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。。。。。。
1 /************************************************************** 2 Problem: 1857 3 User: idy002 4 Language: C++ 5 Result: Accepted 6 Time:32 ms 7 Memory:1272 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cmath> 12 #include <iostream> 13 #define eps 1e-5 14 using namespace std; 15 16 int sg( double x ) { return (x>-eps)-(x<eps); } 17 struct Vector { 18 double x, y; 19 Vector(){} 20 Vector( double x, double y ):x(x),y(y){} 21 Vector operator+( const Vector & b ) const { return Vector(x+b.x,y+b.y); } 22 Vector operator-( const Vector & b ) const { return Vector(x-b.x,y-b.y); } 23 Vector operator*( double k ) const { return Vector(x*k,y*k); } 24 void read() { scanf( "%lf%lf", &x, &y ); } 25 double dis() { return sqrt(x*x+y*y); } 26 }; 27 typedef Vector Point; 28 29 Point A, B, C, D, E1, E2, E, F1, F2, F; 30 double P, Q, R; 31 32 inline double gettime() { 33 return (A-E).dis()/P+(F-D).dis()/Q+(E-F).dis()/R; 34 } 35 double onCD() { 36 double t1, t2; 37 F1 = C, F2 = D; 38 while( (F1-F2).dis() >= eps ) { 39 Vector step = (F2-F1)*(1.0/3.0); 40 F = F1+step; 41 t1 = gettime(); 42 F = F +step; 43 t2 = gettime(); 44 if( t1-t2<0.0 ) { 45 F2 = F2-step; 46 } else { 47 F1 = F1+step; 48 } 49 } 50 F = F1; 51 return gettime(); 52 } 53 double onAB() { 54 double t1, t2; 55 E1 = A, E2 = B; 56 while( (E1-E2).dis() >= eps ) { 57 Vector step = (E2-E1)*(1.0/3.0); 58 E = E1+step; 59 t1 = onCD(); 60 E = E +step; 61 t2 = onCD(); 62 if( t1-t2<0.0 ) { 63 E2 = E2-step; 64 } else { 65 E1 = E1+step; 66 } 67 } 68 E = E1; 69 return onCD(); 70 } 71 int main() { 72 A.read(); 73 B.read(); 74 C.read(); 75 D.read(); 76 scanf( "%lf%lf%lf", &P, &Q, &R ); 77 printf( "%.2lf\n", onAB() ); 78 }
