bzoj 1857

三分，对于单凸的函数（单调的也可以），可以找出最值。

这道题可以感性认识一下。。。。。。

/**************************************************************
    Problem: 1857
    User: idy002
    Language: C++
    Result: Accepted
    Time:32 ms
    Memory:1272 kb
****************************************************************/
 
#include <cstdio>
#include <cmath>
#include <iostream>
#define eps 1e-5
using namespace std;
 
int sg( double x ) { return (x>-eps)-(x<eps); }
struct Vector {
    double x, y;
    Vector(){}
    Vector( double x, double y ):x(x),y(y){}
    Vector operator+( const Vector & b ) const { return Vector(x+b.x,y+b.y); }
    Vector operator-( const Vector & b ) const { return Vector(x-b.x,y-b.y); }
    Vector operator*( double k ) const { return Vector(x*k,y*k); }
    void read() { scanf( "%lf%lf", &x, &y ); }
    double dis() { return sqrt(x*x+y*y); }
};
typedef Vector Point;
 
Point A, B, C, D, E1, E2, E, F1, F2, F;
double P, Q, R;
 
inline double gettime() {
    return (A-E).dis()/P+(F-D).dis()/Q+(E-F).dis()/R;
}
double onCD() {
    double t1, t2;
    F1 = C, F2 = D;
    while( (F1-F2).dis() >= eps ) {
        Vector step = (F2-F1)*(1.0/3.0);
        F = F1+step;
        t1 = gettime();
        F = F +step;
        t2 = gettime();
        if( t1-t2<0.0 ) {
            F2 = F2-step;
        } else {
            F1 = F1+step;
        }
    }
    F = F1;
    return gettime();
}
double onAB() {
    double t1, t2;
    E1 = A, E2 = B;
    while( (E1-E2).dis() >= eps ) {
        Vector step = (E2-E1)*(1.0/3.0);
        E = E1+step;
        t1 = onCD();
        E = E +step;
        t2 = onCD();
        if( t1-t2<0.0 ) {
            E2 = E2-step;
        } else {
            E1 = E1+step;
        }
    }
    E = E1;
    return onCD();
}
int main() {
    A.read();
    B.read();
    C.read();
    D.read();
    scanf( "%lf%lf%lf", &P, &Q, &R );
    printf( "%.2lf
", onAB() );
}