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\n", onAB() );
 }