hdu 4335 Party All the Time(三分求极值，3级)_in the dark forest, there is a fairy kingdom where-优快云博客

本文介绍了一个有趣的问题：在一个精灵王国中，如何找到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。文章提供了一种通过三分法搜索最优解的方法，并附带了完整的代码实现。

Party All the Time

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2161 Accepted Submission(s): 732

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

Sample Output

  
   Case #1: 832

Author

Enterpaise@UESTC_Goldfinger

Source

2012 Multi-University Training Contest 6

Recommend

zhuyuanchen520

思路：三分球极值。为啥可以呢，不知道。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int mm=5e4+9;
const double mx=1e-6;
const double oo=1e40;
class node
{
public:
  double x,y;
}f[mm];
int n;
double fabs(double x)
{
  if(x<0)return -x;
  return x;
}
double get(double mid)
{
  double ret=0.0,z;
  for(int i=0;i<n;++i)
  { z=fabs(mid-f[i].x);
    ret+=z*z*z*f[i].y;
  }
  return ret;
}
int main()
{
  int cas;
  while(~scanf("%d",&cas))
  {
    for(int ca=1;ca<=cas;++ca)
    {
      printf("Case #%d: ",ca);
      scanf("%d",&n);
      double l=oo,r=-oo;
      for(int i=0;i<n;++i)
      {
        scanf("%lf%lf",&f[i].x,&f[i].y);
        l=min(l,f[i].x);r=max(r,f[i].x);
      }
      double ll,rr;
      while(r-l>mx)
      {
        ll=(l+l+r)/3.0;
        rr=(l+r+r)/3.0;
        if(get(ll)>get(rr))
        {
          l=ll;
        }
        else
        {
          r=rr;
        }
      }
      //cout<<get(ll)<<endl;
      printf("%.0lf\n",get(l));
    }
  }
  return 0;
}