題干：

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

解題報(bào)告：

? ? 由于題中給出的a>=0, 所以a有可能為零，（但是這個(gè)題好像并沒有在這里設(shè)坑）此時(shí)曲線為直線，否則曲線為開口向上的拋物線，故為下凸函數(shù)，所以F(x)也為下凸函數(shù)。故可用三分法求F(x)的極值。先算出F(x)的具體值，然后就可直接三分了。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> using namespace std;const int MAX = 10000 + 10; int n, a[MAX], b[MAX], c[MAX];double cal(double x) { //求F(x)double ans = a[0]*x*x + b[0]*x + c[0];for(int i=1; i<n; i++) {ans = max(ans, a[i]*x*x+b[i]*x+c[i]);}return ans; } int main() {int T;scanf("%d",&T);while(T--){scanf("%d", &n);for(int i=0; i<n; i++) {scanf("%d%d%d", &a[i], &b[i], &c[i]);}double l = 0, r = 1000; //三分求極值for(int i=0; i<100; i++) {double mid = l + (r-l)/3;double midd = r - (r-l)/3;if(cal(mid) < cal(midd)) r = midd;else l = mid;}printf("%.4f\n",cal(l));}return 0; }

總結(jié)：

? ?這里還有一個(gè)黑科技！不用定義eps來卡精度！直接卡時(shí)間！還有二分次數(shù)可以多一點(diǎn)，二分區(qū)間為10^6，才二分二十多次，可以二分100次。for循環(huán)！不用while(l<r)！！?

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