HDU/HDOJ 3714 Error Curves 2010年成都赛区

 

Error Curves

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 300    Accepted Submission(s): 113

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

   
   
   
   

    
    
    
    2
1
2 0 0
2
2 0 0
2 -4 2
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0.0000
0.5000
   
   
   
   

 

Author

LIN, Yue

 

Source

The 2010 ACM-ICPC Asia Chengdu Regional Contest

 

Recommend

zhouzeyong

 

我的思路：

可以简单的画一下图发现，几个抛物线取最大的图像只能是一个单峰的函数

而且仅仅只有一个最小值。

画下图能够充分的理解这一点。

然后知道这个函数是单峰的就很好了，我们可以直接三分。把这个最小值给计算出来。

做的时候注意，此题每个函数的最小值可能会小于0。

所以定义的最大值在初始化的时候不能简单的认为就是0了，而要搞成负无穷这里//我WA到si才反应过来的

 

我的代码：

#include<stdio.h>
#include<algorithm>
#include<math.h>

struct node
{
    double a;
    double b;
    double c;
};
node fun[10005];

double max(double a,double b)
{
    if(a>b)
        return a;
    else
        return b;
}

double f(int i,double x)
{
    return fun[i].a*x*x+fun[i].b*x+fun[i].c;
}

double cal(double x,int n)
{
    int i;
    double h=-99999999;
    for(i=1;i<=n;i++)
        h=max(h,f(i,x));
    return h;
}

void solve(int n)
{
	int i;
    double left,right,mid1,mid2,h1,h2;
    left=0,right=1000.0;
    for(i=1;i<=100;i++)
    {
        mid1=(2*left+right)/3;
        mid2=(left+2*right)/3;
        h1=cal(mid1,n);
        h2=cal(mid2,n);
        if(h1>h2)
        {
            left=mid1;
        }
        else
        {
            right=mid2;
        }
    }
    double ans=cal(left,n);
    printf("%.4lf\n",ans);
}

int main()
{
    int i,n,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%lf%lf%lf",&fun[i].a,&fun[i].b,&fun[i].c);
        solve(n);
    }
    return 0;
}