三分法求极值

最新推荐文章于 2022-06-03 21:47:31 发布

Balloons2012

最新推荐文章于 2022-06-03 21:47:31 发布

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分类专栏： BUAA_2012_Summer 文章标签： distance integer input each output go

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该博客介绍了如何运用三分法解决凹凸函数的极值问题，提供了输入、输出的示例。

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三分求解凹凸函数的极值问题。

Party All the Time

Time Limit: 2000ms

Memory Limit: 32768KB

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

#include<stdio.h>
#include<math.h>
const double eps=1e-4;
double x[50001],w[50001]; int N;
double tri(double x)
{
    return x*x*x;
}
double cal(double h)
{
    double res=0.00000;int i;
    for(i=1;i<=N;i++)
    {
        res+=tri(fabs(x[i]-h))*w[i];
    }
    return res;
}
int main()
{
    int T,i; double left,right,mid,midmid; int Ca=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&N);
        for(i=1;i<=N;i++)
            scanf("%lf%lf",x+i,w+i);
        left=-1e6; right=1e6;
        while(fabs(left-right)>eps)
        {
            mid=(left+right)/2;
            midmid=(mid+right)/2;
            if(cal(mid)<cal(midmid))
                right=midmid;
            else
                left=mid;
        }
        printf("Case #%d: %.0lf\n",Ca,cal(left));
        Ca++;
    }
    return 0;
}