The Doors shortest circuit + simple geometry

The Doors

Abstract: Give you some points, and some obstacles to the edge of the line. Ask the shortest geometric distance from the origin to the end point.

Analysis: After the pretreatment see diagram, run the shortest way.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5 using namespacestd;6 Const intINF =0x5fffffff;7 Const intMS =1005;8 Const intMS2 = -;9 Ten structPoint { One     Doublex, y; A    //Point (Double _x = 0, double _y = 0): X (_x), Y (_y) {} -     //the OJ of the compiler is not updated for a long time and may not support pioint{} initialization. You can change to constructor point () initialization.  - }points[ms]; the structEdge { -     intu, v; -     DoubleW; -}edges[ms *MS]; +  - intpSize, esize; + intN; A  at DoubleW[MS], wy[ms][4]; - DoubleDis[ms]; -  - DoubleDConstPoint &a,ConstPoint &b) { -     returnsqrt ((a.x-b.x) * (a.x-b.x) + (A.Y-B.Y) * (A.Y-b.y)); - } in  - DoubleCConstPoint &a,ConstPoint &b,ConstPoint &c) { to     return(C.Y-A.Y) * (a.x-b.x)-(A.Y-B.Y) * (c.x-a.x); + } -  the BOOLIsOk (ConstPoint &a,ConstPoint &b) { *     if(A.x >=b.x) $         return false;Panax Notoginseng     inti =0; -      while(I < n && W[i] <=a.x) thei++; +      while(I < n && W[i] <b.x) { A         if(C (A, B, Point{w[i],0}) * C (A, B, point{w[i],wy[i][0]}) <0) the             return false; +         if(C (A, B, point{w[i],wy[i][1]}) * C (A, B, point{w[i], wy[i][2]}) <0) -             return false; $         if(C (A, B, point{w[i], wy[i][3]}) * C (A, B, Point{w[i],10.0}) <0) $             return false; -i++; -     } the     return true; - }Wuyi  the intinput () { -scanf"%d", &n); Wu     if(n = =-1) -         return 0; AboutPSize =0; $points[psize].x =0; -POINTS[PSIZE++].Y =5; -      for(inti =0; I < n; i++) { -scanf"%LF", &w[i]); A          for(intj =0; J <4; j + +) { +points[psize].x =W[i]; thescanf"%LF", &wy[i][j]); -POINTS[PSIZE++].Y =Wy[i][j]; $         } the     } thepoints[psize].x =Ten; thePOINTS[PSIZE++].Y =5; the  -Esize =0; in      for(inti =0; i < pSize; i++) { the          for(intj = i +1; J < PSize; J + +) { the             if(IsOk (Points[i], points[j])) { AboutEDGES[ESIZE].U =i; theEDGES[ESIZE].V =J; theEDGES[ESIZE++].W =D (Points[i], points[j]); the             } +         } -     } the     return 1;Bayi } the  the voidBellman (intv0) { -      for(inti =0; i < pSize; i++) -Dis[i] = INF;//Fill (Dis,dis + pSize, INF) theDis[v0] =0.0; the     intFlag =1; the      for(intt =1; T < pSize && flag; t++) { theFlag =0; -          for(inti =0; i < esize; i++) { the             if(DIS[EDGES[I].U] + EDGES[I].W <DIS[EDGES[I].V]) { theDIS[EDGES[I].V] = dis[edges[i].u] +EDGES[I].W; theFlag =1;94             } the         } the     } the }98  About intMain () { -      while(Input ()) {101Bellman (0);102printf"%.2lf\n", Dis[psize-1]);103     }104     return 0; the}

The Doors shortest circuit + simple geometry