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POJ 3608 Bridge Across Islands --凸包间距离
题意: 给你两个凸包,求其最短距离。
解法: POJ 我真的是弄不懂了,也不说一声点就是按顺时针给出的,不用调整点顺序。 还是说数据水了,没出乱给点或给逆时针点的数据呢。。反正WA了几发就是了。
求凸包最短距离还是用旋转卡壳的方法,这里采用的是网上给出的一种方法:
英文版: http://cgm.cs.mcgill.ca/~orm/mind2p.html
中文翻译版: http://www.cnblogs.com/bless/archive/2008/08/06/1262438.html
输入的两个凸包须是顺时针。
分别以一个为主卡另外一个,两次取最小值即可。
算法就不分析了, 画个图理解一下就知道了。
#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#define pi acos(-1.0)#define eps 1e-8using namespace std;struct Point{ double x,y; Point(double x=0, double y=0):x(x),y(y) {} void input() { scanf("%lf%lf",&x,&y); }};typedef Point Vector;int dcmp(double x) { if(x < -eps) return -1; if(x > eps) return 1; return 0;}template <class T> T sqr(T x) { return x * x;}Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }double Length(Vector A) { return sqrt(Dot(A, A)); }double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }Vector VectorUnit(Vector x){ return x / Length(x);}Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}double angle(Vector v) { return atan2(v.y, v.x); }double DistanceToSeg(Point P, Point A, Point B) { if(A == B) return Length(P-A); Vector v1 = B-A, v2 = P-A, v3 = P-B; if(dcmp(Dot(v1, v2)) < 0) return Length(v2); if(dcmp(Dot(v1, v3)) > 0) return Length(v3); return fabs(Cross(v1, v2)) / Length(v1);}double SegDistancetoSeg(Point A,Point B,Point C,Point D) { return min(min(DistanceToSeg(C,A,B),DistanceToSeg(D,A,B)),min(DistanceToSeg(A,C,D),DistanceToSeg(B,C,D)));}Point DisP(Point A,Point B) { return Length(B-A); }double MinDisOfTwoConvexHull(Point P[],int n,Point Q[],int m) { int Pymin = 0, Qymax = 0, i,j; for(i=0;i<n;i++) if(dcmp(P[i].y-P[Pymin].y) < 0) Pymin = i; for(i=0;i<m;i++) if(dcmp(Q[i].y-Q[Qymax].y) > 0) Qymax = i; P[n] = P[0], Q[m] = Q[0]; double Mindis = 1e90, Tmp; for(i=0;i<n;i++) { while(dcmp(Tmp = Cross(P[Pymin+1]-P[Pymin],Q[Qymax+1]-P[Pymin])-Cross(P[Pymin+1]-P[Pymin],Q[Qymax]-P[Pymin])) > 0) Qymax = (Qymax+1)%m; if(dcmp(Tmp) < 0) Mindis = min(Mindis,DistanceToSeg(Q[Qymax],P[Pymin],P[Pymin+1])); else Mindis = min(Mindis,SegDistancetoSeg(P[Pymin],P[Pymin+1],Q[Qymax],Q[Qymax+1])); Pymin = (Pymin+1)%n; } return Mindis;}Point P[10005],nP[10005],Q[10005],nQ[10005];int main(){ int n,m,i; while(scanf("%d%d",&n,&m)!=EOF && n+m) { for(i=0;i<n;i++) P[i].input(); for(i=0;i<m;i++) Q[i].input(); printf("%.5f\n",min(MinDisOfTwoConvexHull(P,n,Q,m),MinDisOfTwoConvexHull(Q,m,P,n))); } return 0;}
POJ 3608 Bridge Across Islands --凸包间距离
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