HDU 3714 Error Curves

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3714

这题我死活没读懂什么意思，后来还是别人讲的我才明白。

题目：给出一组一元二次方程，用这一组一元二次方程生成一个新的函数F(x)，求F(x)的最小值。x 对应F(x) 的值=这一组一元二次方程中y_i = a_i * x^2 + b_i * x + c_i y最大的那一个。

分析：凹函数求极值，三分法。

#include <cstdio>
#include <cstring>

struct node
{
    int a, b, c;
};

const int MAXN = 10000 + 10;
const double EXP = 1e-12;
const double INF = 2147483646;

int n;
node D[MAXN];

double calcu( double x )
{
    double ans = -INF;
    double temp;

    for ( int i = 0; i < n; i++ )
    {
        temp = D[i].a * x * x + D[i].b * x + D[i].c;
        if ( temp > ans ) ans = temp;
    }

    return ans;
}

double Solve()
{
    double Left = 0, Right = 1000;
    double mid, midmid;
    double mid_value, midmid_value;

    while ( Left + EXP < Right )
    {
        mid = ( Left + Right ) / 2;
        midmid = ( mid + Right ) / 2;
        mid_value = calcu(mid);
        midmid_value = calcu(midmid);

        if ( mid_value <= midmid_value ) Right = midmid;
        else Left = mid;
    }

    return calcu(Left);
}

int main()
{
    int T;
    scanf( "%d", &T );
    while( T-- )
    {
        scanf( "%d", &n );
        for ( int i = 0; i < n; i++ )
           scanf( "%d%d%d", &D[i].a, &D[i].b, &D[i].c );

        printf( "%.4f\n", Solve() );
    }
    return 0;
}