lrj计算几何模板

整理了一下大白书上的计算几何模板。

  1 #include <cstdio>

  2 #include <algorithm>

  3 #include <cmath>

  4 #include <vector>

  5 using namespace std;

  6 //lrj计算几何模板

  7 struct Point

  8 {

  9     double x, y;

 10     Point(double x=0, double y=0) :x(x),y(y) {}

 11 };

 12 typedef Point Vector;

 13 

 14 Point read_point(void)

 15 {

 16     double x, y;

 17     scanf("%lf%lf", &x, &y);

 18     return Point(x, y);

 19 }

 20 

 21 const double EPS = 1e-10;

 22 

 23 //向量+向量=向量 点+向量=点

 24 Vector operator + (Vector A, Vector B)    { return Vector(A.x + B.x, A.y + B.y); }

 25 

 26 //向量-向量=向量 点-点=向量

 27 Vector operator - (Vector A, Vector B)    { return Vector(A.x - B.x, A.y - B.y); }

 28 

 29 //向量*数=向量

 30 Vector operator * (Vector A, double p)    { return Vector(A.x*p, A.y*p); }

 31 

 32 //向量/数=向量

 33 Vector operator / (Vector A, double p)    { return Vector(A.x/p, A.y/p); }

 34 

 35 bool operator < (const Point& a, const Point& b)

 36 { return a.x < b.x || (a.x == b.x && a.y < b.y); }

 37 

 38 int dcmp(double x)

 39 { if(fabs(x) < EPS) return 0; else return x < 0 ? -1 : 1; }

 40 

 41 bool operator == (const Point& a, const Point& b)

 42 { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }

 43 

 44 /**********************基本运算**********************/

 45 

 46 //点积

 47 double Dot(Vector A, Vector B)

 48 { return A.x*B.x + A.y*B.y; }

 49 //向量的模

 50 double Length(Vector A)    { return sqrt(Dot(A, A)); }

 51 

 52 //向量的夹角，返回值为弧度

 53 double Angle(Vector A, Vector B)

 54 { return acos(Dot(A, B) / Length(A) / Length(B)); }

 55 

 56 //叉积

 57 double Cross(Vector A, Vector B)

 58 { return A.x*B.y - A.y*B.x; }

 59 

 60 //向量AB叉乘AC的有向面积

 61 double Area2(Point A, Point B, Point C)

 62 { return Cross(B-A, C-A); }

 63 

 64 //向量A旋转rad弧度

 65 Vector VRotate(Vector A, double rad)

 66 {

 67     return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));

 68 }

 69 

 70 //将B点绕A点旋转rad弧度

 71 Point PRotate(Point A, Point B, double rad)

 72 {

 73     return A + VRotate(B-A, rad);

 74 }

 75 

 76 //求向量A向左旋转90°的单位法向量,调用前确保A不是零向量

 77 Vector Normal(Vector A)

 78 {

 79     double l = Length(A);

 80     return Vector(-A.y/l, A.x/l);

 81 }

 82 

 83 /**********************点和直线**********************/

 84 

 85 //求直线P + tv 和 Q + tw的交点，调用前要确保两条直线有唯一交点

 86 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w)

 87 {

 88     Vector u = P - Q;

 89     double t = Cross(w, u) / Cross(v, w);

 90     return P + v*t;

 91 }//在精度要求极高的情况下，可以自定义分数类

 92 

 93 //P点到直线AB的距离

 94 double DistanceToLine(Point P, Point A, Point B)

 95 {

 96     Vector v1 = B - A, v2 = P - A;

 97     return fabs(Cross(v1, v2)) / Length(v1);    //不加绝对值是有向距离

 98 }

 99 

100 //点到线段的距离

101 double DistanceToSegment(Point P, Point A, Point B)

102 {

103     if(A == B)    return Length(P - A);

104     Vector v1 = B - A, v2 = P - A, v3 = P - B;

105     if(dcmp(Dot(v1, v2)) < 0)    return Length(v2);

106     else if(dcmp(Dot(v1, v3)) > 0)    return Length(v3);

107     else return fabs(Cross(v1, v2)) / Length(v1);

108 }

109 

110 //点在直线上的射影

111 Point GetLineProjection(Point P, Point A, Point B)

112 {

113     Vector v = B - A;

114     return A + v * (Dot(v, P - A) / Dot(v, v));

115 }

116 

117 //线段“规范”相交判定

118 bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2)

119 {

120     double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1);

121     double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);

122     return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;

123 }

124 

125 //判断点是否在线段上

126 bool OnSegment(Point P, Point a1, Point a2)

127 {

128     Vector v1 = a1 - P, v2 = a2 - P;

129     return dcmp(Cross(v1, v2)) == 0 && dcmp(Dot(v1, v2)) < 0;

130 }

131 

132 //求多边形面积

133 double PolygonArea(Point* P, int n)

134 {

135     double ans = 0.0;

136     for(int i = 1; i < n - 1; ++i)

137         ans += Cross(P[i]-P[0], P[i+1]-P[0]);

138     return ans/2;

139 }

140 

141 int main(void)

142 {

143     Vector a[2];

144     sort(a, a + 2);

145     return 0;

146 }

147 

148 /**********************圆的相关计算**********************/

149 

150 const double PI = acos(-1.0);

151 struct Line

152 {//有向直线

153     Point p;

154     Vector v;

155     double ang;

156     Line()    { }

157     Line(Point p, Vector v): p(p), v(v)    { ang = atan2(v.y, v.x); }

158     Point point(double t)

159     {

160         return p + v*t;

161     }

162     bool operator < (const Line& L) const

163     {

164         return ang < L.ang;

165     }

166 };

167 

168 struct Circle

169 {

170     Point c;    //圆心

171     double r;    //半径

172     Circle(Point c, double r):c(c), r(r)    {}

173     Point point(double a)

174     {//求对应圆心角的点

175         return Point(c.x + r*cos(a), c.y + r*sin(a));

176     }

177 };

178 

179 //两圆相交并返回交点个数 

180 int getLineCircleIntersection(Line L, Circle C, double& t1, double& t2, vector<Point>& sol)

181 {

182     double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;

183     double e = a*a + c*c, f = 2*(a*b + c*d), g = b*b + d*d - C.r*C.r;

184     double delta = f*f - 4*e*g;        //判别式

185     if(dcmp(delta) < 0)    return 0;    //相离

186     if(dcmp(delta) == 0)            //相切

187     {

188         t1 = t2 = -f / (2 * e);

189         sol.push_back(L.point(t1));

190         return 1;

191     }

192     //相交

193     t1 = (-f - sqrt(delta)) / (2 * e);    sol.push_back(L.point(t1));

194     t2 = (-f + sqrt(delta)) / (2 * e);    sol.push_back(L.point(t2));

195     return 2;

196 }

197 

198 //计算向量极角

199 double angle(Vector v)    { return atan2(v.y, v.x); }

200 

201 int getCircleCircleIntersection(Circle C1, Circle C2, vector<Point>& sol)

202 {//圆与圆相交，并返回交点个数

203     double d = Length(C1.c - C2.c);

204     if(dcmp(d) == 0)

205     {

206         if(dcmp(C1.r - C2.r) == 0)    return -1;    //两圆重合

207         return 0;                                //没有交点

208     }

209     if(dcmp(C1.r + C2.r - d) > 0)    return 0;

210     if(dcmp(fabs(C1.r - C2.r) - d) > 0)    return 0;

211 

212     double a = angle(C2.c - C1.c);

213     double da = acos((C1.r*C1.r + d*d - C2.r*C2.r) / (2*C1.r*d));

214     Point p1 = C1.point(a+da), p2 = C1.point(a-da);

215     sol.push_back(p1);

216     if(p1 == p2)    return 1;

217     sol.push_back(p2);

218     return 2;

219 }

220 

221 //过定点作圆的切线并返回切线条数

222 int getTangents(Point p, Circle C, Vector* v)

223 {

224     Vector u = C.c - p;

225     double dist = Length(u);

226     if(dist < C.r)    return 0;

227     else if(dcmp(dist - C.r) == 0)

228     {

229         v[0] = VRotate(u, PI/2);

230         return 1;

231     }

232     else

233     {

234         double ang = asin(C.r / dist);

235         v[0] = VRotate(u, +ang);

236         v[1] = VRotate(u, -ang);

237         return 2;

238     }

239 }

240 

241 //求两个圆的公切线，并返回切线条数

242 //注意，这里的Circle和上面的定义的Circle不一样

243 int getTangents(Circle A, Circle B, Point* a, Point* b)

244 {

245     int cnt = 0;

246     if(A.r < B.r)    { swap(A, B); swap(a, b); }

247     double d2 = (A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y);

248     double rdiff = A.r - B.r;

249     double rsum = A.r + B.r;

250     if(d2 < rdiff*rdiff)    return 0;    //内含

251 

252     double base = atan2(B.y-A.y, B.x-A.x);

253     if(dcmp(d2) == 0 && dcmp(A.r - B.r) == 0)    return -1; //重合

254     if(dcmp(d2 - rdiff*rdiff) == 0)    //内切

255     {

256         a[cnt] = A.point(base); b[cnt] = B.point(base); cnt++;

257         return 1;

258     }

259 

260     //有外公切线

261     double ang = acos((A.r - B.r) / sqrt(d2));

262     a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++;

263     a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++;

264     if(dcmp(rsum*rsum - d2) == 0)

265     {//外切

266         a[cnt] = b[cnt] = A.point(base); cnt++;

267     }

268     else if(dcmp(d2 - rsum*rsum) > 0)

269     {

270         ang = acos((A.r + B.r) / sqrt(d2));

271         a[cnt] = A.point(base + ang); b[cnt] = B.point(PI + base + ang); cnt++;

272         a[cnt] = A.point(base - ang); b[cnt] = B.point(PI + base - ang); cnt++;

273     }

274     return cnt;

275 }

276 

277 //转角发判定点P是否在多边形内部

278 int isPointInPolygon(Point P, Point* Poly, int n)

279 {

280     int wn;

281     for(int i = 0; i < n; ++i)

282     {

283         if(OnSegment(P, Poly[i], Poly[(i+1)%n]))    return -1;    //在边界上

284         int k = dcmp(Cross(Poly[(i+1)%n] - Poly[i], P - Poly[i]));

285         int d1 = dcmp(Poly[i].y - P.y);

286         int d2 = dcmp(Poly[(i+1)%n].y - P.y);

287         if(k > 0 && d1 <= 0 && d2 > 0)    wn++;

288         if(k < 0 && d2 <= 0 && d1 > 0)    wn--;

289     }

290     if(wn != 0)    return 1;    //内部

291     return 0;                //外部

292 }

293 

294 //计算凸包，输入点数组P，个数为n，输出点数组ch。函数返回凸包顶点数。

295 //输入不能有重复点，函数执行后点的顺序会发生变化

296 //如果不希望凸包的边上有输入点，把两个 <= 改成 <

297 //在精度要求高时，可用dcmp比较

298 int ConvexHull(Point* p, int n, Point* ch)

299 {

300     sort(p, p +n);

301     int m = 0;

302     for(int i = 0; i < n; ++i)

303     {

304         while(m > 1 && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) <= 0)    m--;

305         ch[m++] = p[i];

306     }

307     int k = m;

308     for(int i = n-2; i >= 0; --i)

309     {

310         while(m > k && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) <= 0)    m--;

311         ch[m++] = p[i];

312     }

313     if(n > 1)    m--;

314     return m;

315 }

 

旋转卡壳的模板：

int diameter2(vector<Point>& points)

{

    vector<Point> p = ConvexHull(points);

    int n = p.size();

    //for(int i = 0; i < n; ++i)    printf("%d %d\n", p[i].x, p[i].y);

    if(n == 1)    return 0;

    if(n == 2)  return Dist2(p[0], p[1]);

    p.push_back(p[0]);

    int ans = 0;

    for(int u = 0, v = 1; u < n; ++u)

    {// 一条直线贴住边p[u]-p[u+1]

        while(true)

        {

            // 当Area(p[u], p[u+1], p[v+1]) <= Area(p[u], p[u+1], p[v])时停止旋转

            //因为两个三角形有一公共边，所以面积大的那个点到直线距离大 

            // 即Cross(p[u+1]-p[u], p[v+1]-p[u]) - Cross(p[u+1]-p[u], p[v]-p[u]) <= 0

            // 根据Cross(A,B) - Cross(A,C) = Cross(A,B-C)

            // 化简得Cross(p[u+1]-p[u], p[v+1]-p[v]) <= 0

            int diff = Cross(p[u+1]-p[u], p[v+1]-p[v]);

            if(diff <= 0)

            {

                ans = max(ans, Dist2(p[u], p[v]));

                if(diff == 0)    ans = max(ans, Dist2(p[u], p[v+1]));

                break;

            } 

            v = (v+1)%n;

        }

    }

    return ans;

}