hdu4355(三分法)

Party All the Time

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2253    Accepted Submission(s): 765

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 : X_i,W_i, representing the location and the weight of the i-th spirit. ( |x_i|<=10⁶, 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

140.6 53.9 105.1 78.4 10

 

Sample Output

Case #1: 832

 

Author

Enterpaise@UESTC_Goldfinger

 

Source

2012 Multi-University Training Contest 6

 

Recommend

zhuyuanchen520

 

本题题意为在坐标轴上有若干个带有权值wi的点xi，要求在坐标轴上找一个点x使∑|xi-x|^3*wi最小。

若点比较少，可以暴力枚举逐一比较；但本题点比较多且是连续的，故不能用此法。可以联想连续函数的最值问题，本题的函数正好是个凹函数，可以用一定的方法求极值；这个方法就是三分法。

枚举+三分法

#include<iostream>#include<cstdio>using namespace std;const double eps=1e-10;const int MAXN=50000+100;double x[MAXN],w[MAXN];int n;inline double Max(double aa,double bb){return aa<bb?bb:aa;}inline double Three_find(double xx){double ans=0;for(int i=0;i<n;i++){if(x[i]>xx)ans+=(x[i]-xx)*(x[i]-xx)*(x[i]-xx)*w[i];else ans+=(xx-x[i])*(xx-x[i])*(xx-x[i])*w[i];}return ans;}int main(){int cas,i,tag=1;cin>>cas;while(cas--){scanf("%d",&n);for(i=0;i<n;i++)scanf("%lf%lf",&x[i],&w[i]);double tmp,mid,midmid,low=x[0],high=x[n-1];while(low+eps<=high){mid=(low+high)/2;midmid=(mid+high)/2;if(Three_find(mid)+eps>=Three_find(midmid))low=mid;else high=midmid;}printf("Case #%d: %d\n",tag++,(int)(Three_find(high)+0.5));}return 0;}