GDUT_寒假训练题解报告_专题I_K题 个人题解报告
题目：
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
这个题应该是裸三分模板，正好让我这菜鸡练三分，首先题意，在这么多二次函数中，有一个函数F(X)叫做在x处，取所有二次函数的最大值，那么我们就能得到一个F(x)的图像，求F(x)最小值
思路步骤：
1.画图，我们可以知道两个很窄的二次函数，他们能形成一个细缝，能把几乎所有函数值给覆盖，但是也有可能被横穿过去，总之我们通过画图可以得到一个结论：F(x)函数一定一个单峰函数，有最多一个不可导点，最多一个极小值点也有是最小值点，总之是一个单峰的函数，其中最小值点也是我们的答案。
2.第一步说起来短，实际上画图步骤要得出这个结论需要一定的图像处理能力，但是也不算特别难。
------------二分和三分不同，二分定的是：这一边满足条件，那么答案一定在另一边，然后不断缩小范围，使满足条件的最大/最小点得出，但是三分不同。
------------三分用于，在一个范围内寻找最小值或者最大值，三分的思想可以如此来说明：
三分：在一段范围内：，有L，R，寻找他们中间的两个点，这时候把(l,r)分成三份，这里有两个分法；分四份取三份 和 分三份
——>分四份取三份是什么意思呢？这两个点设为mid1，和mid2
mid1=(l+r)/2,mid2=(r+mid)/2或者(l+mid)/2，
——>分三份就容易理解了：mid1=(l2+r)/3;;;mid2=(l+r2)/3
两种都是可以的，看个人写法；
三分的下一步就是判断了：
以一个凸函数为例子，找最小值（两头大中间小）
三种情况：
1.如果答案在mid1左边，那么F(mid1)<F(mid2)
2.如果答案在mid1和mid2之间，那么有F(mid1)<F(mid2)或者F(mid1)>F(mid2)
3.如果答案在mid2右边，有F(mi1)>F(mid2)
我们归类一下：
如果F(mid1)<F(mid2),那么答案在mid2左边，令r=mid2即可；
如果F(mid1)>F(mid2)，那么答案在mid1右边，令l=mid1即可；
然后就是浮点型二分的结束语句while(r-l<eds){}
这个eds就是精度，
代码实现：
#define ULL unsigned long long
using namespace std;
typedef struct element
{
double a,b,c;
}ele;
ele all[10010];
int t,n;
double f(int i,double x)
{//fi(x)的求值函数
return all[i].a*x*x+all[i].b*x+all[i].c;
}
double F(double x)
{
double _MAX=f(0,x);
for(int time=1;time<n;time++)
{
if(f(time,x)>_MAX)_MAX=f(time,x);
}
return _MAX;
}
int main()
{
scanf("%d",&t);
for(int time=0;time<t;time++)
{
scanf("%d",&n);
for(int time1=0;time1<n;time1++)
{
scanf("%lf %lf %lf",&all[time1].a,&all[time1].b,&all[time1].c);
}
double l=0;
double r=1000;
while(r-l>1e-9)
{
double mid1=(l*2+r)/3;
double mid2=(l+r*2)/3;
if(F(mid1)<F(mid2))r=mid2;
else l=mid1;
}
printf("%.4f\n",F(l));
}
return 0;
}