[三分法初探]

[Uva Error Curves]一些二次函数取max的交集还是一个下凸函数。三分即可

#include <bits/stdc++.h>
#define maxn 10010
using namespace std;

int n, a[maxn], b[maxn], c[maxn];

#define F(i, x) a[i] * x * x + b[i] * x + c[i]

inline double cal(double x){
	double ret = -1e12;
	for(int i = 1; i <= n; i ++)
	    ret = max(ret, F(i, x));
	return ret;
}

inline void solve(){
	double l = 0, r = 1000;
	for(int i = 1; i <= 300; i ++){
		double len = (r - l) / 3;
		double m1 = l + len, m2 = r - len;
		if(cal(m1) < cal(m2))r = m2;
		else l = m1;
	}
	printf("%.4lf\n", cal(l));
}

int main(){
    int test;
    scanf("%d", &test);
    while(test --){
		scanf("%d", &n);
		for(int i = 1; i <= n; i ++)
		    scanf("%d%d%d", &a[i], &b[i], &c[i]);
	    solve();
    }

	return 0;
}

[Zoj Light Bulb]列出数学式子，是一个对勾函数(可以直接解)，或者三分，注意影子到地上的判断，不过D-x≤H？？

#include <bits/stdc++.h>

using namespace std;

double H, h, D, P, Q;

inline double check(double x){
	return -x + P / x + Q;
}

void solve(){
	P = D * (h - H), Q = D + H;
	double l = 1e-9, r = min(H, D);
	for(int i = 1; i <= 1000; i ++){
		double len = (r - l) / 3;
		double m1 = l + len, m2 = r - len;
		if(check(m1) > check(m2))r = m2;
		else l = m1;
	}
	printf("%.3lf\n", check(r));
}

int main(){
	int test;
	scanf("%d", &test);
	while(test --){
		scanf("%lf%lf%lf", &H, &h, &D);
		solve();
	}
	return 0;
}

[HDU Line belt]三分套三分

#include <bits/stdc++.h>

using namespace std;

struct Point{
	double x, y;
	inline void read(){scanf("%lf%lf", &x, &y);}
}A, B, C, D;
#define Sqr(x) (x)*(x)
double Len(const Point& a, const Point& b){
	return sqrt(Sqr(a.x-b.x) + Sqr(a.y-b.y));
}
int P, Q, R;
double T(double a, double b){
	Point X, Y;
	X.x = a * (B.x - A.x) + A.x;
	X.y = a * (B.y - A.y) + A.y;
	Y.x = b * (C.x - D.x) + D.x;
	Y.y = b * (C.y - D.y) + D.y;
	return Len(A, X) / P + Len(D, Y) / Q + Len(X, Y) / R;
}

double check(double a){
	double l = 0, r = 1;
	while(r - l > 1e-6){
		double len = (r - l) / 3.0;
		double m1 = l + len, m2 = r - len;
		if(T(a, m1) < T(a, m2))r = m2;
		else l = m1;
	}return T(a, r);
}

int main(){
	int test;
	scanf("%d", &test);
	while(test --){
		A.read(), B.read(), C.read(), D.read();
		scanf("%d%d%d", &P, &Q, &R);
		double l = 0, r = 1;
		while(r - l > 1e-6){
			double len = (r - l) / 3.0;
			double m1 = l + len, m2 = r - len;
			if(check(m1) < check(m2))r = m2;
			else l = m1;
		}
		printf("%.2lf\n", check(r));
	}
	return 0;
}