给定一个起始点,一个矩形,一个圆,三者互不相交。求从起始点->圆->矩形的最短距离。
自己画一画就知道距离和会是凹函数,不过不是一个凹函数。按与水平向量夹角为圆心角求圆上某点坐标,[0, PI] , [PI, 2*pi]两个区间的点会有两个凹函数。所以要做两次三分才行。
- #include<algorithm>
- #include<iostream>
- #include<fstream>
- #include<sstream>
- #include<cstring>
- #include<cstdlib>
- #include<string>
- #include<vector>
- #include<cstdio>
- #include<queue>
- #include<stack>
- #include<cmath>
- #include<map>
- #include<set>
- #define FF(i, a, b) for(int i=a; i<b; i++)
- #define FD(i, a, b) for(int i=a; i>=b; i--)
- #define REP(i, n) for(int i=0; i<n; i++)
- #define CLR(a, b) memset(a, b, sizeof(a))
- #define LL long long
- #define PB push_back
- #define eps 1e-10
- #define debug puts("**debug**");
- using namespace std;
- const double PI = acos(-1);
- struct Point
- {
- double x, y;
- Point(double x=0, double y=0):x(x), y(y){}
- };
- typedef Point Vector;
- 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);
- }
- int dcmp(double x)
- {
- if(fabs(x) < eps) return 0; return x < 0 ? -1 : 1;
- }
- 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 Cross(Vector a, Vector b) { return a.x*b.y - a.y*b.x; }
- double DistanceToSegment(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);
- else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
- else return fabs(Cross(v1, v2)) / Length(v1);
- }
- struct Circle
- {
- Point c;
- double r;
- Circle(){}
- Circle(Point c, double r):c(c), r(r){}
- Point point(double a) //根据圆心角求点坐标
- {
- return Point(c.x+cos(a)*r, c.y+sin(a)*r);
- }
- }o;
- Point p, p1, p2, p3, p4, s;
- double a, b, c, d;
- double Calc(double x)
- {
- p = o.point(x);
- double d1 = DistanceToSegment(p, p1, p2),
- d2 = DistanceToSegment(p, p2, p3),
- d3 = DistanceToSegment(p, p3, p4),
- d4 = DistanceToSegment(p, p4, p1);
- //点p到矩形最近距离加上s到p距离
- return min(min(d1, d2), min(d3, d4)) + Length(s-p);
- }
- double solve()
- {
- double L, R, m, mm, mv, mmv;
- L = 0; R = PI;
- while (L + eps < R)
- {
- m = (L + R) / 2;
- mm = (m + R) / 2;
- mv = Calc(m);
- mmv = Calc(mm);
- if (mv <= mmv) R = mm; //三分法求最大值时改为mv>=mmv
- else L = m;
- }
- double ret = Calc(L);
- L = PI; R = 2*PI;
- while (L + eps < R)
- {
- m = (L + R) / 2;
- mm = (m + R) / 2;
- mv = Calc(m);
- mmv = Calc(mm);
- if (mv <= mmv) R = mm;
- else L = m;
- }
- return min(ret, Calc(L));
- }
- int main()
- {
- while(scanf("%lf%lf", &s.x, &s.y))
- {
- if(s.x == 0 && s.y == 0) break;
- scanf("%lf%lf%lf", &o.c.x, &o.c.y, &o.r);
- scanf("%lf%lf%lf%lf", &a, &b, &c, &d);
- //确定矩形四个点坐标,左上点开始 逆时针
- double maxx, maxy, minx, miny;
- maxx = max(a, c); maxy = max(b, d);
- minx = min(a, c); miny = min(b, d);
- p1 = Point(minx, maxy);
- p2 = Point(minx, miny);
- p3 = Point(maxx, miny);
- p4 = Point(maxx, maxy);
- double ans = solve();
- printf("%.2f\n", ans);
- }
- return 0;
- }
本文探讨了一种使用三分法解决从起始点到圆再到矩形的最短路径问题的方法,涉及几何计算、函数凹凸性质和算法优化。通过分析路径的几何特性,将问题转化为求解凹函数的最大值问题,并采用三分法进行高效求解。
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