Uva5009 Error Curves

已知n条二次曲线si(x) = ai*x^2 + bi*x + ci(ai ≥ 0),定义F(x) = max{si(x)},求出F(x)在[0,1000]上的最小值.

链接：传送门

分析：最大值最小，我们可以利用二分来解，但是有一个更牛的方法叫：“三分法”，这个方法的应用范围是凸函数，可以看一个图像：

L和R是边界，m1,m2是三等分点，如果f(m1) < f(m2),那么最小值肯定在[l,m2]内，注意，不是[l,m1]因为如果m1在最低点右边，那么就会矛盾，同理，如果f(m2) < f(m1),那么最小值肯定在[m1,r]之间，剩下的操作和二分法基本上就是一样的了。

对于本题而言，可以看出也是一个凸函数，所以我们可以利用三分法来快速求最小值.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#include<map>
#include<string>using namespace std;const int maxn = 10010;int n, a[maxn], b[maxn], c[maxn],t;double f(double x)
{double ans = a[1] * x * x + b[1] * x + c[1];for (int i = 2; i <= n; i++)ans = max(ans, a[i] * x*x + b[i] * x + c[i]);return ans;
}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 L = 0.0, R = 1000.0;while (R - L > 0.000000001){double m1 = L + (R - L) / 3, m2 = R - (R - L) / 3;if (f(m1) < f(m2))R = m2;elseL = m1;}printf("%.4lf\n", f(L));}return 0;
}

转载于:https://www.cnblogs.com/zbtrs/p/7413234.html

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