HDOJ4355-Party All the Time(三分)

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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.

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

Code:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 static const int MAXN = 5e4 + 10;
 4 static const double OO = 1e9 + 0.0;
 5 static const double EPS = 1e-5;
 6 typedef long long LL;
 7 int t , n;
 8 double point[MAXN];
 9 double weight[MAXN];
10 double Cal(double p)
11 {
12     double ret = 0;
13     for(int i = 1 ; i <= n ; ++i)
14     {
15         double s = fabs(point[i] - p);
16         ret += s * s * s * weight[i];
17     }
18     return ret;
19 }
20 double Search(double l , double r)
21 {
22     while(r - l > EPS)
23     {
24         double ll = (2 * l + r) / 3.0;
25         double rr = (2 * r + l) / 3.0;
26         double tp1 = Cal(ll);
27         double tp2 = Cal(rr);
28         if(tp1 > tp2)//abandon left
29         {
30             l = ll;
31         }
32         else
33             r = rr;
34     }
35     return l;
36 }
37 int main()
38 {
39     scanf("%d" , &t);
40     for(int c = 1 ; c <= t ; ++c)
41     {
42         double l = OO , r = -OO;
43         scanf("%d" , &n);
44         for(int i = 1 ; i <= n ; ++i)
45         {
46             scanf("%lf%lf" , &point[i] , &weight[i]);
47             l = min(point[i] , l);
48             r = max(point[i] , r);
49         }
50         double ans = Search(l , r);
51         printf("Case #%d: %.0lf\n" , c , Cal(ans));
52     }
53 }

View Code

HDOJ4355-Party All the Time(三分)

标签：onclick   head   ini   min   splay   idt   ast   oca   printf   

原文地址：http://www.cnblogs.com/jianglingxin/p/6814662.html