Party All the Time
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5260 Accepted Submission(s): 1624
Problem Description
In the Dark forest, there is a Fairy kingdom where all the spirits will go together and Celebrate the harvest every year. But there is one thing you may not know that they hate walking so much that they would prefer to stay at home if they need to walk a long
way.According to our observation,a spirit weighing W will increase its unhappyness for S3*W units if it walks a distance of S kilometers.
Now give you every spirit's weight and location,find the best place to celebrate the harvest which make the sum of unhappyness of every spirit the least.
Now give you every spirit's weight and location,find the best place to celebrate the harvest which make the sum of unhappyness of every spirit the least.
Input
The first line of the input is the number T(T<=20), which is the number of cases followed. The first line of each case consists of one integer N(1<=N<=50000), indicating the number of spirits. Then comes N lines in the order that x[i]<=x[i+1] for
all i(1<=i<N). The i-th line contains two real number : Xi,Wi, representing the location and the weight of the i-th spirit. ( |xi|<=106, 0<wi<15 )
Output
For each test case, please output a line which is "Case #X: Y", X means the number of the test case and Y means the minimum sum of unhappyness which is rounded to the nearest integer.
Sample Input
1 4 0.6 5 3.9 10 5.1 7 8.4 10
Sample Output
Case #1: 832
对于菜鸟们来说,根本就不存在水题和难题的差别,反正都做得一脸蒙蔽。。。
这道题被生硬地做成了一道卡掉本菜m+1次的难题,并且弱鸡一直没找到超时的原因,后来发现有人说千万不要用pow()来做幂方,然后我就看到自己的程序里基本上都是用现成的数学函数,索性把所有的数学函数都手写出来。
代码如下:
本文介绍了一个有趣的问题:在一个精灵王国中,如何选择一个位置来庆祝丰收,使得所有精灵因行走而产生的不快乐总和最小。文章提供了一个具体的编程实现案例。
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