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));
}
}
HDU-3714 Error Curves(凸函数求极值)的更多相关文章
- LA 5009 (HDU 3714) Error Curves (三分)
Error Curves Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu SubmitStatusPr ...
- hdu 3714 Error Curves(三分)
Error Curves Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Tot ...
- HDU 3714 Error Curves
Error Curves 思路:这个题的思路和上一个题的思路一样,但是这个题目卡精度,要在计算时,卡到1e-9. #include<cstdio> #include<cstring& ...
- hdu 3714 Error Curves(三分)
http://acm.hdu.edu.cn/showproblem.php?pid=3714 [题意]: 题目意思看了很久很久,简单地说就是给你n个二次函数,定义域为[0,1000], 求x在定义域中 ...
- nyoj 1029/hdu 3714 Error Curves 三分
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3714 懂了三分思想和F(x)函数的单调性质,这题也就是水题了 #include "stdio ...
- 三分 HDOJ 3714 Error Curves
题目传送门 /* 三分:凹(凸)函数求极值 */ #include <cstdio> #include <algorithm> #include <cstring> ...
- HDU-4717 The Moving Points(凸函数求极值)
The Moving Points Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others ...
- Error Curves HDU - 3714
Josephina is a clever girl and addicted to Machine Learning recently. She pays much attention to a m ...
- HDU 3714/UVA1476 Error Curves
Error Curves Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Tota ...
随机推荐
- [VIJOS2053][SDOI2019]世界地图:最小生成树+虚树
分析 可以发现第一列和最后一列永远不会被删除,于是我们可以想到维护前后缀最小生成树,但是直接维护的话显然时间空间两爆炸.(通过上网找题解)可以发现我们关心的只是最左边和最右边两列,而不关心内部的连边情 ...
- JS框架_(JQuery.js)Tooltip弹出式按钮插件
百度云盘 传送门 密码:7eh5 弹出式按钮效果 <!DOCTYPE html> <html > <head> <meta charset="UTF ...
- 分布式-信息方式-ActiveMQ静态网络连接多线程的consumer(消费者)访问集群
操作如下: 1:把整个conf文件夹复制一份,比如叫做conf22:修改里面的 activemq.xml文件(1)里面的 brokerName不能跟原来的重复(2)数据存放的文件名称不能重复,比如:& ...
- 191112Django项目常用配置
创建项目 >django-admin startproject project01 创建应用 >python manage.py startapp app01 settings.py 配置 ...
- note:memcache 調試遇到的一些問題
1. gdb調試時提示:<value optimized out> ,原因是memcache的編譯选项做了优化,修改configure中CFLAGS中的-O3为-O0,可以解决gdb调试过 ...
- 如何用ORM自定义排序
ORM连接的数据库不同,方法也不同 postgresql: a.extra(select={"new_order":"position(status in '2,3,4, ...
- webshell 常见 Bypass waf 技巧总结
在渗透学习的过程中,总会遇到各种情况,例如 php 大马被 waf 拦截的时候,那么如何制作免杀 php webshell 呢,接下来就由我带各位小伙伴们一起踏上大马免杀之路,不喜勿喷. 一篇好的文章 ...
- Java反序列化漏洞从入门到深入(转载)
前言 学习本系列文章需要的Java基础: 了解Java基础语法及结构(菜鸟教程) 了解Java面向对象编程思想(快速理解请上知乎读故事,深入钻研建议买本<疯狂Java讲义>另外有一个刘意老 ...
- chrome浏览器爬虫WebDriverException解决采用python + selenium + chrome + headless模式
WebDriverException: Message: unknown error: Chrome failed to start: crashed 第一种:如果出现下面情况: chrome浏览器有 ...
- JavaEE-实验四 HTML与JSP基础编程
1.使用HTML的表单以及表格标签,完成以下的注册界面(验证码不做) html代码(css写于其中) <!DOCTYPE html> <html> <head> & ...