HDU 3714 Error Curves

Error Curves

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3172    Accepted Submission(s): 1209

Problem Description

Josephina is a clever girl and addicted to Machine Learning recently. She
pays much attention to a method called Linear Discriminant Analysis, which
has many interesting properties.
In order to test the algorithm's efficiency, she collects many datasets.
What's more, each data is divided into two parts: training data and test
data. She gets the parameters of the model on training data and test the
model on test data. To her surprise, she finds each dataset's test error curve is just a parabolic curve. A parabolic curve corresponds to a quadratic function. In mathematics, a quadratic function is a polynomial function of the form f(x) = ax2 + bx + c. The quadratic will degrade to linear function if a = 0.



It's very easy to calculate the minimal error if there is only one test error curve. However, there are several datasets, which means Josephina will obtain many parabolic curves. Josephina wants to get the tuned parameters that make the best performance on all datasets. So she should take all error curves into account, i.e., she has to deal with many quadric functions and make a new error definition to represent the total error. Now, she focuses on the following new function's minimum which related to multiple quadric functions. The new function F(x) is defined as follows: F(x) = max(Si(x)), i = 1...n. The domain of x is [0, 1000]. Si(x) is a quadric function. Josephina wonders the minimum of F(x). Unfortunately, it's too hard for her to solve this problem. As a super programmer, can you help her?

 

Input

The input contains multiple test cases. The first line is the number of cases T (T < 100). Each case begins with a number n (n ≤ 10000). Following n lines, each line contains three integers a (0 ≤ a ≤ 100), b (|b| ≤ 5000), c (|c| ≤ 5000), which mean the corresponding coefficients of a quadratic function.

 

Output

For each test case, output the answer in a line. Round to 4 digits after the decimal point.

 

Sample Input

212 0 022 0 02 -4 2

 

Sample Output

0.00000.5000

 

Author

LIN, Yue

 

Source

2010 Asia Chengdu Regional Contest 

读题有点难，题目的意思是就是给你n个二次函数，定义域为[0，1000]，求x在定义域中每个x所在的n个函数的最大值的最小值。对于一个x，你必须求出它在多个函数中的最大值，然后你可以改变x的值，然后在这些最大值里面找到一个最小值。

#include<stdio.h>#define eps 1e-15double a[10005],b[10005],c[10005];int n;double max(double a,double b)//max函数{    if(a>b)        return a;    else        return b;}double min(double a,double b)//min函数{    if(a>b)        return b;    else        return a;}double ssum(double x){    double ans=a[0]*x*x+b[0]*x+c[0];    for(int i = 1; i < n; i++)    {        ans = max(ans,a[i]*x*x+b[i]*x+c[i]);//对于一个x求它在若干个二次方程中的最大值    }    return ans;}double lss()//三分法{    double left,right,mm,m,ssmm,ssm;    left = 0;    right = 1000;    while(left+eps<right)    {        m = (left+right)/2;        mm = (m+right)/2;        ssm = ssum(m);        ssmm = ssum(mm);        if(ssm > ssmm)            left=m;        else            right=mm;    }    return ssum(right);}int main(){    int N,i;    scanf("%d",&N);    while(N--)    {        scanf("%d",&n);        for(i = 0; i < n; i++)        {            scanf("%lf%lf%lf",&a[i],&b[i],&c[i]);        }        printf("%.4lf\n",lss());    }    return 0;}