2. 直线与线段
2.0 预备函数
// 结构定义与宏定义 #include<stdio.h> #include<string.h> #include<stdlib.h> #include <math.h> #define eps 1e-8 #define zero(x) (((x)>0?(x):-(x))<eps) struct point { double x,y; }; struct line { point a,b; }; // 计算 s cross t product (P1-P0)x(P2-P0) double xmult(point p1,point p2,point p0) { return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); } double xmult(double x1,double y1,double x2,double y2,double x0,double y0) { return (x1-x0)*(y2-y0)-(x2-x0)*(y1-y0); } // 计算 t dot t product (P1-P0).(P2-P0) double dmult(point p1,point p2,point p0) { return (p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y); } double dmult(double x1,double y1,double x2,double y2,double x0,double y0) { return (x1-x0)*(x2-x0)+(y1-y0)*(y2-y0); } // 两点距离 double distance(point p1,point p2) { return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); } double distance(double x1,double y1,double x2,double y2) { return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); }
2.1 判三点是否共线
int dots_inline(point p1,point p2,point p3) { return zero(xmult(p1,p2,p3)); }
2.2 判点是否在线段上
// 判点是否在线段上, , , , 包括端点（下面为两种接口模式） int dot_online_in(point p,line l) { return zero(xmult(p,l.a,l.b))&&(l.a.x-p.x)*(l.b.x-p.x)<eps&&(l.a.y-p.y)*(l.b.y-p.y)<eps; } int dot_online_in(point p,point l1,point l2) { return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps; } // 判点是否在线段上, , , , 不包括端点 int dot_online_ex(point p,line l) { return dot_online_in(p,l)&&(!zero(p.x-l.a.x)||!zero(p.y-l.a.y)) &&(!zero(p.x-l.b.x)||!zero(p.y-l.b.y)); }
2.3 判断两点在线段的同一侧
// 判两点在线段同侧, , , , 点在线段上返回 0 0 0 0 int same_side(point p1,point p2,line l) { return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)>eps; } int same_side(point p1,point p2,point l1,point l2) { return xmult(l1,p1,l2)*xmult(l1,p2,l2)>eps; }
2.4 判断两点是否在线段的异侧
// 判两点在线段异侧,点在线段上返回 0 int opposite_side(point p1,point p2,line l) { return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)<-eps; } int opposite_side(point p1,point p2,point l1,point l2) { return xmult(l1,p1,l2)*xmult(l1,p2,l2)<-eps; }
2.5 求点关于直线的对称点
// 点 点关于直线的对称点 / / / // y by lyt // 缺点：用了斜率 // 也可以利用" " " " 点到直线上的最近点" " " " 来做，避免使用斜率。 point symmetric_point(point p1, point l1, point l2) { point ret; if (l1.x > l2.x - eps && l1.x < l2.x + eps) { ret.x = (2 * l1.x - p1.x); ret.y = p1.y; } else { double k = (l1.y - l2.y ) / (l1.x - l2.x); ret.x = (2*k*k*l1.x + 2*k*p1.y - 2*k*l1.y - k*k*p1.x + p1.x) / (1 + k*k); ret.y = p1.y - (ret.x - p1.x ) / k; } return ret; }
2.7 判断两线段是否相交
2.7.1 常用版
//定义点 struct Point { double x; double y; }; typedef struct Point point; //叉积 double multi(point p0, point p1, point p2) { return ( p1.x - p0.x )*( p2.y - p0.y )-( p2.x - p0.x )*( p1.y - p0.y ); } //相交返回 true,否则为 false, 接口为两线段的端点 bool isIntersected(point s1,point e1, point s2,point e2) { return (max(s1.x,e1.x) >= min(s2.x,e2.x)) && (max(s2.x,e2.x) >= min(s1.x,e1.x)) && (max(s1.y,e1.y) >= min(s2.y,e2.y)) && (max(s2.y,e2.y) >= min(s1.y,e1.y)) && (multi(s1,s2,e1)*multi(s1,e1,e2)>0) && (multi(s2,s1,e2)*multi(s2,e2,e1)>0); }
2.7.2 不常用版
// 判两线段相交,包括端点和部分重合 int intersect_in(line u,line v) { if (!dots_inline(u.a,u.b,v.a)||!dots_inline(u.a,u.b,v.b)) return !same_side(u.a,u.b,v)&&!same_side(v.a,v.b,u); return dot_online_in(u.a,v)||dot_online_in(u.b,v)||dot_online_in(v.a,u)||dot_online_in(v.b,u); } int intersect_in(point u1,point u2,point v1,point v2) { if (!dots_inline(u1,u2,v1)||!dots_inline(u1,u2,v2)) return !same_side(u1,u2,v1,v2)&&!same_side(v1,v2,u1,u2); return dot_online_in(u1,v1,v2)||dot_online_in(u2,v1,v2)||dot_online_in(v1,u1,u2)||dot_online_in(v2,u1,u 2); } // 判两线段相交,不包括端点和部分重合 int intersect_ex(line u,line v) { return opposite_side(u.a,u.b,v)&&opposite_side(v.a,v.b,u); } int intersect_ex(point u1,point u2,point v1,point v2) { return opposite_side(u1,u2,v1,v2)&&opposite_side(v1,v2,u1,u2); }
2.8 求两条直线的交点
// 计算两直线交点, , , , 注意事先判断直线是否平行 ! ! ! ! // 线段交点请另外判线段相交( ( ( ( 同时还是要判断是否平行) !) point intersection(point u1,point u2,point v1,point v2) { point ret=u1; double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x)) /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x)); ret.x+=(u2.x-u1.x)*t; ret.y+=(u2.y-u1.y)*t; return ret; }
2.9 点到直线的最近距离
point ptoline(point p,point l1,point l2) { point t=p; t.x+=l1.y-l2.y,t.y+=l2.x-l1.x; return intersection(p,t,l1,l2); }
2.10 点到线段的最近距离
point ptoseg(point p,point l1,point l2) { point t=p; t.x+=l1.y-l2.y,t.y+=l2.x-l1.x; if (xmult(l1,t,p)*xmult(l2,t,p)>eps) return distance(p,l1)<distance(p,l2)?l1:l2; return intersection(p,t,l1,l2); }
3. 多边形
3.0 预备浮点函数
#include <stdlib.h> #include<stdio.h> #include<string.h> #include <math.h> #define MAXN 1000 //offset 为多变形坐标的最大绝对值 #define offset 10000 #define eps 1e-8 // 浮点数判 0 0 0 0 #define zero(x) (((x)>0?(x):-(x))<eps) // 浮点数判断符 #define _sign(x) ((x)>eps?1:((x)<-eps?2:0)) // 定义点 struct point { double x,y; } pt[MAXN ]; // 定义线段 struct line { point a,b; }; // 叉积 double xmult(point p1,point p2,point p0) { return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); }
3.1 判定是否是凸多边形
// 判定凸多边形,顶点按顺时针或逆时针给出, 允许相邻边共线, 是凸多边形返回 1 否则返回 0 int is_convex(int n,point* p) { int i,s[3]= {1,1,1}; for (i=0; i<n&&s[1]|s[2]; i++) s[_sign(xmult(p[(i+1)%n],p[(i+2)%n],p[i]))]=0; return s[1]|s[2]; } // 判凸行，顶点按顺时针或逆时针给出, 不允许相邻边共线, 是凸多边形返回 1 ， 否则返回 0 int is_convex_v2(int n,point* p) { int i,s[3]= {1,1,1}; for (i=0; i<n&&s[0]&&s[1]|s[2]; i++) s[_sign(xmult(p[(i+1)%n],p[(i+2)%n],p[i]))]=0; return s[0]&&s[1]|s[2]; }
3.2 判定点是否在多边形内
// 判点在凸多边形内或多边形边上时返回 1 1 1 1 ，严格在凸多边形外返回 0 0 0 0 int inside_convex(point q,int n,point* p) { int i,s[3]= {1,1,1}; for (i=0; i<n&&s[1]|s[2]; i++) s[_sign(xmult(p[(i+1)%n],q,p[i]))]=0; return s[1]|s[2]; } / / / // 判点严格在凸多边形内返回 , 1, 在边上或者严格在外返回 0 0 0 0 int inside_convex_v2(point q,int n,point* p) { int i,s[3]= {1,1,1}; for (i=0; i<n&&s[0]&&s[1]|s[2]; i++) s[_sign(xmult(p[(i+1)%n],q,p[i]))]=0; return s[0]&&s[1]|s[2]; } // 判点在任意多边形内, 顶点按顺时针或逆时针给出 //on_edge 表示点在多边形边上时的返回值,offset 为多边形坐标上限,严格在内返回1 ，严格在外返回0 int inside_polygon(point q,int n,point* p,int on_edge=2) { point q2; int i=0,count; while (i<n) for (count=i=0,q2.x=rand()+offset,q2.y=rand()+offset; i<n; i++) { if (zero(xmult(q,p[i],p[(i+1)%n]))&&(p[i].x-q.x)*(p[(i+1)%n].x-q.x)<eps &&(p[i].y-q.y)*(p[(i+1)%n].y-q.y)<eps) return on_edge; else if (zero(xmult(q,q2,p[i]))) break; else if (xmult(q,p[i],q2)*xmult(q,p[(i+1)%n],q2)<-eps&& xmult(p[i],q,p[(i+1)%n])*xmult(p[i],q2,p[(i+1)%n])<-eps) count++; } return count&1; }
3.3 判定一条线段是否在一个任意多边形内
// 预备函数 inline int opposite_side(point p1,point p2,point l1,point l2) { return xmult(l1,p1,l2)*xmult(l1,p2,l2)<-eps; } inline int dot_online_in(point p,point l1,point l2) { return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps; } //判线段在任意多边形内,顶点按顺时针或逆时针给出,与边界相交返回1 int inside_polygon(point l1,point l2,int n,point* p) { point t[MAXN],tt; int i,j,k=0; if (!inside_polygon(l1,n,p)||!inside_polygon(l2,n,p)) return 0; for (i=0; i<n; i++) { if (opposite_side(l1,l2,p[i],p[(i+1)%n])&&opposite_side(p[i],p[(i+1)%n],l1,l2)) return 0; else if (dot_online_in(l1,p[i],p[(i+1)%n])) t[k++]=l1; else if (dot_online_in(l2,p[i],p[(i+1)%n])) t[k++]=l2; else if (dot_online_in(p[i],l1,l2)) t[k++]=p[i]; } for (i=0; i<k; i++) for (j=i+1; j<k; j++) { tt.x=(t[i].x+t[j].x)/2; tt.y=(t[i].y+t[j].y)/2; if (!inside_polygon(tt,n,p)) return 0; } return 1; }
. 4. 三角形
4.0 预备函数
#include <math.h> #include <string.h> #include <stdlib.h> #include<stdio.h> / / / // 定义点 struct point { double x,y; }; typedef struct point point; // 定义直线 struct line { point a,b; }; typedef struct line line; // 两点距离 double distance(point p1,point p2) { return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); } // 两直线求交点 point intersection(line u,line v) { point ret=u.a; double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x)) /((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x)); ret.x+=(u.b.x-u.a.x)*t; ret.y+=(u.b.y-u.a.y)*t; return ret; }
4.1 求三角形的外心
point circumcenter(point a,point b,point c) { line u,v; u.a.x=(a.x+b.x)/2; u.a.y=(a.y+b.y)/2; u.b.x=u.a.x-a.y+b.y; u.b.y=u.a.y+a.x-b.x; v.a.x=(a.x+c.x)/2; v.a.y=(a.y+c.y)/2; v.b.x=v.a.x-a.y+c.y; v.b.y=v.a.y+a.x-c.x; return intersection(u,v); }
4.2 求三角形内心
point incenter(point a,point b,point c) { line u,v; double m,n; u.a=a; m=atan2(b.y-a.y,b.x-a.x); n=atan2(c.y-a.y,c.x-a.x); u.b.x=u.a.x+cos((m+n)/2); u.b.y=u.a.y+sin((m+n)/2); v.a=b; m=atan2(a.y-b.y,a.x-b.x); n=atan2(c.y-b.y,c.x-b.x); v.b.x=v.a.x+cos((m+n)/2); v.b.y=v.a.y+sin((m+n)/2); return intersection(u,v); }
4.3 求三角形垂心
point perpencenter(point a,point b,point c) { line u,v; u.a=c; u.b.x=u.a.x-a.y+b.y; u.b.y=u.a.y+a.x-b.x; v.a=b; v.b.x=v.a.x-a.y+c.y; v.b.y=v.a.y+a.x-c.x; return intersection(u,v); }
. 5. 圆
5.0 预备函数
#include <math.h> #include <stdlib.h> #include <stdio.h> #include <string.h> #define eps 1e-8 struct point { double x,y; }; typedef struct point point; double xmult(point p1,point p2,point p0) { return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); } double distance(point p1,point p2) { return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); } // 点到直线的距离 double disptoline(point p,point l1,point l2) { return fabs(xmult(p,l1,l2))/distance(l1,l2); } // 求两直线交点 point intersection(point u1,point u2,point v1,point v2) { point ret=u1; double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x)) /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x)); ret.x+=(u2.x-u1.x)*t; ret.y+=(u2.y-u1.y)*t; return ret; }
5.1 判定直线是否与圆相交
//判直线和圆相交,包括相切 int intersect_line_circle(point c,double r,point l1,point l2) { return disptoline(c,l1,l2)<r+eps; }
5.2 判定线段与圆相交
int intersect_seg_circle(point c,double r, point l1,point l2) { double t1=distance(c,l1)-r,t2=distance(c,l2)-r; point t=c; if (t1<eps||t2<eps) return t1>-eps||t2>-eps; t.x+=l1.y-l2.y; t.y+=l2.x-l1.x; return xmult(l1,c,t)*xmult(l2,c,t)<eps&&disptoline(c,l1,l2)-r<eps; }
5.3 判圆和圆相交
int intersect_circle_circle(point c1,double r1,point c2,double r2) { return distance(c1,c2)<r1+r2+eps&&distance(c1,c2)>fabs(r1-r2)-eps; }
5.4 计算圆上到点p最近点
//当p为圆心时，返回圆心本身 point dot_to_circle(point c,double r,point p) { point u,v; if (distance(p,c)<eps) return p; u.x=c.x+r*fabs(c.x-p.x)/distance(c,p); u.y=c.y+r*fabs(c.y-p.y)/distance(c,p)*((c.x-p.x)*(c.y-p.y)<0?-1:1); v.x=c.x-r*fabs(c.x-p.x)/distance(c,p); v.y=c.y-r*fabs(c.y-p.y)/distance(c,p)*((c.x-p.x)*(c.y-p.y)<0?-1:1); return distance(u,p)<distance(v,p)?u:v; }
5.5 计算直线与圆的交点
// 计算直线与圆的交点,保证直线与圆有交点 // 计算线段与圆的交点可用这个函数后判点是否在线段上 void intersection_line_circle(point c,double r,point l1,point l2,point& p1,point& p2) { point p=c; double t; p.x+=l1.y-l2.y; p.y+=l2.x-l1.x; p=intersection(p,c,l1,l2); t=sqrt(r*r-distance(p,c)*distance(p,c))/distance(l1,l2); p1.x=p.x+(l2.x-l1.x)*t; p1.y=p.y+(l2.y-l1.y)*t; p2.x=p.x-(l2.x-l1.x)*t; p2.y=p.y-(l2.y-l1.y)*t; }
5.6 计算两个圆的交点
// 计算圆与圆的交点,保证圆与圆有交点,圆心不重合 void intersection_circle_circle(point c1,double r1,point c2,double r2,point& p1,point& p2) { point u,v; double t; t=(1+(r1*r1-r2*r2)/distance(c1,c2)/distance(c1,c2))/2; u.x=c1.x+(c2.x-c1.x)*t; u.y=c1.y+(c2.y-c1.y)*t; v.x=u.x+c1.y-c2.y; v.y=u.y-c1.x+c2.x; intersection_line_circle(c1,r1,u,v,p1,p2); }
. 6. 球面
6.0 给出地球经度纬度，计算圆心角
#include <math.h> const double pi=acos(-1); // 计算圆心角lat表示纬度,-90<=w<=90,lng 表示经度 // 返回两点所在大圆劣弧对应圆心角 ,0<=angle<=pi double angle(double lng1,double lat1,double lng2,double lat2) { double dlng=fabs(lng1-lng2)*pi/180; while (dlng>=pi+pi) dlng-=pi+pi; if (dlng>pi) dlng=pi+pi-dlng; lat1*=pi/180,lat2*=pi/180; return acos(cos(lat1)*cos(lat2)*cos(dlng)+sin(lat1)*sin(lat2)); }
6.1 已知经纬度，计算地球上两点直线距离
// 计算距离,r为球半径 double line_dist(double r,double lng1,double lat1,double lng2,double lat2) { double dlng=fabs(lng1-lng2)*pi/180; while (dlng>=pi+pi) dlng-=pi+pi; if (dlng>pi) dlng=pi+pi-dlng; lat1*=pi/180,lat2*=pi/180; return r*sqrt(2-2*(cos(lat1)*cos(lat2)*cos(dlng)+sin(lat1)*sin(lat2))); }
6.2 已知经纬度，计算地球上两点球面距离
// 计算球面距离,r为球半径 inline double sphere_dist(double r,double lng1,double lat1,double lng2,double lat2) { return r*angle(lng1,lat1,lng2,lat2); }
. 7. 三维几何的若干模板
7.0 预备函数
// 三维几何函数库 #include <math.h> #define eps 1e-8 #define zero(x) (((x)>0?(x):-(x))<eps) struct point3 { double x,y,z; }; struct line3 { point3 a,b; }; struct plane3 { point3 a,b,c; }; // 计算cross product U x V point3 xmult(point3 u,point3 v) { point3 ret; ret.x=u.y*v.z-v.y*u.z; ret.y=u.z*v.x-u.x*v.z; ret.z=u.x*v.y-u.y*v.x; return ret; } // 计算dot product U . V double dmult(point3 u,point3 v) { return u.x*v.x+u.y*v.y+u.z*v.z; } //差 矢量差 U-V point3 subt(point3 u,point3 v) { point3 ret; ret.x=u.x-v.x; ret.y=u.y-v.y; ret.z=u.z-v.z; return ret; } // 取平面法向量 point3 pvec(plane3 s) { return xmult(subt(s.a,s.b),subt(s.b,s.c)); } point3 pvec(point3 s1,point3 s2,point3 s3) { return xmult(subt(s1,s2),subt(s2,s3)); } //两点距离,单参数取向量大小 double distance(point3 p1,point3 p2) { return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)+(p1.z-p2.z)*(p1.z-p2.z)); } // 向量大小 double vlen(point3 p) { return sqrt(p.x*p.x+p.y*p.y+p.z*p.z); }
7.1 判定三点是否共线
// 判三点共线 int dots_inline(point3 p1,point3 p2,point3 p3) { return vlen(xmult(subt(p1,p2),subt(p2,p3)))<eps; }
7.2 判定四点是否共面
// 判四点共面 int dots_onplane(point3 a,point3 b,point3 c,point3 d) { return zero(dmult(pvec(a,b,c),subt(d,a))); }
7.1 判定点是否在线段上
// 判点是否在线段上,包括端点和共线 int dot_online_in(point3 p,line3 l) { return zero(vlen(xmult(subt(p,l.a),subt(p,l.b))))&&(l.a.x-p.x)*(l.b.x-p.x)<eps&& (l.a.y-p.y)*(l.b.y-p.y)<eps&&(l.a.z-p.z)*(l.b.z-p.z)<eps; } int dot_online_in(point3 p,point3 l1,point3 l2) { return zero(vlen(xmult(subt(p,l1),subt(p,l2))))&&(l1.x-p.x)*(l2.x-p.x)<eps&& (l1.y-p.y)*(l2.y-p.y)<eps&&(l1.z-p.z)*(l2.z-p.z)<eps; } // 判点是否在线段上,不包括端点 int dot_online_ex(point3 p,line3 l) { return dot_online_in(p,l)&&(!zero(p.x-l.a.x)||!zero(p.y-l.a.y)||!zero(p.z-l.a.z))&& (!zero(p.x-l.b.x)||!zero(p.y-l.b.y)||!zero(p.z-l.b.z)); } int dot_online_ex(point3 p,point3 l1,point3 l2) { return dot_online_in(p,l1,l2)&&(!zero(p.x-l1.x)||!zero(p.y-l1.y)||!zero(p.z-l1.z))&& (!zero(p.x-l2.x)||!zero(p.y-l2.y)||!zero(p.z-l2.z)); }
7.2 判断点是否在空间三角形上
// 判点是否在空间三角形上,包括边界,三点共线无意义 int dot_inplane_in(point3 p,plane3 s) { return zero(vlen(xmult(subt(s.a,s.b),subt(s.a,s.c)))-vlen(xmult(subt(p,s.a),subt(p,s.b)))- vlen(xmult(subt(p,s.b),subt(p,s.c)))-vlen(xmult(subt(p,s.c),subt(p,s.a)))); } int dot_inplane_in(point3 p,point3 s1,point3 s2,point3 s3) { return zero(vlen(xmult(subt(s1,s2),subt(s1,s3)))-vlen(xmult(subt(p,s1),subt(p,s2)))- vlen(xmult(subt(p,s2),subt(p,s3)))-vlen(xmult(subt(p,s3),subt(p,s1)))); } // 判点是否在空间三角形上,不包括边界,三点共线无意义 int dot_inplane_ex(point3 p,plane3 s) { return dot_inplane_in(p,s)&&vlen(xmult(subt(p,s.a),subt(p,s.b)))>eps&& vlen(xmult(subt(p,s.b),subt(p,s.c)))>eps&&vlen(xmult(subt(p,s.c),subt(p,s.a)))>eps; } int dot_inplane_ex(point3 p,point3 s1,point3 s2,point3 s3) { return dot_inplane_in(p,s1,s2,s3)&&vlen(xmult(subt(p,s1),subt(p,s2)))>eps&& vlen(xmult(subt(p,s2),subt(p,s3)))>eps&&vlen(xmult(subt(p,s3),subt(p,s1)))>eps; }
7.3 判断两点是否在线段同侧
// 判两点在线段同侧,点在线段上返回0,不共面无意义 int same_side(point3 p1,point3 p2,line3 l) { return dmult(xmult(subt(l.a,l.b),subt(p1,l.b)),xmult(subt(l.a,l.b),subt(p2,l.b)))>eps; } int same_side(point3 p1,point3 p2,point3 l1,point3 l2) { return dmult(xmult(subt(l1,l2),subt(p1,l2)),xmult(subt(l1,l2),subt(p2,l2)))>eps; }
7.4 判断两点是否在线段异侧
//判两点在线段异侧,点在线段上返回0,不共面无意义 int opposite_side(point3 p1,point3 p2,line3 l) { return dmult(xmult(subt(l.a,l.b),subt(p1,l.b)),xmult(subt(l.a,l.b),subt(p2,l.b)))<-eps; } int opposite_side(point3 p1,point3 p2,point3 l1,point3 l2) { return dmult(xmult(subt(l1,l2),subt(p1,l2)),xmult(subt(l1,l2),subt(p2,l2)))<-eps; }
7.5 判断两点是否在平面同侧
// 判两点在平面同侧,点在平面上返回0 int same_side(point3 p1,point3 p2,plane3 s) { return dmult(pvec(s),subt(p1,s.a))*dmult(pvec(s),subt(p2,s.a))>eps; } int same_side(point3 p1,point3 p2,point3 s1,point3 s2,point3 s3) { return dmult(pvec(s1,s2,s3),subt(p1,s1))*dmult(pvec(s1,s2,s3),subt(p2,s1))>eps; }
7.6 判断两点是否在平面异侧
// 判两点在平面异侧,点在平面上返回0 int opposite_side(point3 p1,point3 p2,plane3 s) { return dmult(pvec(s),subt(p1,s.a))*dmult(pvec(s),subt(p2,s.a))<-eps; } int opposite_side(point3 p1,point3 p2,point3 s1,point3 s2,point3 s3) { return dmult(pvec(s1,s2,s3),subt(p1,s1))*dmult(pvec(s1,s2,s3),subt(p2,s1))<-eps; }
7.7 判断两空间直线是否平行
// 判两直线平行 int parallel(line3 u,line3 v) { return vlen(xmult(subt(u.a,u.b),subt(v.a,v.b)))<eps; } int parallel(point3 u1,point3 u2,point3 v1,point3 v2) { return vlen(xmult(subt(u1,u2),subt(v1,v2)))<eps; }
7.8 判断两平面是否平行
// 判两平面平行 int parallel(plane3 u,plane3 v) { return vlen(xmult(pvec(u),pvec(v)))<eps; } int parallel(point3 u1,point3 u2,point3 u3,point3 v1,point3 v2,point3 v3) { return vlen(xmult(pvec(u1,u2,u3),pvec(v1,v2,v3)))<eps; }
7.9 判断直线是否与平面平行
// 判直线与平面平行 int parallel(line3 l,plane3 s) { return zero(dmult(subt(l.a,l.b),pvec(s))); } int parallel(point3 l1,point3 l2,point3 s1,point3 s2,point3 s3) { return zero(dmult(subt(l1,l2),pvec(s1,s2,s3))); }
7.10 判断两直线是否垂直
// 判两直线垂直 int perpendicular(line3 u,line3 v) { return zero(dmult(subt(u.a,u.b),subt(v.a,v.b))); } int perpendicular(point3 u1,point3 u2,point3 v1,point3 v2) { return zero(dmult(subt(u1,u2),subt(v1,v2))); }
7.11 判断两平面是否垂直
//判两平面垂直 int perpendicular(plane3 u,plane3 v) { return zero(dmult(pvec(u),pvec(v))); } int perpendicular(point3 u1,point3 u2,point3 u3,point3 v1,point3 v2,point3 v3) { return zero(dmult(pvec(u1,u2,u3),pvec(v1,v2,v3))); }
7.12 判断两条空间线段是否相交
// 判两线段相交,包括端点和部分重合 int intersect_in(line3 u,line3 v) { if (!dots_onplane(u.a,u.b,v.a,v.b)) return 0; if (!dots_inline(u.a,u.b,v.a)||!dots_inline(u.a,u.b,v.b)) return !same_side(u.a,u.b,v)&&!same_side(v.a,v.b,u); return dot_online_in(u.a,v)||dot_online_in(u.b,v)||dot_online_in(v.a,u)||dot_online_in(v.b,u); } int intersect_in(point3 u1,point3 u2,point3 v1,point3 v2) { if (!dots_onplane(u1,u2,v1,v2)) return 0; if (!dots_inline(u1,u2,v1)||!dots_inline(u1,u2,v2)) return !same_side(u1,u2,v1,v2)&&!same_side(v1,v2,u1,u2); return dot_online_in(u1,v1,v2)||dot_online_in(u2,v1,v2)||dot_online_in(v1,u1,u2)||dot_online_in(v2,u1,u2); } // 判两线段相交,不包括端点和部分重合 int intersect_ex(line3 u,line3 v) { return dots_onplane(u.a,u.b,v.a,v.b)&&opposite_side(u.a,u.b,v)&&opposite_side(v.a,v.b,u); } int intersect_ex(point3 u1,point3 u2,point3 v1,point3 v2) { return dots_onplane(u1,u2,v1,v2)&&opposite_side(u1,u2,v1,v2)&&opposite_side(v1,v2,u1,u2); }
7.13 判断线段是否与空间三角形相交
// 判线段与空间三角形相交,包括交于边界和(部分)包含 int intersect_in(line3 l,plane3 s) { return !same_side(l.a,l.b,s)&&!same_side(s.a,s.b,l.a,l.b,s.c)&& !same_side(s.b,s.c,l.a,l.b,s.a)&&!same_side(s.c,s.a,l.a,l.b,s.b); } int intersect_in(point3 l1,point3 l2,point3 s1,point3 s2,point3 s3) { return !same_side(l1,l2,s1,s2,s3)&&!same_side(s1,s2,l1,l2,s3)&& !same_side(s2,s3,l1,l2,s1)&&!same_side(s3,s1,l1,l2,s2); } // 判线段与空间三角形相交,不包括交于边界和(部分)包含 int intersect_ex(line3 l,plane3 s) { return opposite_side(l.a,l.b,s)&&opposite_side(s.a,s.b,l.a,l.b,s.c)&& opposite_side(s.b,s.c,l.a,l.b,s.a)&&opposite_side(s.c,s.a,l.a,l.b,s.b); } int intersect_ex(point3 l1,point3 l2,point3 s1,point3 s2,point3 s3) { return opposite_side(l1,l2,s1,s2,s3)&&opposite_side(s1,s2,l1,l2,s3)&& opposite_side(s2,s3,l1,l2,s1)&&opposite_side(s3,s1,l1,l2,s2); }
7.14 计算两条直线的交点
// 计算两直线交点,注意事先判断直线是否共面和平行! // 线段交点请另外判线段相交(同时还是要判断是否平行!) point3 intersection(line3 u,line3 v) { point3 ret=u.a; double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x)) /((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x)); ret.x+=(u.b.x-u.a.x)*t; ret.y+=(u.b.y-u.a.y)*t; ret.z+=(u.b.z-u.a.z)*t; return ret; } point3 intersection(point3 u1,point3 u2,point3 v1,point3 v2) { point3 ret=u1; double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x)) /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x)); ret.x+=(u2.x-u1.x)*t; ret.y+=(u2.y-u1.y)*t; ret.z+=(u2.z-u1.z)*t; return ret; }
7.15 计算直线与平面的交点
// 计算直线与平面交点,注意事先判断是否平行,并保证三点不共线! // 线段和空间三角形交点请另外判断 point3 intersection(line3 l,plane3 s) { point3 ret=pvec(s); double t=(ret.x*(s.a.x-l.a.x)+ret.y*(s.a.y-l.a.y)+ret.z*(s.a.z-l.a.z))/ (ret.x*(l.b.x-l.a.x)+ret.y*(l.b.y-l.a.y)+ret.z*(l.b.z-l.a.z)); ret.x=l.a.x+(l.b.x-l.a.x)*t; ret.y=l.a.y+(l.b.y-l.a.y)*t; ret.z=l.a.z+(l.b.z-l.a.z)*t; return ret; } point3 intersection(point3 l1,point3 l2,point3 s1,point3 s2,point3 s3) { point3 ret=pvec(s1,s2,s3); double t=(ret.x*(s1.x-l1.x)+ret.y*(s1.y-l1.y)+ret.z*(s1.z-l1.z))/ (ret.x*(l2.x-l1.x)+ret.y*(l2.y-l1.y)+ret.z*(l2.z-l1.z)); ret.x=l1.x+(l2.x-l1.x)*t; ret.y=l1.y+(l2.y-l1.y)*t; ret.z=l1.z+(l2.z-l1.z)*t; return ret; }
7.16 计算两平面的交线
// 计算两平面交线,注意事先判断是否平行,并保证三点不共线! line3 intersection(plane3 u,plane3 v) { line3 ret; ret.a=parallel(v.a,v.b,u.a,u.b,u.c)?intersection(v.b,v.c,u.a,u.b,u.c):intersection(v.a,v.b,u.a,u.b,u.c); ret.b=parallel(v.c,v.a,u.a,u.b,u.c)?intersection(v.b,v.c,u.a,u.b,u.c):intersection(v.c,v.a,u.a,u.b,u.c); return ret; } line3 intersection(point3 u1,point3 u2,point3 u3,point3 v1,point3 v2,point3 v3) { line3 ret; ret.a=parallel(v1,v2,u1,u2,u3)?intersection(v2,v3,u1,u2,u3):intersection(v1,v2,u1,u2,u3); ret.b=parallel(v3,v1,u1,u2,u3)?intersection(v2,v3,u1,u2,u3):intersection(v3,v1,u1,u2,u3); return ret; }
7.17 点到直线的距离
// 点到直线距离 double ptoline(point3 p,line3 l) { return vlen(xmult(subt(p,l.a),subt(l.b,l.a)))/distance(l.a,l.b); } double ptoline(point3 p,point3 l1,point3 l2) { return vlen(xmult(subt(p,l1),subt(l2,l1)))/distance(l1,l2); }
7.18 计算点到平面的距离
// 点到平面距离 double ptoplane(point3 p,plane3 s) { return fabs(dmult(pvec(s),subt(p,s.a)))/vlen(pvec(s)); } double ptoplane(point3 p,point3 s1,point3 s2,point3 s3) { return fabs(dmult(pvec(s1,s2,s3),subt(p,s1)))/vlen(pvec(s1,s2,s3)); }
7.19 计算直线到直线的距离
// 直线到直线距离 double linetoline(line3 u,line3 v) { point3 n=xmult(subt(u.a,u.b),subt(v.a,v.b)); return fabs(dmult(subt(u.a,v.a),n))/vlen(n); } double linetoline(point3 u1,point3 u2,point3 v1,point3 v2) { point3 n=xmult(subt(u1,u2),subt(v1,v2)); return fabs(dmult(subt(u1,v1),n))/vlen(n); }
7.20 空间两直线夹角的cos值
// 两直线夹角cos值 double angle_cos(line3 u,line3 v) { return dmult(subt(u.a,u.b),subt(v.a,v.b))/vlen(subt(u.a,u.b))/vlen(subt(v.a,v.b)); } double angle_cos(point3 u1,point3 u2,point3 v1,point3 v2) { return dmult(subt(u1,u2),subt(v1,v2))/vlen(subt(u1,u2))/vlen(subt(v1,v2)); }
7.21 两平面夹角的cos值
// 两平面夹角cos值 double angle_cos(plane3 u,plane3 v) { return dmult(pvec(u),pvec(v))/vlen(pvec(u))/vlen(pvec(v)); } double angle_cos(point3 u1,point3 u2,point3 u3,point3 v1,point3 v2,point3 v3) { return dmult(pvec(u1,u2,u3),pvec(v1,v2,v3))/vlen(pvec(u1,u2,u3))/vlen(pvec(v1,v2,v3)); }
7.22 直线与平面夹角sin值
//直线平面夹角sin值 double angle_sin(line3 l,plane3 s) { return dmult(subt(l.a,l.b),pvec(s))/vlen(subt(l.a,l.b))/vlen(pvec(s)); } double angle_sin(point3 l1,point3 l2,point3 s1,point3 s2,point3 s3) { return dmult(subt(l1,l2),pvec(s1,s2,s3))/vlen(subt(l1,l2))/vlen(pvec(s1,s2,s3)); }
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1. 最远曼哈顿距离
#include <stdio.h> #define INF 9999999999999.0 struct Point { double x[5]; } pt[100005]; double dis[32][100005], coe[5], minx[32], maxx[32]; // 去掉绝对值后有 2^D种可能 void GetD(int N, int D) { int s, i, j, tot=(1<<D); for (s=0; s<tot; s++) { for (i=0; i<D; i++) if (s&(1<<i)) coe[i]=-1.0; else coe[i]=1.0; for (i=0; i<N; i++) { dis[s][i]=0.0; for (j=0; j<D; j++) dis[s][i]=dis[s][i]+coe[j]*pt[i].x[j]; } } } // 取每种可能中的最大差距 void Solve(int N, int D) { int s, i, tot=(1<<D); double tmp, ans; for (s=0; s<tot; s++) { minx[s]=INF; maxx[s]=-INF; for (i=0; i<N; i++) { if (minx[s]>dis[s][i]) minx[s]=dis[s][i]; if (maxx[s]<dis[s][i]) maxx[s]=dis[s][i]; } } ans=0.0; for (s=0; s<tot; s++) { tmp=maxx[s]-minx[s]; if (tmp>ans) ans=tmp; } printf("%.2lf ", ans); } int main (void) { int n, i; while (scanf("%d",&n)==1) { for (i=0; i<n; i++) scanf("%lf%lf%lf%lf%lf",&pt[i].x[0],&pt[i].x[1],&pt[i].x[2],&pt[i].x[3],&pt[i].x[4]); GetD(n, 5); Solve(n, 5); } return 0; }
. 2. 最近点对
#include <stdio.h> #include <math.h> #include <stdlib.h> #define Max(x,y) (x)>(y)?(x):(y) struct Q { double x, y; } q[100001], sl[10], sr[10]; int cntl, cntr, lm, rm; double ans; int cmp(const void*p1, const void*p2) { struct Q*a1=(struct Q*)p1; struct Q*a2=(struct Q*)p2; if (a1->x<a2->x)return -1; else if (a1->x==a2->x)return 0; else return 1; } double CalDis(double x1, double y1, double x2, double y2) { return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); } void MinDis(int l, int r) { if (l==r) return; double dis; if (l+1==r) { dis=CalDis(q[l].x,q[l].y,q[r].x,q[r].y); if (ans>dis) ans=dis; return; } int mid=(l+r)>>1, i, j; MinDis(l,mid); MinDis(mid+1,r); lm=mid+1-5; if (lm<l) lm=l; rm=mid+5; if (rm>r) rm=r; cntl=cntr=0; for (i=mid; i>=lm; i--) { if (q[mid+1].x-q[i].x>=ans)break; sl[++cntl]=q[i]; } for (i=mid+1; i<=rm; i++) { if (q[i].x-q[mid].x>=ans)break; sr[++cntr]=q[i]; } for (i=1; i<=cntl; i++) for (j=1; j<=cntr; j++) { dis=CalDis(sl[i].x,sl[i].y,sr[j].x,sr[j].y); if (dis<ans) ans=dis; } } int main (void) { int n, i; while (scanf("%d",&n)==1&&n) { for (i=1; i<=n; i++) scanf("%lf %lf", &q[i].x,&q[i].y); qsort(q+1,n,sizeof(struct Q),cmp); ans=CalDis(q[1].x,q[1].y,q[2].x,q[2].y); MinDis(1,n); printf("%.2lf ",ans/2.0); } return 0; }
. 4. 最小包围圆
#include<stdio.h> #include<string.h> #include<math.h> struct Point { double x; double y; } pt[1005]; struct Traingle { struct Point p[3]; }; struct Circle { struct Point center; double r; } ans; // 计算两点距离 double Dis(struct Point p, struct Point q) { double dx=p.x-q.x; double dy=p.y-q.y; return sqrt(dx*dx+dy*dy); } // 计算三角形面积 double Area(struct Traingle ct) { return fabs((ct.p[1].x-ct.p[0].x)*(ct.p[2].y-ct.p[0].y)-(ct.p[2].x-ct.p[0].x)*(ct.p[1].y-ct.p[0].y))/2.0; } // 求三角形的外接圆，返回圆心和半径( 存在结构体"圆"中) struct Circle CircumCircle(struct Traingle t) { struct Circle tmp; double a, b, c, c1, c2; double xA, yA, xB, yB, xC, yC; a = Dis(t.p[0], t.p[1]); b = Dis(t.p[1], t.p[2]); c = Dis(t.p[2], t.p[0]); //根据 S = a * b * c / R / 4;求半径 R tmp.r = (a*b*c)/(Area(t)*4.0); xA = t.p[0].x; yA = t.p[0].y; xB = t.p[1].x; yB = t.p[1].y; xC = t.p[2].x; yC = t.p[2].y; c1 = (xA*xA+yA*yA - xB*xB-yB*yB) / 2; c2 = (xA*xA+yA*yA - xC*xC-yC*yC) / 2; tmp.center.x = (c1*(yA - yC)-c2*(yA - yB)) / ((xA - xB)*(yA - yC)-(xA - xC)*(yA - yB)); tmp.center.y = (c1*(xA - xC)-c2*(xA - xB)) / ((yA - yB)*(xA - xC)-(yA - yC)*(xA - xB)); return tmp; } // 确定最小包围圆 struct Circle MinCircle(int num, struct Traingle ct) { struct Circle ret; if (num==0) ret.r = 0.0; else if (num==1) { ret.center = ct.p[0]; ret.r = 0.0; } else if (num==2) { ret.center.x = (ct.p[0].x+ct.p[1].x)/2.0; ret.center.y = (ct.p[0].y+ct.p[1].y)/2.0; ret.r = Dis(ct.p[0], ct.p[1])/2.0; } else if(num==3) ret = CircumCircle(ct); return ret; } // 递归实现增量算法 void Dfs(int x, int num, struct Traingle ct) { int i, j; struct Point tmp; ans = MinCircle(num, ct); if (num==3) return; for (i=1; i<=x; i++) if (Dis(pt[i], ans.center)>ans.r) { ct.p[num]=pt[i]; Dfs(i-1, num+1, ct); tmp=pt[i]; for (j=i; j>=2; j--) pt[j]=pt[j-1]; pt[1]=tmp; } } void Solve(int n) { struct Traingle ct; Dfs(n, 0, ct); } int main (void) { int n, i; while (scanf("%d", &n)!=EOF && n) { for (i=1; i<=n; i++) scanf("%lf %lf", &pt[i].x, &pt[i].y); Solve(n); printf("%.2lf %.2lf %.2lf ", ans.center.x, ans.center.y, ans.r); } return 0; }
5. 求两个圆的交点
#include<stdio.h> #include<string.h> #include<math.h> #include<stdlib.h> const double eps = 1e-8; const double PI = acos(-1.0); struct Point { double x; double y; }; typedef struct Point point; struct Line { double s, t; }; typedef struct Line Line; struct Circle { Point center; double r; Line line[505]; int cnt; bool covered; } circle[105]; double distance(point p1, point p2) { double dx = p1.x-p2.x; double dy = p1.y-p2.y; return sqrt(dx*dx + dy*dy); } point intersection(point u1,point u2, point v1,point v2) { point ret = u1; double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x)) / ((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x)); ret.x += (u2.x-u1.x)*t; ret.y += (u2.y-u1.y)*t; return ret; } void intersection_line_circle(point c,double r,point l1,point l2,point& p1,point& p2) { point p=c; double t; p.x+=l1.y-l2.y; p.y+=l2.x-l1.x; p=intersection(p,c,l1,l2); t=sqrt(r*r-distance(p,c)*distance(p,c))/distance(l1,l2); p1.x=p.x+(l2.x-l1.x)*t; p1.y=p.y+(l2.y-l1.y)*t; p2.x=p.x-(l2.x-l1.x)*t; p2.y=p.y-(l2.y-l1.y)*t; } // 计算圆与圆的交点,保证圆与圆有交点,圆心不重合 void intersection_circle_circle(point c1,double r1,point c2,double r2,point& p1,point& p2) { point u,v; double t; t=(1+(r1*r1-r2*r2)/distance(c1,c2)/distance(c1,c2))/2; u.x=c1.x+(c2.x-c1.x)*t; u.y=c1.y+(c2.y-c1.y)*t; v.x=u.x+c1.y-c2.y; v.y=u.y-c1.x+c2.x; intersection_line_circle(c1,r1,u,v,p1,p2); }
6. 求三角形外接圆圆心
struct Point { double x; double y; } pt[1005]; struct Traingle { struct Point p[3]; }; struct Circle { struct Point center; double r; } ans; // 计算两点距离 double Dis(struct Point p, struct Point q) { double dx=p.x-q.x; double dy=p.y-q.y; return sqrt(dx*dx+dy*dy); } // 计算三角形面积 double Area(struct Traingle ct) { return fabs((ct.p[1].x-ct.p[0].x)*(ct.p[2].y-ct.p[0].y)-(ct.p[2].x-ct.p[0].x)*(ct.p[1].y-ct.p[0].y))/2.0; } // 求三角形的外接圆，返回圆心和半径( 存在结构体"圆"中) struct Circle CircumCircle(struct Traingle t) { struct Circle tmp; double a, b, c, c1, c2; double xA, yA, xB, yB, xC, yC; a = Dis(t.p[0], t.p[1]); b = Dis(t.p[1], t.p[2]); c = Dis(t.p[2], t.p[0]); //根据 S = a * b * c / R / 4;求半径 R tmp.r = (a*b*c)/(Area(t)*4.0); xA = t.p[0].x; yA = t.p[0].y; xB = t.p[1].x; yB = t.p[1].y; xC = t.p[2].x; yC = t.p[2].y; c1 = (xA*xA+yA*yA - xB*xB-yB*yB) / 2; c2 = (xA*xA+yA*yA - xC*xC-yC*yC) / 2; tmp.center.x = (c1*(yA - yC)-c2*(yA - yB)) / ((xA - xB)*(yA - yC)-(xA - xC)*(yA - yB)); tmp.center.y = (c1*(xA - xC)-c2*(xA - xB)) / ((yA - yB)*(xA - xC)-(yA - yC)*(xA - xB)); return tmp; }
7. 求凸包
#include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> #define INF 999999999.9 #define PI acos(-1.0) struct Point { double x, y, dis; } pt[1005], Stack[1005], p0; int top, tot; // 计算几何距离 double Dis(double x1, double y1, double x2, double y2) { return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); } //极角比较，返回1: p0p1在p0p2的右侧，返回0:p0,p1,p2共线 int Cmp_PolarAngel(struct Point p1, struct Point p2, struct Point pb) { double delta=(p1.x-pb.x)*(p2.y-pb.y)-(p2.x-pb.x)*(p1.y-pb.y); if (delta<0.0) return 1; else if (delta==0.0) return 0; else return -1; } // 判断向量p2p3是否对p1p2构成左旋 bool Is_LeftTurn(struct Point p3, struct Point p2, struct Point p1) { int type=Cmp_PolarAngel(p3, p1, p2); if (type<0) return true; return false; } // 先按极角排，再按距离由小到大排 int Cmp(const void*p1, const void*p2) { struct Point*a1=(struct Point*)p1; struct Point*a2=(struct Point*)p2; int type=Cmp_PolarAngel(*a1, *a2, p0); if (type<0) return -1; else if (type==0) { if (a1->dis<a2->dis) return -1; else if (a1->dis==a2->dis) return 0; else return 1; } else return 1; } // 求凸包 void Solve(int n) { int i, k; p0.x=p0.y=INF; for (i=0; i<n; i++) { scanf("%lf %lf",&pt[i].x, &pt[i].y); if (pt[i].y < p0.y) { p0.y=pt[i].y; p0.x=pt[i].x; k=i; } else if (pt[i].y==p0.y) { if (pt[i].x<p0.x) { p0.x=pt[i].x; k=i; } } } pt[k]=pt[0]; pt[0]=p0; for (i=1; i<n; i++) pt[i].dis=Dis(pt[i].x,pt[i].y, p0.x,p0.y); qsort(pt+1, n-1, sizeof(struct Point), Cmp); // 去掉极角相同的点 tot=1; for (i=2; i<n; i++) if (Cmp_PolarAngel(pt[i], pt[i-1], p0)) pt[tot++]=pt[i-1]; pt[tot++]=pt[n-1]; // 求凸包 top=1; Stack[0]=pt[0]; Stack[1]=pt[1]; for (i=2; i<tot; i++) { while (top>=1 && Is_LeftTurn(pt[i], Stack[top], Stack[top-1])==false) top--; Stack[++top]=pt[i]; } } int main (void) { int n; while (scanf("%d",&n)==2) { Solve(n); } return 0; }
. 8. 凸包卡壳旋转求出所有对踵点、最远点对
#include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> #define INF 999999999.9 #define PI acos(-1.0) struct Point { double x, y, dis; } pt[6005], Stack[6005], p0; int top, tot; // 计算几何距离 double Dis(double x1, double y1, double x2, double y2) { return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); } //极角比较返回-1:p0p1在p0p2的右侧，返回0:p0,p1,p2共线 int Cmp_PolarAngel(struct Point p1, struct Point p2, struct Point pb) { double delta=(p1.x-pb.x)*(p2.y-pb.y)-(p2.x-pb.x)*(p1.y-pb.y); if (delta<0.0) return 1; else if (delta==0.0) return 0; else return -1; } // 判断向量p2p3是否对p1p2 bool Is_LeftTurn(struct Point p3, struct Point p2, struct Point p1) { int type=Cmp_PolarAngel(p3, p1, p2); if (type<0) return true; return false; } // 先按极角排，再按距离由小到大排 int Cmp(const void*p1, const void*p2) { struct Point*a1=(struct Point*)p1; struct Point*a2=(struct Point*)p2; int type=Cmp_PolarAngel(*a1, *a2, p0); if (type<0) return -1; else if (type==0) { if (a1->dis<a2->dis) return -1; else if (a1->dis==a2->dis) return 0; else return 1; } else return 1; } // 求凸包 void Hull(int n) { int i, k; p0.x=p0.y=INF; for (i=0; i<n; i++) { scanf("%lf %lf",&pt[i].x, &pt[i].y); if (pt[i].y < p0.y) { p0.y=pt[i].y; p0.x=pt[i].x; k=i; } else if (pt[i].y==p0.y) { if (pt[i].x<p0.x) { p0.x=pt[i].x; k=i; } } } pt[k]=pt[0]; pt[0]=p0; for (i=1; i<n; i++) pt[i].dis=Dis(pt[i].x,pt[i].y, p0.x,p0.y); qsort(pt+1, n-1, sizeof(struct Point), Cmp); // 去掉极角相同的点 tot=1; for (i=2; i<n; i++) if (Cmp_PolarAngel(pt[i], pt[i-1], p0)) pt[tot++]=pt[i-1]; pt[tot++]=pt[n-1]; // 求凸包 top=1; Stack[0]=pt[0]; Stack[1]=pt[1]; for (i=2; i<tot; i++) { while (top>=1 && Is_LeftTurn(pt[i], Stack[top], Stack[top-1])==false) top--; Stack[++top]=pt[i]; } } // 计算叉积 double CrossProduct(struct Point p1, struct Point p2, struct Point p3) { return (p1.x-p3.x)*(p2.y-p3.y)-(p2.x-p3.x)*(p1.y-p3.y); } // 卡壳旋转，求出凸多边形所有对踵点 void Rotate(struct Point*ch, int n) { int i, p=1; double t1, t2, ans=0.0, dif; ch[n]=ch[0]; for (i=0; i<n; i++) { // 如果下一个点与当前边构成的三角形的面积更大，则说明此时不构成对踵点 while (fabs(CrossProduct(ch[i],ch[i+1],ch[p+1])) > fabs(CrossProduct(ch[i],ch[i+1],ch[p]))) p=(p+1)%n; dif=fabs(CrossProduct(ch[i],ch[i+1],ch[p+1])) - fabs(CrossProduct(ch[i],ch[i+1],ch[p])); // 如果当前点和下一个点分别构成的三角形面积相等，则说明两条边即为平行线 ，对角线两端都可能是对踵点 if (dif==0.0) { t1=Dis(ch[p].x, ch[p].y, ch[i].x, ch[i].y); t2=Dis(ch[p+1].x, ch[p+1].y, ch[i+1].x, ch[i+1].y); if (t1>ans)ans=t1; if (t2>ans)ans=t2; } // 说明 p，i是对踵点 else if (dif<0.0) { t1=Dis(ch[p].x, ch[p].y, ch[i].x, ch[i].y); if (t1>ans)ans=t1; } } printf("%.2lf ",ans); } int main (void) { int n; while (scanf("%d",&n)==1) { Hull(n); Rotate(Stack, top+1); } return 0; }
9. 凸包+ 旋转卡壳求平面面积最大三角
#include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> #define INF 99999999999.9 #define PI acos(-1.0) struct Point { double x, y, dis; } pt[50005], Stack[50005], p0; int top, tot; double Dis(double x1, double y1, double x2, double y2) { return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); } int Cmp_PolarAngel(struct Point p1, struct Point p2, struct Point pb) { double delta=(p1.x-pb.x)*(p2.y-pb.y)-(p2.x-pb.x)*(p1.y-pb.y); if (delta<0.0) return 1; else if (delta==0.0) return 0; else return -1; } bool Is_LeftTurn(struct Point p3, struct Point p2, struct Point p1) { int type=Cmp_PolarAngel(p3, p1, p2); if (type<0) return true; return false; } int Cmp(const void*p1, const void*p2) { struct Point*a1=(struct Point*)p1; struct Point*a2=(struct Point*)p2; int type=Cmp_PolarAngel(*a1, *a2, p0); if (type<0) return -1; else if (type==0) { if (a1->dis<a2->dis) return -1; else if (a1->dis==a2->dis) return 0; else return 1; } else return 1; } void Hull(int n) { int i, k; p0.x=p0.y=INF; for (i=0; i<n; i++) { scanf("%lf %lf",&pt[i].x, &pt[i].y); if (pt[i].y < p0.y) { p0.y=pt[i].y; p0.x=pt[i].x; k=i; } else if (pt[i].y==p0.y) { if (pt[i].x<p0.x) { p0.x=pt[i].x; k=i; } } } pt[k]=pt[0]; pt[0]=p0; for (i=1; i<n; i++) pt[i].dis=Dis(pt[i].x,pt[i].y, p0.x,p0.y); qsort(pt+1, n-1, sizeof(struct Point), Cmp); tot=1; for (i=2; i<n; i++) if (Cmp_PolarAngel(pt[i], pt[i-1], p0)) pt[tot++]=pt[i-1]; pt[tot++]=pt[n-1]; top=1; Stack[0]=pt[0]; Stack[1]=pt[1]; for (i=2; i<tot; i++) { while (top>=1 && Is_LeftTurn(pt[i], Stack[top], Stack[top-1])==false) top--; Stack[++top]=pt[i]; } } double TArea(struct Point p1, struct Point p2, struct Point p3) { return fabs((p1.x-p3.x)*(p2.y-p3.y)-(p2.x-p3.x)*(p1.y-p3.y)); } void Rotate(struct Point*ch, int n) { if (n<3) { printf("0.00 "); return; } int i, j, k; double ans=0.0, tmp; ch[n]=ch[0]; for (i=0; i<n; i++) { j=(i+1)%n; k=(j+1)%n; while ((j!=k) && (k!=i)) { while (TArea(ch[i],ch[j],ch[k+1])>TArea(ch[i],ch[j],ch[k])) k=(k+1)%n; tmp=TArea(ch[i],ch[j], ch[k]); if (tmp>ans) ans=tmp; j=(j+1)%n; } } printf("%.2lf ",ans/2.0); } int main (void) { int n; while (scanf("%d",&n)==1) { if (n==-1)break; Hull(n); Rotate(Stack, top+1); } return 0; }
10. Pick 定理
// Pick 定理求整点多边形内部整点数目 //(1) 给定顶点座标均是整点（或正方形格点）的简单多边形 ，皮克定理说明了其面积A和 //内部格点数目i、边上格点数目b的关系：A= i+b/2-1； //(2) 在两点(x1 ,y1)， (x2,y2)连线之间的整点个数(包含一个端点)为:gcd ( |x1 －x2| ,|y1 － y2|) ； //(3) 求三角形面积用叉乘 #include<stdio.h> #include<stdlib.h> #include<math.h> #include<string.h> long long x[3], y[3], area, b; long long My_Abs(long long t) { if (t<0) return -t; return t; } long long Gcd(long long x, long long y) { if (y==0) return x; long long mod=x%y; while (mod) { x=y; y=mod; mod=x%y; } return y; } int main (void) { int i; while (1) { for (i = 0; i < 3; i ++) scanf("%lld %lld", &x[i], &y[i]); if(x[0]==0&&y[0]==0&&x[1]==0&&y[1]==0&&x[2]==0&&y[2]==0) break; area = (x[1]-x[0])*(y[2]-y[0])-(x[2]-x[0])*(y[1]-y[0]); area = My_Abs(area); b=0; b=Gcd(My_Abs(x[1]-x[0]), My_Abs(y[1]-y[0])) + Gcd(My_Abs(x[2]-x[0]), My_Abs(y[2]-y[0])) + Gcd(My_Abs(x[1]-x[2]), My_Abs(y[1]-y[2])); printf("%lld ", (area-b+2)/2); } return 0; }
11. 求多边形面积和重心
#include <stdio.h> #include <math.h> int x[1000003], y[1000003]; double A, tx, ty, tmp; int main (void) { int cases, n, i; scanf ("%d", &cases); while (cases --) { scanf ("%d", &n); A = 0.0; x[0] = y[0] = 0; for (i = 1; i <= n; i ++) { scanf ("%d %d", &x[i], &y[i]); A += (x[i-1]*y[i] - x[i]*y[i-1]); } A += x[n]*y[1] - x[1]*y[n]; A =A / 2.0; tx = ty = 0.0; for (i = 1; i < n; i ++) { tmp = x[i]*y[i+1] - x[i+1]*y[i]; tx += (x[i]+x[i+1]) * tmp; ty += (y[i]+y[i+1]) * tmp; } tmp = x[n]*y[1] - x[1]*y[n]; tx += (x[n]+x[1])*tmp; ty += (y[n]+y[1])*tmp; printf ("%.2lf %.2lf ", tx/(6.0*A), ty/(6.0*A)); } return 0; }
12. 判断一个简单多边形是否有核
#include <stdio.h> #include <string.h> const int INF = (1<<30); struct Point { int x, y; } pt[150]; typedef struct Point Point; bool turn_right[150]; int det(Point s1, Point t1, Point s2, Point t2) { int d1x = t1.x-s1.x; int d1y = t1.y-s1.y; int d2x = t2.x-s2.x; int d2y = t2.y-s2.y; return d1x*d2y - d2x*d1y; } void Swap(int &a, int &b) { if (a>b) { int t=a; a=b; b=t; } } int main (void) { int n, i, cross, maxx, minx, maxy, miny, maxn, minn, countn=0; while (scanf("%d", &n)==1&&n) { maxx=maxy=-INF; minx=miny=INF; // 点按顺时针给出 for (i=1; i<=n; i++) { scanf("%d %d", &pt[i].x, &pt[i].y); if (maxx<pt[i].x) maxx=pt[i].x; if (maxy<pt[i].y) maxy=pt[i].y; if (minx>pt[i].x) minx=pt[i].x; if (miny>pt[i].y) miny=pt[i].y; } pt[n+1]=pt[1]; pt[n+2]=pt[2]; pt[n+3]=pt[3]; pt[n+4]=pt[4]; // 求每条线段的转向 for (i=1; i<=n+1; i ++) { cross = det(pt[i],pt[i+1], pt[i+1], pt[i+2]); if (cross<0) turn_right[i+1]=true; else turn_right[i+1]=false; } // 两条边连续右转的为凸处，只有此时才可影响“核”肯恩存在的范围 for (i=2; i<= n+1; i++) if (turn_right[i] && turn_right[i+1]) { if (pt[i].x==pt[i+1].x) { minn=pt[i].y; maxn=pt[i+1].y; Swap(minn, maxn); if (minn>miny) miny=minn; if (maxn<maxy) maxy=maxn; } else { minn=pt[i].x; maxn=pt[i+1].x; Swap(minn, maxn); if (minn>minx) minx=minn; if (maxn<maxx) maxx=maxn; } } if (minx<=maxx && miny<=maxy) printf("Floor #%d Surveillance is possible. ", ++countn); else printf("Floor #%d Surveillance is impossible. ", ++countn); } return 0; }
13. 模拟退火
#include <stdio.h> #include <stdlib.h> #include <math.h> #define Lim 0.999999 #define EPS 1e-2 #define PI acos(-1.0) double Temp, maxx, minx, maxy, miny, lx, ly, dif; int nt, ns, nc; struct Target { double x, y; } T[105]; struct Solution { double x, y; double f; } S[25], P, A; double Dis(double x1, double y1, double x2, double y2) { return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); } void Seed(void) { int i, j; for (i=0; i<ns; i++) { S[i].x=minx+((double)(rand()%1000+1)/1000.0)*lx; S[i].y=miny+((double)(rand()%1000+1)/1000.0)*ly; S[i].f=0.0; for (j=0; j<nt; j++) S[i].f=S[i].f+Dis(S[i].x,S[i].y, T[j].x, T[j].y); } } void Trans(void) { int i, j, k; double theta; for (i=0; i<ns; i++) { P=S[i]; for (j=0; j<nc; j++) { theta=(((double)(rand()%1000+1))/1000.0)*2.0*PI; A.x=P.x+Temp*cos(theta); A.y=P.y+Temp*sin(theta); if (A.x<minx||A.x>maxx||A.y<miny||A.y>maxy) continue; A.f=0.0; for (k=0; k<nt; k++) A.f=A.f+Dis(A.x,A.y,T[k].x,T[k].y); dif=A.f-S[i].f; if (dif<0.0)S[i]=A; else { dif=exp(-dif/Temp); if (dif>Lim) S[i]=A; } } } } int main (void) { int i, k; while (scanf("%d",&nt)==1&&nt) { maxx=maxy=0; minx=miny=(1<<20); for (i=0; i<nt; i++) { scanf("%lf %lf",&T[i].x,&T[i].y); if (maxx<T[i].x)maxx=T[i].x; if (minx>T[i].x)minx=T[i].x; if (maxy<T[i].y)maxy=T[i].y; if (miny>T[i].y)miny=T[i].y; } lx=maxx-minx; ly=maxy-miny; Temp=sqrt(lx*lx+ly*ly)/3.0; ns=5, nc=10; Seed(); while (Temp>EPS) { Trans(); Temp=Temp*0.40; } k=0; for (i=1; i<ns; i++) if (S[k].f>S[i].f) k=i; printf ("%.0lf ", S[k].f); } return 0; }
14. 六边形坐标系
// 第一种六边形坐标系 #include<stdio.h> #include<math.h> #include<string.h> #include<stdlib.h> double Dis(double x1, double y1, double x2, double y2) { double dx=x1-x2; double dy=y1-y2; return sqrt(dx*dx+dy*dy); } void Get_KL(double L, double x, double y, int &k, int &l, double &cd) { k=floor((2.0*x)/(3.0*L)); l=floor((2.0*y)/(sqrt(3.0)*L)); double d1, d2, x1, y1, x2, y2; if ((k+l)&1) { x1=k*L*1.5; y1=(l+1.0)*L*sqrt(3.0)*0.5; x2=(k+1.0)*L*1.5; y2=l*L*sqrt(3.0)*0.5; d1=Dis(x1,y1, x,y); d2=Dis(x2,y2, x,y); if (d1>d2) { k++; cd=d2; } else { l++; cd=d1; } } else { x1=k*L*1.5; y1=l*L*sqrt(3.0)*0.5; x2=(k+1.0)*L*1.5; y2=(l+1.0)*L*sqrt(3.0)*0.5; d1=Dis(x1,y1, x,y); d2=Dis(x2,y2, x,y); if (d1>d2) { k++,l++; cd=d2; } else cd=d1; } } int My_Abs(int x) { if (x<0) return -x; return x; } int main (void) { double L, x1, y1, x2, y2, ans, cd1, cd2; int k1, l1, k2, l2; while (scanf("%lf %lf %lf %lf %lf",&L,&x1,&y1,&x2,&y2)==5) { if (L==0.0&&x1==0.0&&y1==0.0&&x2==0.0&&y2==0.0) break; Get_KL(L, x1, y1, k1, l1, cd1); Get_KL(L, x2, y2, k2, l2, cd2); if (k1==k2&&l1==l2) printf("%.3lf ", Dis(x1,y1, x2,y2)); else { ans=cd1+cd2; if (My_Abs(k1-k2) > My_Abs(l1-l2)) ans=ans+sqrt(3.0)*L*My_Abs(k1-k2); else ans=ans+sqrt(3.0)*L*My_Abs(k1-k2)+sqrt(3.0)*L*(double)(My_Abs(l1-l2)-My_Abs(k1-k2))/2.0 ; printf("%.3lf ", ans); } } return 0; }
// 第二种六边形坐标系 #include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> struct A { int x, y, num; } a[10001]; const int dec[6][2] = {{-1,1},{-1,0},{0,-1},{1,-1},{1,0},{0,1}}; bool adj(int x1, int y1, int x2, int y2) { if (x1 == x2 && abs(y1-y2) == 1) return true; if (y1 == y2 && abs(x1-x2) == 1) return true; if (x1 == x2 + 1 && y1 == y2 -1) return true; if (x1 == x2 - 1 && y1 == y2 +1) return true; return false; } bool flag[10001]; int main (void) { int i, j, k, x, u, v, cut, minn, cnt[6]; memset(cnt, 0, sizeof(cnt)); a[1].num = 1, cnt[1] = 1; a[1].x = a[1].y = 0; for (i = 2; i < 10001; i ++) { k = (int)((3.0+sqrt(12.0*i - 3.0))/6.0+0.0000001); if (i == 3*(k-1)*(k-1)+3*(k-1)+1) k --; j = i - (3*(k-1)*(k-1)+3*(k-1)+1); // 当前的六边形是第k层的第j个六边形 if (j == 1) a[i].x = a[i-1].x, a[i].y = a[i-1].y + 1; else { x = (j-1) / k; a[i].x = a[i-1].x + dec[x][0], a[i].y = a[i-1].y + dec[x][1]; } memset(flag, false, sizeof(flag)); x = 12*k-6, cut = 0; for (u = i-1, v = 0; u>=1&&v<x; u --, v ++) if (adj(a[u].x, a[u].y, a[i].x, a[i].y)) { cut ++; flag[a[u].num] = true; if (cut == 3) break; } minn = 10001; for (u = 1; u < 6; u ++) if ((!flag[u])&&minn > cnt[u]) { minn = cnt[u]; x = u; } a[i].num = x; cnt[x] ++; } scanf ("%d", &x); while (x --) { scanf ("%d", &i); printf ("%d ", a[i].num); } return 0; }
15. 用一个给定半径的圆覆盖最多的点
// 同半径圆的圆弧表示 #include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> #define PI acos(-1.0) struct Point { double x, y; } pt[2005]; double dis[2005][2005]; struct List { double a; bool flag; int id; } List[8005]; int cnt; double Dis(int i, int j) { double dx=pt[i].x-pt[j].x; double dy=pt[i].y-pt[j].y; return sqrt(dx*dx+dy*dy); } int Cmp(const void*p1, const void*p2) { struct List*a1=(struct List*)p1; struct List*a2=(struct List*)p2; if (a1->a<a2->a)return -1; else if (a1->a==a2->a) return a1->id-a2->id; else return 1; } int main (void) { int n, i, j, ans, num; double r, theta, delta, a1, a2; while (scanf("%d %lf",&n,&r)==2) { if (n==0&&r==0.0) break; r=r+0.001; r=r*2.0; for (i=1; i<=n; i++) scanf("%lf %lf", &pt[i].x, &pt[i].y); for (i=1; i<n; i++) for (j=i+1; j<=n; j++) { dis[i][j]=Dis(i, j); dis[j][i]=dis[i][j]; } ans=0; for (i=1; i<=n; i++) { cnt=0; for (j=1; j<=n; j++) if ((j!=i)&&(dis[i][j]<=r)) { theta=atan2(pt[j].y-pt[i].y, pt[j].x-pt[i].x); if (theta<0.0) theta=theta+2.0*PI; delta=acos(dis[i][j]/r); a1=theta-delta; a2=theta+delta; List[++cnt].a=a1; List[cnt].flag=true; List[cnt].id=cnt; List[++cnt].a=a2; List[cnt].flag=false; List[cnt].id=cnt; } qsort(List+1,cnt,sizeof(struct List),Cmp); num=0; for (j=1; j<=cnt; j++) if (List[j].flag) { num++; if (num>ans) ans=num; } else num--; } printf("It is possible to cover %d points. ", ans+1); } return 0; }
16. 不等大的圆的圆弧表示
intersection_circle_circle(circle[i].center, circle[i].r, circle[j].center, circle[j].r, p1, p2); a1= atan2(p1.y-circle[j].center.y, p1.x-circle[j].center.x); if (a1<0.0) a1=a1+2.0*PI; a2= atan2(p2.y-circle[j].center.y, p2.x-circle[j].center.x); if (a2<0.0) a2=a2+2.0*PI; if (a1>a2) { tmp=a1; a1=a2; a2=tmp; } mid=(a1+a2)/2.0; xtest = circle[j].center.x +circle[j].r*cos(mid); ytest = circle[j].center.y +circle[j].r*sin(mid); if (!point_in_circle(xtest, ytest, i)) { circle[j].cnt++; circle[j].line[circle[j].cnt].s=0; circle[j].line[circle[j].cnt].t=a1; circle[j].cnt++; circle[j].line[circle[j].cnt].s=a2; circle[j].line[circle[j].cnt].t=2.0*PI; } else { circle[j].cnt++; circle[j].line[circle[j].cnt].s=a1; circle[j].line[circle[j].cnt].t=a2; }
17. 矩形面积并
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> struct Node { int l, r, cnt; double cover; } node[80005]; struct Point { double x; double y1, y2; int id_y1, id_y2, id_x; bool flag; } pt[20005]; double y[20005]; int total, cnty; int cmp1(const void*p1, const void*p2) { double*a1=(double*)p1; double*a2=(double*)p2; if (*a1<*a2) return -1; else if (*a1==*a2) return 0; else return 1; } int cmp2(const void*p1, const void*p2) { struct Point*a1=(struct Point*)p1; struct Point*a2=(struct Point*)p2; if (a1->x<a2->x) return -1; else if (a1->x==a2->x) { if (a1->id_x<a2->id_x) return -1; else if (a1->id_x==a2->id_x) return 0; else return 1; } else return 1; } int find(double target) { int head=1, tail=cnty, mid; while (head<=tail) { mid=(head+tail)>>1; if (y[mid]==target) return mid; else if (y[mid]<target) head=mid+1; else tail=mid-1; } return 0; } void Build(int l, int r, int s) { node[s].l=l; node[s].r=r; node[s].cnt=0; node[s].cover=0.0; if (l+1<r) { int mid=(l+r)>>1; Build(l,mid,s<<1); Build(mid,r,(s<<1)+1); } } void Update(int s) { if (node[s].cnt>0) node[s].cover=y[node[s].r]-y[node[s].l]; else if(node[s].l+1==node[s].r) node[s].cover=0.0; else node[s].cover=node[s<<1].cover+node[(s<<1)+1].cover; } void Insert(int l, int r, int s) { if (l<=node[s].l&&node[s].r<=r) { node[s].cnt++; Update(s); return; } if (node[s].l+1<node[s].r) { int mid=(node[s].l+node[s].r)>>1; if (l<mid) Insert(l,r,s<<1); if (r>mid) Insert(l,r,(s<<1)+1); Update(s); } } void Delete(int l, int r, int s) { if (l<=node[s].l&&node[s].r<=r) { if (node[s].cnt>0) node[s].cnt--; Update(s); return; } if (node[s].l+1<node[s].r) { int mid=(node[s].l+node[s].r)>>1; if (l<mid) Delete(l,r,s<<1); if (r>mid) Delete(l,r,(s<<1)+1); Update(s); } } int main (void) { int n, i, j, countn=0; double ans; while (scanf("%d", &n)==1 && n) { cnty=total=0; for (i=1; i<=n; i++) { total++; scanf("%lf %lf", &pt[total].x, &pt[total].y1); pt[total].flag=true; pt[total].id_x=total; y[++cnty]=pt[total].y1; total++; scanf("%lf %lf", &pt[total].x, &pt[total].y2); pt[total].flag=false; pt[total].id_x=total; y[++cnty]=pt[total].y2; pt[total].y1=pt[total-1].y1; pt[total-1].y2=pt[total].y2; } qsort(y+1, cnty, sizeof(double), cmp1); j=cnty; cnty=1; for (i=2; i<=j; i++) if (y[i]!=y[i-1]) y[++cnty]=y[i]; for (i=1; i<=total; i++) { pt[i].id_y1=find(pt[i].y1); pt[i].id_y2=find(pt[i].y2); } qsort(pt+1, total, sizeof(struct Point), cmp2); ans=0.0; Build(1,cnty,1); Insert(pt[1].id_y1, pt[1].id_y2, 1); for (i=2; i<=total; i++) { ans=ans+(pt[i].x-pt[i-1].x)*node[1].cover; if (pt[i].flag) Insert(pt[i].id_y1, pt[i].id_y2, 1); else Delete(pt[i].id_y1, pt[i].id_y2, 1); } printf("%.0lf ", ans+1e-10); } return 0; }
18. 矩形的周长并
#include <stdio.h> #include <string.h> #include <stdlib.h> struct Point { int x, y; } plist[10001]; struct Line { int x, b, e, flag; } llist[10001]; struct Item { int y, id, idx; } ilist[10001]; struct Node { int l, r, c, m, line; bool lf, rf; } node[40005]; int ys[10001]; int cmp1(const void*p1, const void*p2) { struct Item *a1 = (struct Item*)p1; struct Item *a2 = (struct Item*)p2; return a1->y - a2->y; } int cmp2(const void*p1, const void*p2) { struct Item *a1 = (struct Item*)p1; struct Item *a2 = (struct Item*)p2; return a1->id - a2->id; } int cmp3(const void*p1, const void*p2) { struct Line *a1 = (struct Line*)p1; struct Line *a2 = (struct Line*)p2; return a1->x - a2->x; } void getm(int s) { if (node[s].c > 0) { node[s].m = ys[node[s].r-1] - ys[node[s].l-1]; node[s].line = 1; node[s].rf = node[s].lf = true; } else if (node[s].r - node[s].l <= 1) { node[s].m = node[s].line = 0; node[s].rf = node[s].lf = false; } else { node[s].m = node[s<<1].m + node[(s<<1)+1].m; node[s].line = node[s<<1].line + node[(s<<1)+1].line; if (node[s<<1].rf && node[(s<<1)+1].lf) node[s].line --; node[s].lf = node[s<<1].lf; node[s].rf = node[(s<<1)+1].rf; } } void build(int l, int r, int s) { node[s].l = l; node[s].r = r; node[s].c = node[s].m = node[s].line; if (node[s].r - node[s].l > 1) { int mid = (node[s].l + node[s].r)>>1; build(l,mid,s<<1); build(mid,r,(s<<1)+1); } } void insert(int l, int r, int s) { if (l <= node[s].l && node[s].r <= r) { node[s].c ++; getm(s); } if (node[s].r - node[s].l > 1) { int mid = (node[s].l + node[s].r)>>1; if (l < mid) insert(l, r, s<<1); if (mid < r) insert(l, r, (s<<1)+1); getm(s); } } void delet(int l, int r, int s) { if (l <= node[s].l && node[s].r <= r) { node[s].c --; getm(s); } if (node[s].r - node[s].l > 1) { int mid = (node[s].l + node[s].r)>>1; if (l < mid) delet(l, r, s<<1); if (mid < r) delet(l, r, (s<<1)+1); getm(s); } } int main (void) { int n, i, j, l, r, x1, y1, x2, y2, tot, p, ans; while (scanf ("%d", &n) == 1 && n) { for (i = 0; i < n; i ++) { scanf ("%d %d %d %d", &x1, &y1, &x2, &y2); l = 2*i; r = l + 1; plist[l].x = x1; plist[l].y = y1; plist[r].x = x2; plist[r].y = y2; ilist[l].y = y1; ilist[l].id = l; ilist[r].y = y2; ilist[r].id = r; } tot = 2*n; qsort(ilist, tot, sizeof(struct Item), cmp1); ys[0] = ilist[0].y; ilist[0].idx = 0; j = 0; for (i = 1; i < tot; i ++) { if (ilist[i].y != ilist[i-1].y) { j ++; ys[j] = ilist[i].y; } ilist[i].idx = j; } p = j + 1; qsort(ilist, tot, sizeof(struct Item), cmp2); for (i = 0; i < n; i ++) { l = 2*i; r = l + 1; llist[l].x = plist[l].x; llist[l].b = ilist[l].idx; llist[l].e = ilist[r].idx; llist[l].flag = 1; llist[r].x = plist[r].x; llist[r].b = ilist[l].idx; llist[r].e = ilist[r].idx; llist[r].flag = 0; } qsort(llist, tot, sizeof(struct Line), cmp3); build(1,p,1); insert(llist[0].b+1, llist[0].e+1,1); int now_m = node[1].m, now_line = node[1].line; ans = now_m; for (i = 1; i < tot; i ++) { if (llist[i].flag) insert(llist[i].b+1, llist[i].e+1, 1); else delet(llist[i].b+1, llist[i].e+1, 1); ans += (abs(node[1].m - now_m) + 2*(llist[i].x - llist[i-1].x)*now_line); now_m = node[1].m; now_line = node[1].line; } printf ("%d ", ans); } return 0; }
19. 最近圆对
#include<iostream> #include <cstdio> #include<stdlib.h> #include<string.h> #include<set> #include <math.h> using namespace std; set <int>tree; set <int>::iterator iter; struct Point { double x; int id, flag; } p1[100001], p2[100001]; int tot1, tot2; struct Q { double x,y, r; } q[50001]; int cmp(const void*p1, const void*p2) { struct Point*a1=(struct Point*)p1; struct Point*a2=(struct Point*)p2; if (a1->x<a2->x) return -1; else if (a1->x==a2->x) return a2->flag-a1->flag; else return 1; } int cmp1(const void*p1, const void*p2) { struct Q*a1=(struct Q*)p1; struct Q*a2=(struct Q*)p2; if (a1->y<a2->y)return -1; else if (a1->y==a2->y)return 0; else return 1; } double dis(double x1, double y1, double x2, double y2) { return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); } bool judge(int i, int j, double d) { if (dis(q[i].x, q[i].y, q[j].x, q[j].y)<=q[i].r+q[j].r+2.0*d) return true; return false; } bool insert(int v,double d) { iter = tree.insert(v).first; if (iter != tree.begin()) { if (judge(v, *--iter,d)) { return true; } ++iter; } if (++iter != tree.end()) { if (judge(v, *iter,d)) { return true; } } return false; } bool remove(int v,double d) { iter = tree.find(v); if (iter != tree.begin() && iter != --tree.end()) { int a = *--iter; ++iter; int b = *++iter; if (judge(a, b,d)) { return true; } } tree.erase(v); return false; } bool check(double d) { int i=1, j=1; while (i<=tot1&&j<=tot2) { if (p1[i].x-d<=p2[j].x+d) { if (insert(p1[i++].id, d)) return true; } else { if (remove(p2[j++].id, d)) return true; } } while (i<=tot1) { if (insert(p1[i++].id, d)) return true; } while (j<=tot2) { if (remove(p2[j++].id, d)) return true; } return false; } int main (void) { int cases, n, i; scanf("%d",&cases); while (cases--) { scanf("%d",&n); tot1=tot2=0; for (i=1; i<=n; i++) scanf("%lf %lf %lf",&q[i].x,&q[i].y, &q[i].r); qsort(q+1,n,sizeof(struct Q),cmp1); for (i=1; i<=n; i++) { tot1++; p1[tot1].x=q[i].x-q[i].r; p1[tot1].id=i; p1[tot1].flag=1; tot2++; p2[tot2].x=q[i].x+q[i].r; p2[tot2].id=i; p2[tot2].flag=-1; } qsort(p1+1,tot1,sizeof(struct Point),cmp); qsort(p2+1,tot2,sizeof(struct Point),cmp); double head=0.0, tail=dis(q[1].x,q[1].y,q[2].x,q[2].y)+1.0, mid; while (tail-head>1e-8) { tree.clear(); mid=(head+tail)/2.0; if (check(mid)) { tail=mid; } else head=mid; } printf ("%.6lf ",2.0*head); } return 0; }
20. 求两个圆的面积交
double area_of_overlap(point c1, double r1, point c2, double r2) { double a = distance(c1, c2), b = r1, c = r2; double cta1 = acos((a * a + b * b - c * c) / 2 / (a * b)), cta2 = acos((a * a + c * c - b * b) / 2 / (a * c)); double s1 = r1*r1*cta1 - r1*r1*sin(cta1)*(a * a + b * b - c * c) / 2 / (a * b); double s2 = r2*r2*cta2 - r2*r2*sin(cta2)*(a * a + c * c - b * b) / 2 / (a * c); return s1 + s2; }