidean space is the point minimizing the sum of distances to the sample points.
一组样本点的离散集的"几何中位数"是使得样本点到其距离和最小的点。
For the 1-dimensional case, the geometric median coincides with the median. This is because the univariate median also minimizes the sum of distances from the points.
对于一维空间,几何中位数与中位数重合。因为一元中位数使得各点到其距离之和最小。
def minimizeCost(self, positions, lengths): size = len(positions) segments = [( positions[x], positions[x] + lengths[x] ) for x in range(size)] segments = sorted( segments, key = lambda x : x[0] ) ans = 0 mid = size / 2 start = segments[mid][0] for x in range(mid - 1, -1, -1): ans += start - segments[x][1] start -= segments[x][1] - segments[x][0] end = segments[mid][1] for x in range(mid + 1, size): ans += segments[x][0] - end end += segments[x][1] - segments[x][0] return ans
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