1476 Error Curves
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 nds 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 denition
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 dened 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 rst 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 (jbj 5000), c (jcj 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.
Universidad de Valladolid OJ: 1476 { Error Curves 2/2
Sample Input
2
1
2 0 0
2
2 0 0
2 -4 2
Sample Output
0.0000
0.5000
题意：
已知n条二次曲线Si(x) = ai * x ^ 2 + bi * x + ci(ai >= 0)，定义F(x) = max{Si(x) | i = 1,2,…,n}。求F(x)在[0,1000]上的最小值。
分析：
不论有多少条曲线，F(x)总是一个下凸函数，于是我们考虑使用三分法。在区间[L,R]上选取两个三分点m1和m2，比较F(m1)和F(m2)的大小，如果F(m1) < F(m2)，说明解在
[L,m2]上否则在[m1,R]上。
注意三分搜索的时候需要确定迭代次数，一般来说，迭代次数是通过实验确定的，如果答案不正确则加大迭代次数，如果超时则减少。如果都不行，则把终止条件改写成解区间长度小于某阈值。
1 #include <cstdio>
2 #include <algorithm>
3 using namespace std;
4 const int maxn = 10000;
5 int n,a[maxn + 1],b[maxn + 1],c[maxn + 1];
6 double F(double x){
7 double ans = a[0] * x * x + b[0] * x + c[0];
8 for(int i = 1 ; i < n ; i++)
9 ans = max(ans,a[i] * x * x + b[i] * x + c[i]);
10 return ans;
11 }
12 int main(){
13 int T; scanf("%d",&T);
14 while(T--){
15 scanf("%d",&n);
16 for(int i = 0 ; i < n ; i++) scanf("%d%d%d",&a[i],&b[i],&c[i]);
17 double L = 0.0,R = 1000.0;
18 // 三分
19 for(int i = 0 ; i < 100 ; i++){
20 double m1 = L + (R - L) / 3;
21 double m2 = R - (R - L) / 3;
22 if(F(m1) < F(m2)) R = m2; else L = m1;
23 }
24 printf("%.4lf\n",F(L));
25 }
26 return 0;
27 }