Error Curves

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 6241    Accepted Submission(s): 2341

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
 
题目的意思就是给出多个开口向上的一元二次方程,求出极大值的最小值,抛物线肯定是凸函数,直接三分就行了
#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

  1. LA 5009 (HDU 3714) Error Curves (三分)

    Error Curves Time Limit:3000MS    Memory Limit:0KB    64bit IO Format:%lld & %llu SubmitStatusPr ...

  2. hdu 3714 Error Curves(三分)

    Error Curves Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Tot ...

  3. 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& ...

  4. 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 ...

  5. 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 ...

  6. 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> ...

  7. 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 ...

  8. Error Curves HDU - 3714

    Josephina is a clever girl and addicted to Machine Learning recently. She pays much attention to a m ...

  9. 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

  1. [HG]提高组 题解

    首先很容易想到暴力DP 设状态f[i][j]表示当前放了第i个数,最大的数为j的方案数. 然后根据转移推出实际上是在下图走路的方案数 \[ \left( \left( \begin{matrix} x ...

  2. Codeforces Round #201 (Div. 2). E--Number Transformation II(贪心)

    Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Description You ar ...

  3. JSP中解决session超时跳转到登陆页面并跳出iframe框架或局部区域的方法

    当session会话超时,页面请求被重新定位到了登陆界面.但登录界面在iframe中的解决方案:在登录页面中加入下面的js代码: <script type="text/javascri ...

  4. 把execel表数据导入mysql数据库

    今天,是我来公司第二周的第一天. 作为新入职的实习生,目前还没适合我的实质项目工作,今天的学习任务是: 把execel表数据导入到mysql数据库,再练习下java操作JDBC. 先了解下execel ...

  5. LeetCode 47. 全排列 II(Permutations II)

    题目描述 给定一个可包含重复数字的序列,返回所有不重复的全排列. 示例: 输入: [1,1,2] 输出: [ [1,1,2], [1,2,1], [2,1,1] ] 解题思路 类似于LeetCode4 ...

  6. openssl-1.0.1u静态库编译

    不管Windows还是linux都是需要安装好perl环境的 Windows步骤 1.解压openssl-1.0.1u.tar.gz 2.使用Vs2005命令行工具进入解压后的目录 3.执行如下命令 ...

  7. 异步上传&预览图片-不压缩图片

    本例使用ajaxFileUpload异步上传预览图片 <bean id="multipartResolver" class="org.springframework ...

  8. linux如何杀掉进程(kill)

    方法/步骤1: 使用“ps -e|grep mysql”命令,查看mysql程序的对应的pid号.结果如下图:   方法/步骤2: 使用“kill -9 2891”命令,可以结束掉mysqld_saf ...

  9. java里poi操作excel的工具类(兼容各版本)

    转: java里poi操作excel的工具类(兼容各版本) 下面是文件内具体内容,文件下载: import java.io.FileNotFoundException; import java.io. ...

  10. 阶段3 2.Spring_10.Spring中事务控制_11 spring5新特性的介绍

    jdk1.7和1.8的差别 准备好的一个maven工程 反射创建对象10亿次 ,用的时间 替换jdk的版本 选择为1.7 切换了1.7的版本以后呢执行的速度就变的非常的慢 两个版本的对比 响应式编程风 ...