loj#2015. 「SCOI2016」妖怪 凸函数/三分

题目链接

loj#2015. 「SCOI2016」妖怪

题解

对于每一项展开

的到\(atk+\frac{dnf}{b}a + dnf + \frac{atk}{a} b\)

令$T = \frac{a}{b} $

原式$=atk+Tdnf + dnf + \frac{atk}{T} $

这就是那个单峰的对勾函数，

把单峰函数复合为求最值，发现也是个单峰函数(下凸壳)

三分就好了

或者维护一个最大值得下凸壳

代码

#include<cstdio>
#include<algorithm> 
inline int read() {
	int x = 0,f = 1;
	char c = getchar();
	while(c < '0' || c > '9') c = getchar();
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar();
	return x * f;
}
#define db double
const int maxn = 5000007; 
db atk[maxn],dnf[maxn];
int n;
db calc(db T) {
	double ret = 0.0;
	for(int i = 1;i <= n;++ i) {
		db tmp = dnf[i] + atk[i] + dnf[i] / T + atk[i] * T;
		ret = std::max(tmp,ret);
	}
	return ret;
}
int main() {
	n = read();
	for(int i = 1;i <= n;++ i)
		scanf("%lf%lf",atk + i,dnf + i);
	db l = 0,r = 10.0;
	for(;r - l >= (1e-12);) {
		db m1 = l + (r - l) / 3.0,m2 = r - (r - l) / 3.0;
		db c1 = calc(m1),c2 = calc(m2);
		if(c1 > c2) l = m1;
		else r = m2;
	}
	printf("%.4lf\n",std::min(calc(r),calc(l))) ;
	return 0;
}