## The 2010 ACM-ICPC Asia Chengdu Regional Contest Error Curves 三分法求凸函数极值

Error Curves

Time Limit: 2 Seconds      Memory Limit: 65536 KB

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 minimal which related to multiple quadric functions.

The new function F(x) is defined as follow:

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.

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

#### Sample Output

```0.0000
0.5000
```
`简明题意：求一堆开口向上的二次函数在[0,1000]范围上函数值最大值的最小值。`
`二次函数的子集仍然为凸函数，所以可以用三分法求极值。精度实在很蛋疼，这题要求值域精确到1e-4，但是定义域没说精确到多少，结果死wa，卡到1e-10终于过了。。`
`贴代码`
``` 1# include <cstdio> 2# include <cmath> 3using namespace std; 4int n; 5int data[10001][3]; 6# define max(a,b) ((a)>(b)?(a):(b)) 7double cal(double mid) 8{ 9   double res=-1e26;10   for(int i=0;i<n;i++)11     res=max(res,data[i][0]*mid*mid+data[i][1]*mid+data[i][2]);12   return res;13}14int main()15{16    int test;17    scanf("%d",&test);18    while(test--)19    {20       scanf("%d",&n);21       for(int i=0;i<n;i++)22         scanf("%d%d%d",&data[i][0],&data[i][1],&data[i][2]);23       double s=0.0,e=1000.0;24       double last=s;25       while(fabs(e-s)>1e-10)26       {27       28         double m1=(s+e)/2.0,m2=(m1+e)/2.0;29         if(cal(m1)<cal(m2))30           e=m2;31         else 32           s=m1;33       }34       printf("%.4lf\n",cal(e));35    }36    return 0;37}3839
```

posted on 2010-11-16 00:50 yzhw 阅读(757) 评论(0)  编辑 收藏 引用 所属分类: numberic

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