HDU-4355-三分

Party All the Time

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5791    Accepted Submission(s): 1841

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

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

 　　　　求Y=Σw=Σ(wi*|xi-X|³) 使得Y达到最小的那个Y_min.

　　　 注意到这个函数不是单调的，所以二分不可行，要用三分来做，mid=(l+r)/2,mid2=(mid+r)/2,然后f(x)较大的那个点作为新的l/r边界，慢慢缩小范围。

 　　　　

#include<bits/stdc++.h>
using namespace std;
#define sf(a) scanf("%d",&a) 
double x[50010],w[50010];
int N;
double cal(double X)
{
    double r=0;
    for(int i=1;i<=N;++i){
        double t=fabs(X-x[i]);
        r+=w[i]*t*t*t;
    }
    return r;
}
int main()
{
    int t,i,j,k;
    int cas=0;
    cin>>t;
    while(t--){
        cin>>N;
        for(i=1;i<=N;++i)
            scanf("%lf%lf",x+i,w+i);
        double l=-1000000,r=1000000;
        for(i=1;i<=200;++i){
            double mid=(l+r)/2;
            double mid2=(mid+r)/2;
            if(cal(mid)<cal(mid2)){
                r=mid2;
            }
            else{
                l=mid;
            }
        }
        printf("Case #%d: %.0f
",++cas,cal(l));
    }
    return 0;
}