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
 
题意:给出多个二次函数的A,B,C系数,求出0到1000范围内x对应的所有二次函数的最大值的最小值。
 
解析:三分即可(貌似题目保证了x对应的函数类似二次函数)
 
代码

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<utility>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#include<iterator>
#include<stack>
using namespace std;
const int INF=1e9+;
const double eps=1e-;
const int maxn=;
int N,A[maxn],B[maxn],C[maxn];
double Cal(double mid)
{
double ret=-;
for(int i=;i<N;i++)
{
ret=max(ret,A[i]*mid*mid+B[i]*mid+C[i]);
}
return ret;
}
double solve()
{
double x=,y=,mid,midmid;
int cnt=;
while(cnt--)
{
mid=(x+y)/;
midmid=(mid+y)/;
if(Cal(mid)<Cal(midmid)) y=midmid;
else x=mid;
}
return Cal(x);
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&N);
for(int i=;i<N;i++) scanf("%d%d%d",&A[i],&B[i],&C[i]);
printf("%.4f\n",solve());
}
return ;
}


Hdu3714-Error Curves(三分)的更多相关文章

  1. UVA - 1476 Error Curves 三分

                                           Error Curves Josephina is a clever girl and addicted to Machi ...

  2. 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)Tota ...

  3. hdu3714 Error Curves

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

  4. 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 ...

  5. 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 ...

  6. UVALive 5009 Error Curves 三分

    //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include& ...

  7. HDU3714 Error Curves (单峰函数)

    大意: 给你n个二次函数Si(x),F(x) = max{Si(x)} 求F(x)在[0,1000]上的最小值. S(x)=ax^2+bx+c       (0<=a<=100, |b|, ...

  8. 【单峰函数,三分搜索算法(Ternary_Search)】UVa 1476 - Error Curves

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

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

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

  10. hdu 3714 Error Curves(三分)

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

随机推荐

  1. hdu2089:不要62(基础数位dp)

    题意:规定一个合法的号码不能含有4或者是连续的62 给定区间[n,m] 问此区间内合法的号码的个数 分析:数位dp dp[i][j]代表 最高位为 j 的 i 位数有多少个合法的 然后按题目规则进行转 ...

  2. iPhone 5,6,6 plus 尺寸

  3. 关于C#中的弱引用

    本文前部分来自:http://www.cnblogs.com/mokey/archive/2011/11/24/2261605.html 分割线后为作者补充部分. 一:什么是弱引用 了解弱引用之前,先 ...

  4. install samba on crux without net.

    1. 解压文件 tar -xvfz samba-.tar.gz 2. 进入目录 cd samba- 3. 配置 ./configure 4. 编译 make 5. 安装 make install

  5. GridControl 设置焦点单元格

    gdvNew.Focus(); //GridControl 控件获取焦点 gdvNew.FocusedRowHandle = _smtReport.Count - 1; //设置焦点行 gdvNew. ...

  6. .net中的特性

    本文来之:http://hi.baidu.com/sanlng/item/afa31eed0a383e0e570f1d3e 在一般的应用中,特性(Attribute,以称为属性)好像被使用的不是很多. ...

  7. Lanucherr 默认显示第几屏

    Launcher.java static final int SCREEN_COUNT = 5;static final int DEFAULT_SCREEN = 2;//第一页是从0开始计数,这里是 ...

  8. oracle 数据库 分割字符串返回结果集函数

    CREATE OR REPLACE FUNCTION "UFN_SPLIT" (      p_list varchar2,      p_sep varchar2 := ',' ...

  9. hdu Intelligent IME

    算法:字典树 题意:手机9键拼音:2:abc  3:def 4:ghi 5:jkl 6:mno 7:pqrs 8:tuv 9:wxyz: 第一行读入一个T,代表测试组数: 之后输入两个整数n和m, 有 ...

  10. No1_7.类和对象_Java学习笔记

    一.面向对象的特点:a.封装:封装是面向对象的核心思想,将对象的属性和行为封装起来的载体就是类,类通常对客户隐藏其实现细节,这就是封装的思想: 保证了类内部数据的完整性,应用该类的用户不能轻易直接操纵 ...