UVA - 1476 Error Curves 3분

Error Curves
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 isdivided into two parts: training data and test data. She gets the parameters of the model on trainingdata 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 curvecorresponds to a quadratic function. In mathematics, a quadratic function is a polynomial function ofthe 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, thereare several datasets, which means Josephina will obtain many parabolic curves. Josephina wants toget the tuned parameters that make the best performance on all datasets. So she should take all errorcurves into account, i.e., she has to deal with many quadric functions and make a new error definitionto represent the total error. Now, she focuses on the following new function’s minimal which related tomultiple quadric functions.The new function F(x) is defined as follow: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?InputThe input contains multiple test cases. The first line is the number of cases T (T < 100). Each casebegins 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.OutputFor each test case, output the answer in a line. Round to 4 digits after the decimal point.Sample Input212 0 022 0 02 -4 2Sample Output0.00000.5000
 
제목:
n개의 2차 커브 S(x)를 지정하고 F(x) = max(Si(x)를 정의하여 F(x)가 0~1000에서 가장 작은 값을 구합니다.
문제:
기초 문제 3점, 하철 3점.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long ll;
const int  N = 10000 + 10;
int T,a[N],b[N],c[N],n;
double f(double x) {
    double ans = a[1] * x * x + b[1] * x + c[1];
    for(int i = 1; i <= n; i++) {
        ans = max(ans, a[i] * x * x + b[i] * x + c[i]);
    }
    return ans;
}
double three_search(double l,double r) {
    for(int i = 0 ;i < 100; i++) {
        double mid = l + (r - l) / 3;
        double mid2 = r - (r - l) / 3;
        if(f(mid) > f(mid2)) l = mid;
        else r = mid2;
    }
    return f(l);
}
int main() {
    scanf("%d",&T);
    while(T--) {
        scanf("%d",&n);
        for(int i = 1; i <= n; i++) scanf("%d%d%d",&a[i],&b[i],&c[i]);
        double ans = three_search(0,1000);
        printf("%.4f
",ans);
    }
    return 0;
}