LA5009 Error Curves

题目大意：给定n个二次函数fi(x)，规定F(x) = max(fi(x))，求F在[0,1000]上的max

 

其实F（x）也是一个单峰函数，因为如果出现双峰，显然有一个可以继续往上走把另一个峰遮住，即另一个峰不可能是max

三分即可

被精度卡了半天

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
#include <string> 
#include <cmath> 
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
template<class T>
inline void swap(T &a, T &b)
{
    T tmp = a;a = b;b = tmp;
}
inline void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const double INF = 0x3f3f3f3f;
const double eps = 0.0000000001;


int t,n,a[10100],b[10100],c[10010];

double f(double x)
{
    double ma = x * x * a[1] + x * b[1] + c[1], t;
    for(int i = 2;i <= n;++ i) 
    {
        t = x * x * a[i] + x * b[i] + c[i];
        //if(!cmp(t, ma)) ma = t;
        ma = max(ma, t);
    }
    return ma;
}

int main()
{
    read(t);
    for(;t;--t)
    {
        read(n);
        for(int i = 1;i <= n;++ i) read(a[i]), read(b[i]), read(c[i]);
        double l = 0, r = 1000, ans, mid1, mid2;
        while(r - l - eps >= 0)
        {
            mid1 = l + (r - l) / 3.0;mid2 = r - (r - l) / 3.0;
            if(f(mid1) - f(mid2) >= 0) l = mid1;
            else r = mid2;
        }
        printf("%.4lf\n", f(l));
    }
    return 0;
}