hdu 4355 Party All the Time （2012 Multi-University Training Contest 6 ） 三分搜索

Party All the Time

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 926    Accepted Submission(s): 341

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 S ³*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 : X _i,W _i, representing the location and the weight of the i-th spirit. ( |x _i|<=10 ⁶, 0<w _i<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

 

 

Author

Enterpaise@UESTC_Goldfinger

 

 

Source

2012 Multi-University Training Contest 6

 

 

 

因为 s^3 *w     s加了 绝对值 所以 是一个 凸性 函数 可以用 三分做

三分搜索

 1 #include<stdio.h>

 2 #include<iostream>

 3 #include<algorithm>

 4 #include<cstring>

 5 #include<cmath>

 6 #include<queue>

 7 #include<set>

 8 #include<map>

 9 #define Min(a,b)  a>b?b:a

10 #define Max(a,b)  a>b?a:b

11 #define CL(a,num)  memset(a,num,sizeof(a));

12 #define inf 9999999

13 #define maxn 50010

14 #define eps  1e-5

15 #define ll long long

16 using namespace std;

17 double a[maxn],w[maxn];

18 int  n;

19 double f(double pos)

20 {

21     int i;

22     double ans  = 0;

23     for( i =  0 ; i < n ; ++i)

24     {

25         double  k = fabs(a[i] - pos);

26 

27         ans += k*k*k*w[i];

28     }

29      return ans ;

30 }

31 double  solve(double l,double r)

32 {

33 

34         double mid,midmid;

35         while(r - l > eps)

36         {

37               mid = (l + r)/2.0;

38              midmid = (r + mid)/2.0;

39 

40 

41             if(f(mid) <= f(midmid))// f 就算函数值

42                 r = midmid;

43             else  l = mid;

44         }

45     return f(l);

46 }

47 int main()

48 {

49     int t,i;

50     scanf("%d",&t);

51     int cas = 0;

52     while(t--)

53     {

54         scanf("%d",&n);

55         for(i = 0 ;i < n;++i )

56         {

57             scanf("%lf %lf",&a[i],&w[i]);

58         }

59         double ans = solve(a[0],a[n - 1]);

60         printf("Case #%d: %.0lf\n",++cas,ans);

61     }

62 }