HDU-3714 Error Curves(凸函数求极值)
Error Curves
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 6241 Accepted Submission(s): 2341
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?
1
2 0 0
2
2 0 0
2 -4 2
0.5000
#pragma GCC diagnostic error "-std=c++11"
#include<bits/stdc++.h>
#define _ ios_base::sync_whit_stdio(0);cin.tie(0); using namespace std;
const int N = + ;
const int INF = (<<);
const double eps = 1e-; double a[N], b[N], c[N];
int n; double fun(double x){
double res = - INF;
for(int i = ; i < n; i++)
res = max(res, a[i] * x * x + b[i] * x + c[i]);
return res;
} double ternary_search(double L, double R){
double mid1, mid2;
while(R - L > eps){
mid1 = ( * L + R) / ;
mid2 = (L + * R) / ;
if(fun(mid1) >= fun(mid2)) L = mid1;
else R = mid2;
}
return (L + R) * 0.5;
} int main(){
int T;
scanf("%d", &T);
while(T--){
scanf("%d", &n);
for(int i = ; i < n; i++){
scanf("%lf %lf %lf", &a[i], &b[i], &c[i]);
}
double x = ternary_search(, );
printf("%.4f\n", fun(x));
}
}
最新文章
- python 汇总
- (旧)子数涵数&#183;Flash——路径补间
- Oracle数据库之PL/SQL基础
- 1、java基础回顾与加强
- Android开发之发送邮件功能的实现(源代码分享)
- 运用JavaScript构建你的第一个Metro式应用程序(onWindows 8)(三)
- 关于 WP上应用调试时报错“指定的通信资源(端口)”已由另一个应用程序使用 问题
- View如何设置16进制颜色值
- iOS子线程更新UI的两种方法
- vs项目和msql不兼容解决方案
- Echarts数据可视化地理坐标系geo,开发全解+完美注释
- JMeter接口测试系列-关联参数
- 阿里云SLB出现502 Bad Gateway 错误排查解决方法
- 关于如何在Visual Studio上仿真调试安卓的U3D应用
- Python机器学习笔记 使用scikit-learn工具进行PCA降维
- 线性表->;应用->;一元多项式
- golang 学习笔记 ---内存分配与管理
- Android开发中adb命令的常用方法
- openshift rhc
- Keras实践:模型可视化