BZOJ1857 [Scoi2010]传送带 【三分法】

题目链接

BZOJ1857

题解

画画图就发现实际上是在(AB)上和(CD)上分别选两个点(E),(F)，使得(t_{AE} + t_{EF} + t_{FD})最小
然后猜想到当(E)固定时，这个值的函数关于(|CF|)是下凸的
当(F)总取最优时，关于(|AE|)也是下凸的
感觉十分的对

两层三分即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 100005,maxm = 100005,INF = 1000000000;
struct point{
	double x,y;
}A,B,C,D,AB,CD;
inline point operator -(const point& a,const point& b){
    return (point){a.x - b.x,a.y - b.y};
}
inline point operator +(const point& a,const point& b){
    return (point){a.x + b.x,a.y + b.y};
}
inline point operator *(const double& a,const point& b){
	return (point){a * b.x,a * b.y};
}
double dis(const point& a,const point& b){
	return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
double P,Q,R;
double cal2(double lam1,double lam2){
	point E = A + lam1 * AB,F = C + lam2 * CD;
	return dis(A,E) / P + dis(E,F) / R + dis(F,D) / Q;
}
double cal(double lam){
	double l = 0,r = 1,lmid,rmid,len,cl,cr;
	while (r - l > 0.00001){
		len = r - l;
		lmid = l + len / 3;
		rmid = r - len / 3;
		cl = cal2(lam,lmid); cr = cal2(lam,rmid);
		if (cl > cr) l = lmid;
		else r = rmid;
	}
	return cal2(lam,(r + l) / 2);
}
int main(){
	scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf",&A.x,&A.y,&B.x,&B.y,&C.x,&C.y,&D.x,&D.y,&P,&Q,&R);
	AB = B - A; CD = D - C;
	double l = 0,r = 1,lmid,rmid,len,cl,cr;
	while (r - l > 0.00001){
		len = r - l;
		lmid = l + len / 3;
		rmid = r - len / 3;
		cl = cal(lmid); cr = cal(rmid);
		if (cl > cr) l = lmid;
		else r = rmid;
	}
	printf("%.2lf
",cal((l + r) / 2));
	return 0;
}