hdu3714 三分找最值

Error Curves

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4928    Accepted Submission(s): 1867

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

 

题意:

有n个二次函数，现在要在区间[0，1000]找这些二次函数的最大值最小。

思路:

由于a大于0，所以这些函数都是开口向上的，所以对于区间内的这些二次函数的最大值构成的这个函数其实也是开口向上的，

我们要找的就是它的最小值了。可以用三分解决。

/*
 * Author:  sweat123
 * Created Time:  2016/7/15 14:54:08
 * File Name: main.cpp
 */
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<time.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1<<30
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define key_value ch[ch[root][1]][0]
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
#define eps 1e-9
using namespace std;
const int MAXN = 100010;
double a[MAXN],b[MAXN],c[MAXN];
int n;
double getval(double x){
    double ans = -INF;
    for(int i = 1; i <= n; i++){
        ans = max(ans,a[i] * x * x + b[i] * x + c[i]);
    }
    return ans;
}
double solve(){
    double l,r,m,mm;
    l = 0,r = 1000;
    while(r - l > eps){
        //m = (l + r) / 2;
        //mm = (m + r) / 2;
        m = (2*l + r) / 3;
        mm = (2*r + l) / 3;
        if(getval(m) > getval(mm)){
            l = m;
        } else {
            r = mm;
        }
    }
    return getval(l);
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i = 1; i <= n; i++){
            scanf("%lf%lf%lf",&a[i],&b[i],&c[i]);
        }
        printf("%.4lf
",solve());
    }
    return 0;
}