HDU-4454 stealing a cake tri-Enumeration

Given a point, a circle, and a rectangle, how short is the shortest distance from a point to a circle to a rectangle? To reach a target, you can just join or pass through it.

Solution: currently, we only know that dividing from one point to the Circle based on the two half circles of [0, Pi], [Pi, 2 * Pi] is indeed a concave function to satisfy the distance, however, if the distance to the rectangle is still three points, you don't know how to get it. The specific method is to first translate the entire coordinate system so that the center of the circle falls at (0, 0), and then enumerate the points on the circle by three points. The distance consists of two parts, one of which is the distance from the point to the point on the circle, one part is the distance from the point to the rectangle, and the latter is converted to the distance from the point to the four line segments.

CodeAs follows:

# Include <cstdlib> # Include <Cstring> # Include <Cstdio> # Include <Algorithm> # Include <Iostream> # Include <Cmath> Using   Namespace  STD; Const   Double EPS = 1E- 6  ;  Const   Double Pi = ACOs (- 1.0  );  Double CX, Cy, CR; //  Circle center and radius  Inline  Int Sign ( Double  X ){  Return X <-EPS? - 1 : X> EPS? 1 : 0  ;}  Struct  Point {  Double  X, Y; point (  Double Xx = 0 , Double YY = 0  ): X (XX), y (yy) {} Point  Operator -(Const Point & ot) Const { //  Difference between two vertices to obtain the Vector          Return Point (X-ot. X, Y- Ot. Y );}  Double   Operator *( Const Point & ot) Const { //  Heavy load point Product          Return X * ot. x + y * Ot. Y ;} Double   Operator ^ ( Const Point & ot) Const { //  Heavy Load Cross Product          Return X * ot. Y-y * Ot. X ;}  Bool   Operator <( Const Point & ot) Const  {  If (Sign (X-Ot. X )){  Return Sign (X-ot. X) < 0 ; //  First, sort by abscissa } Else { //  Then sort by ordinate              Return Sign (Y-ot. Y) < 0  ;}}  Void  Show () {printf (  " X = %. 2f, y = %. 2f \ n  "  , X, y) ;}}; point it, RP [  4  ];  Double Dist ( Const Point &, Const Point & B ){  Return SQRT (A. x-b.x) * (A. x-b.x) + (A. y-b.y) * (A. Y- B. Y ));}  Double Dtol ( Const Point & NP,Const Point & St, Const Point & ED ){ //  Distance from a point to a line segment      Double  RET;  If (Sign (ED-St) * (NP-St)> 0 & Sign (St-ed) * (NP-ed)> 0 )){ //  Used to determine the projection of a point on the Line Segment Ret = FABS (St-np) ^ (ED-np ))/ Dist (St, Ed );}  Else {RET = Min (Dist (NP, St), dist (NP, Ed )); //  A small value from point to two endpoints  }  Return  RET ;}  Double Dtor ( Const Point & NP ){ //  Shortest distance to four line segments      Double D1 = min (dtol (NP, RP [ 0 ], RP [ 1 ]), Dtol (NP, RP [ 0 ], RP [ 2  ]);  Double D2 = min (dtol (NP, RP [ 1 ], RP [ 3 ]), Dtol (NP, RP [ 2 ], RP [ 3  ]);  Return  Min (D1, D2 );}  Double Tsearch ( Double L, Double  R ){ Double  Delta;  While (R-l> = EPS) {Delta = (R-l )/ 3.0 ; //  Delta indicates 1/3 of the Interval Length. Point LP (Cr * Cos (L + delta), Cr * sin (L + Delta); point RP (Cr * Cos (R-delta), Cr * sin (R- Delta ));  Double D1 = dist (it, LP) + Dtor (LP );  Double D2 = dist (it, RP) +Dtor (RP );  If (Sign (D1-D2)> 0  ) {L + = Delta ;}  Else  {R -= Delta ;}} point FP (Cr * Cos (R), Cr * Sin (R ));  Return Dist (it, FP) + Dtor (FP );}  Void  Solve (){  // First 3 points [0, Pi], angle within 3 points [Pi, 2 * Pi] Printf ( "  %. 2f \ n  " , Min (tsearch ( 0 , Pi), tsearch (PI, 2 * Pi )));}  Int  Main (){  While (Scanf ( "  % Lf  " , & It. X, & it. Y), sign (it. x) | Sign (it. y) {scanf ( "  % Lf  " , & CX, & cy, & Cr ); //  Read center coordinates and radius It. X-= Cx, it. Y-= Cy; //  Translate the center of the circle to the source point, and change other coordinates. Scanf ( "  % Lf  " , & RP [ 0 ]. X, & RP [ 0  ]. Y); RP [  0 ]. X-= Cx, RP [ 0 ]. Y-= Cy; scanf (  "  % Lf  " , & RP [ 1 ]. X, & RP [ 1  ]. Y); RP [  1 ]. X-= Cx, RP [ 1 ]. Y-= Cy; RP [  2 ]. X = RP [ 0 ]. X, RP [ 2 ]. Y = RP [ 1  ]. Y; RP [  3 ]. X = RP [ 1 ]. X, RP [ 3 ]. Y = RP [ 0  ]. Y; sort (RP, RP + 4 ); //  Obtains and sorts the four vertices of a rectangle to obtain the relationship between a line segment and a vertex.          Double T = SQRT ( 2.0 )/ 2  ; Solve ();} Return   0  ;}