计算几何题目总结

模板  （只用这一个）   点  线  圆  

//#include<bits/stdc++.h>
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>
using namespace std;
const double eps=1e-8;
const double pi=acos(-1.0);
inline double sqr(double x){ return x*x;                                  }
inline int dcmp(double x)  { if(fabs(x)<eps) return 0;return (x>0? 1: -1);}
struct Point{
    double x,y;
    Point(){ x=0,y=0; }
    Point(double _x,double _y):x(_x),y(_y){}
    void input(){ scanf("%lf%lf",&x,&y); }
    void output(){ printf("%.2f %.2f
",x,y); }
    friend istream &operator >>(istream &os,Point &b){os>>b.x>>b.y;return os;             }
    friend ostream &operator <<(ostream &os,Point &b){os<<b.x<<' '<<b.y;return os;        }
    bool operator ==(const Point &b)const{return  ( dcmp(x-b.x)==0&&dcmp(y-b.y)==0);      }
    bool operator !=(const Point &b)const{return !((dcmp(x-b.x)==0&&dcmp(y-b.y)==0));     }
    bool operator <(const Point &b)const {return (dcmp(x-b.x)==0? dcmp(y-b.y)<0 : x<b.x); }
    double operator ^(const Point &b)const{     return x*b.y-y*b.x;}//叉积
    double operator *(const Point &b)const{     return x*b.x+y*b.y;} //点积
    Point operator +(const Point &b)const {     return Point(x+b.x,y+b.y);}
    Point operator -(const Point &b)const {     return Point(x-b.x,y-b.y);}
    Point operator *(double a)            {     return Point(x*a,y*a);    }
    Point operator /(double a)            {     return Point(x/a,y/a);    }
    double len2()                         {     return sqr(x)+sqr(y);     }//长度平方
    double len()                          {      return sqrt(len2());     }//长度
    double polar(){  return atan2(y,x);  }//向量的极角   //返回与x轴正向夹角(-pi~pi]
    Point change_len(double r){           //转化为长度为r的向量
        double l=len();
        if(dcmp(l)==0)  return *this;  //零向量
        return Point(x*r/l,y*r/l);
    }
    Point rotate_left() {  return Point(-y,x); }//逆时针旋转90度
    Point rotate_right(){  return Point(y,-x); }//顺时针旋转90度
    Point rotate(Point p,double ang){           //绕点p逆时针旋转ang度
        Point v=(*this)-p;  double c=cos(ang),s=sin(ang);
        return Point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
    }
    Point normal(){  return Point(-y/len(),x/len()); }//单位化，逆时针旋转90°
};
inline double cross(Point a,Point b)    {   return a.x*b.y-a.y*b.x;                             } //叉积
inline double dot(Point a,Point b)      {   return a.x*b.x+a.y*b.y;                             } //点积
double rad(Point a,Point b)             {   return fabs(atan2(fabs(cross(a,b)),dot(a,b)));      }  //两个向量的夹角
double area2(Point A, Point B, Point C) {   return cross(B-A, C-A);                             }  // 四边形面积
bool   is_parallel(Point a,Point b)       {   double p=rad(a,b);return dcmp(p)==0||dcmp(p-pi)==0; }//判断向量是否平行

typedef Point Vector;
double Angle(Vector A, Vector B)        { return acos(dot(A, B)/A.len()/B.len());           }  // 角度
double dis(Point A,Point B)             { return sqrt( (A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));}
Vector Normal(Vector A)                 { double L =A.len(); return Vector(-A.y/L, A.x/L);   }  //rad为弧度 且为逆时针旋转的角
bool ToLeftTest(Point a,Point b,Point c){ return cross(b - a, c - b) > 0;                       }  //向量A左转90°的单位法向量
struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e):s(_s),e(_e){} //两点确定直线
    Line(Point p,double ang){   //一个点和斜率(弧度制)确定直线
        s=p;
        if(dcmp(ang-pi/2)==0){ e=s+Point(0,1);        }
        else                 { e=s+Point(1,tan(ang)); }
    }
    void input(){ s.input(); e.input(); }
    Point operator &(const Line &b)const{    //求两直线交点
        Point res=s;
        double t=((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
        res.x+=(e.x-s.x)*t;
        res.y+=(e.y-s.y)*t;
        return res;
    }
};
struct Circle{
    Point c;
    double r;
    Circle() {c.x=0; c.y=0; r=0; }
    Circle(Point c,double r):c(c),r(r) {}
    Point point(double a){
        return Point (c.x+cos(a)*r,c.y+sin(a)*r);
    }
}A,B;
Point Circle_Cirle(Circle c1, Circle c2){
    double d=dis(c1.c,c2.c);   Point k=c2.c-c1.c;
    double a=k.polar();
    double da=acos( (c1.r*c1.r+d*d-c2.r*c2.r)/(2*c1.r*d) );
    Point p1=c1.point(a-da);  //printf("%.10f %.10f
",p1.x,p1.y);
    Point p2=c1.point(a+da); // printf("%.10f %.10f
",p2.x,p2.y)
    return p1;
}
double PolygonArea(vector<Point> p){    //多边形的有向面积，加上绝对值就是面积  正值表示输入点按照逆时针 否则为顺时针
    int n=p.size(); double area=0;
    for(int i=1;i<n-1;i++)  area+=cross(p[i]-p[0],p[i+1]-p[0]);
    return fabs(area/2); // 可改
}
double PolygonLength(vector<Point> vp){  //多边形周长
    int m=vp.size(); double area=0;
    for(int i=0;i<m;i++) area+=dis(vp[i],vp[(i+1)%m]);
    return fabs(area/2);
}
bool up(Point a,Point b)  { if(a.x!=b.x) return a.x<b.x;  return a.y<b.y; }
bool cmp2(Point a,Point b){ Point c; return cross(a-c,b-c)>0;     }
bool cmp1(Point a,Point b) // (-pi,pi]
{
    if( abs(atan2(a.y,a.x)-atan2(b.y,b.x))>eps )
        return atan2(a.y,a.x)<atan2(b.y,b.x);
    else return a.x<b.x;
}
int Quadrant(Point a)//象限排序，注意包含四个坐标轴
{
    if(a.x>0&&a.y>=0)  return 1;
    if(a.x<=0&&a.y>0)  return 2;
    if(a.x<0&&a.y<=0)  return 3;
    if(a.x>=0&&a.y<0)  return 4;
}
bool cmp3(Point a,Point b)  //先按象限从小到大排序 再按极角从小到大排序
{
    if(Quadrant(a)==Quadrant(b))//返回值就是象限
        return cmp1(a,b);
    else Quadrant(a)<Quadrant(b);
}
bool cmp(Point a, Point b)//先按象限排序，再按极角排序，再按远近排序
{
    if (a.y == 0 && b.y == 0 && a.x*b.x <= 0)return a.x>b.x;
    if (a.y == 0 && a.x >= 0 && b.y != 0)return true;
    if (b.y == 0 && b.x >= 0 && a.y != 0)return false;
    if (b.y*a.y <= 0)return a.y>b.y;
    Point one;
    one.y = one.x = 0;
    return cross(one-a,one-b) > 0 || (cross(one-a,one-b) == 0 && a.x < b.x);
}
void tubao(vector<Point> vs){
    Point pa[60];    Point pt[60];
    int n=vs.size();
    for(int i=0;i<n;i++) pa[i]=vs[i];
    sort(pa,pa+n,up);
    int m=0;
    for(int i=0;i<n;i++)
    {
        while(m>1 && cross(pt[m-1]-pt[m-2],pa[i]-pt[m-2])<=0) m--;
        pt[m++]=pa[i];
    }
    int k=m;
    for(int i=n-2;i>=0;i--)
    {
        while(m>k && cross(pt[m-1]-pt[m-2],pa[i]-pt[m-2])<=0) m--;
        pt[m++]=pa[i];
    }
    if(n>1) m--;  //  1-m
}
const int maxn=60;
Point pa[maxn];
int main(){
    Point s; scanf("%lf %lf",&s.x,&s.y);
    int tot=1;
    while(~scanf("%lf %lf",&pa[tot].x,&pa[tot].y) && pa[tot].x!=1100) {++tot; }
    //cout<<tot<<endl;
    sort(pa+1,pa+tot,cmp2);
    printf("(%.0f,%.0f)
",s.x,s.y);
    for(int i=1;i<tot;i++){
        printf("(%.0f,%.0f)
",pa[i].x,pa[i].y);
    }
}
View Code

平面直线图   ( 求直线交的多边形第K大模板  )

struct PSLG{    //平面直线图 处理平面内所有直线围成的所有多边形 传入直线交点之间的每条线段
    struct Edge{
        int from,to;
        double ang;
        Edge(){ ang=from=to=0; }
        Edge(int s,int t,double a){ from=s,to=t,ang=a; }
    };
    int n,m,face_cnt;   //平面个数 包括外面最大的多边形
    double area[MAXN];  //每个多边形面积
    Point point[MAXN];  //平面内所有的点
    vector<Edge>edge;
    vector<int>G[MAXN];
    vector<vector<Point> >face;
    int vis[2*MAXN],left[2*MAXN],pre[2*MAXN];   //left表示这条边的左侧属于哪个面
    void Init(){
        face.clear();
        edge.clear();
        for(int i=0;i<n;i++)    G[i].clear();
        n=m=0;
    }
    PSLG(){ Init(); }
    void AddEdge(int from, int to){             //需要建立反向边帮助寻找下一条边
       edge.pb(Edge(from,to,(point[to]-point[from]).polar()));
       edge.pb(Edge(to,from,(point[from]-point[to]).polar()));
       m=edge.size();
       G[from].pb(m-2);
       G[to].pb(m-1);
    }
    void Build(){
        for(int u=0;u<n;u++){
            int d=G[u].size();
            for(int i=0;i<d;i++)
                for(int j=i+1;j<d;j++)
                    if(edge[G[u][i]].ang>edge[G[u][j]].ang)
                        swap(G[u][i],G[u][j]);
            for(int i=0;i<d;i++)    pre[G[u][(i+1)%d]]=G[u][i]; //从u出发的i条边顺时针旋转的第一条边是pre[i]
        }
        face_cnt=0; memset(vis,0,sizeof(vis));
        for(int u=0;u<n;u++){
            for(int i=0;i<G[u].size();i++){
                int e=G[u][i];
                if(!vis[e]){
                    face_cnt++;
                    vector<Point> polygon;
                    while(1){
                        vis[e]=1;
                        left[e]=face_cnt;
                        int from=edge[e].from;
                        polygon.pb(point[from]);
                        e=pre[e^1];         //逆时针旋转最多的一条边即为顺时针转动的第一条边
                        if(e==G[u][i])  break;
                    }
                    face.pb(polygon);
               }
            }
        }
        for(int i=0;i<face_cnt;i++)  area[i]=polygon_area(face[i]);
        
        
        /*   包括外面最大的
        int n;  scanf("%d",&n);
        for(int i=0;i<n;i++)    line[i].input();
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++)
                if(!is_parallel(line[i].e-line[i].s,line[j].e-line[j].s)){
                    Point inter=line[i]&line[j];
                    pslg.point[pslg.n++]=inter;
                    point[i].pb({dot(inter-line[i].s,line[i].e-line[i].s),pslg.n-1});
                    point[j].pb({dot(inter-line[j].s,line[j].e-line[j].s),pslg.n-1});
                }
            sort(point[i].begin(),point[i].end());
            for(int j=1;j<point[i].size();j++)  pslg.AddEdge(point[i][j-1].se,point[i][j].se);
        }
    */
    }
};
View Code

直线相交检测     2-重合 1-相交 0-不相交  两点式  double 类型    

int check_line_line(Line A,Line B)
{
    Point a=A.a; Point b=A.b;
    Point c=B.a; Point d=B.b;
    double A1=b.y-a.y,B1=-(b.x-a.x),C1=b.y*a.x-b.x*a.y;
    double A2=d.y-c.y,B2=-(d.x-c.x),C2=d.y*c.x-d.x*c.y;
    double k=A1*B2-A2*B1;
    if(fabs(k)<eps)
    {
        if(fabs( C2*A1-C1*A2)<eps && fabs(B1*C2-C1*B2)<eps ) return 2
        else                   return 0;
    }
    return 1;
}
直线相交检测

直线相交求交点  1-相交情况求交点  两点式  double 类型  

Point line_line(Line A,Line B)
{
    Point a=A.a; Point b=A.b;
    Point c=B.a; Point d=B.b;
    double A1=b.y-a.y,B1=-(b.x-a.x),C1=b.y*a.x-b.x*a.y;
    double A2=d.y-c.y,B2=-(d.x-c.x),C2=d.y*c.x-d.x*c.y;
    double k=A1*B2-A2*B1;
    double x=-(B1*C2-C1*B2)*1.000000000/k;
    double y= (A1*C2-C1*A2)*1.00000000/k;
    Point ans;
    ans.x=y;
    ans.y=y;
    return ans;
}
直线相交求交点

 poj 1269  两条直线相交 判断  ( 重合  相交  不相交 )

void XX(Line A,Line B)
{
    Point a=A.a; Point b=A.b;
    Point c=B.a; Point d=B.b;
    double A1=b.y-a.y,B1=-(b.x-a.x),C1=b.y*a.x-b.x*a.y;
    double A2=d.y-c.y,B2=-(d.x-c.x),C2=d.y*c.x-d.x*c.y;
    double k=A1*B2-A2*B1;
    if(fabs(k)<eps)
    {
        if( fabs( C2*A1-C1*A2)<eps && fabs(B1*C2-C1*B2)<eps ) printf("LINE
");
        else                   printf("NONE
");
    }
    else
    {
            double x=-(B1*C2-C1*B2)*1.000000000/k;
            double y=(A1*C2-C1*A2)*1.00000000/k;
            printf("POINT %.2f %.2f
",x,y);
    }
}
直线相交

 线段相交    1-相交 0-不相交   (非严格相交 严格相交)

bool check_se_se_uff(Segment A,Segment B)
{
    Point a=A.a; Point b=A.b;Point c=B.a; Point d=B.b;
    return Cross(a-c,a-d)*Cross(b-c,b-d)<=0 && Cross(c-a,c-b)*Cross(d-a,d-b)<=0 ;

}
bool check_se_se_ff(Segment A,Segment B){
    Point a=A.a; Point b=A.b;Point c=B.a; Point d=B.b;
    return
    max(A.a.x,A.b.x) >= min(B.a.x,B.b.x) &&
    max(B.a.x,B.b.x) >= min(A.a.x,A.b.x) &&
    max(A.a.y,A.b.y) >= min(B.a.y,B.b.y) &&
    max(B.a.y,B.b.y) >= min(A.a.y,A.b.y) &&
    Cross(a-c,a-d)*Cross(b-c,b-d)<=0 &&
    Cross(c-a,c-b)*Cross(d-a,d-b)<=0 ;
}
线段相交检测

线段交点求法同直线一样 

hdu  3272  枚举中间ABCD   P(ABCD)P

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
const double pi=acos(-1.0);
const double D_MAX=1e100;
const double D_MIN=-1e100;
const double eps=1e-9;
int sgn(double x){ if(fabs(x) < eps)return 0;  if(x >0) return 1; return -1; }
int dcmp(double x, double y){ if(fabs(x - y) < eps) return 0; if(x > y) return 1;return -1;}
void usehanshu(){double x;}//floor(x)向下取整函数ceil(x)向上取整函数round(x)四舍五入函数
struct Point  { int  x,y;   Point(int a=0,int b=0)                     { x=a;y=b; };   };
struct Segment{ Point a,b;  Segment(Point x=Point(0,0),Point y=Point(0,0) )  { a=x;b=y; };   };
struct Line   { Point a,b;  Line(Point x=Point(0,0),Point y=Point(0,0) )     { a=x;b=y; };   };
double dis(Point A,Point B) { return sqrt( 1.0*(A.x-B.x)*(A.x-B.x)+1.0*(A.y-B.y)*(A.y-B.y) );  }

/*-----------show time ---------------------*/
int a[5];
double make_x(Point p,Point c){
    if(p.y*c.y>0) p.y=-p.y;
    return dis(p,c);
}
double make_y(Point p,Point c){
    if(p.x*c.x>0) p.x=-p.x;
    return dis(p,c);
}
double make_xy(Point p,Point c){
    if(p.x*c.x>0) p.x=-p.x;
    if(p.y*c.y>0) p.y=-p.y;
    return dis(p,c);
}
int main(){
    int T; scanf("%d",&T);
    while(T--){
        double num=1e9+10;
        Point c; Point d; Point p; scanf("%d %d %d %d %d %d",&c.x,&c.y,&d.x,&d.y,&p.x,&p.y);
        for(int i=1;i<=4;i++) a[i]=i;
        do{
            int t=0; Point w=p;  double ans=0;
            for(int i=1;i<=4;i++){
                if(a[i]==1 || a[i]==2 ) t+=a[i];
                else if(a[i]==3){
                    if(t==0)      ans+=dis(w,c);
                    else if(t==1) ans+=make_x(w,c);
                    else if(t==2) ans+=make_y(w,c);
                    else          ans+=make_xy(w,c);
                    w=c; t=0;
                }
                else if(a[i]==4){
                    if(t==0)      ans+=dis(w,d);
                    else if(t==1) ans+=make_x(w,d);
                    else if(t==2) ans+=make_y(w,d);
                    else          ans+=make_xy(w,d);
                    w=d; t=0;
                }
            }
            if(t==0)      ans+=dis(w,p);
            else if(t==1) ans+=make_x(w,p);
            else if(t==2) ans+=make_y(w,p);
            else          ans+=make_xy(w,p);
            num=min(ans,num);
        }
        while(next_permutation(a+1,a+1+4));
        printf("%.2f
",num);
    }
}
HDU 3272

poj 1873 凸包

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>
#define ll long long
using namespace std;
const int maxn=1e5+10;
struct Point  { double x,y;   Point(double a=0,double b=0) { x=a;y=b; };   };
typedef Point Vector;
Vector operator - (Point  A, Point  B){ return Vector(A.x-B.x, A.y-B.y); } // 向量生成
double Cross(Vector A, Vector B) { return A.x*B.y-A.y*B.x;   }  // 叉积
double dis(Point A,Point B) { return sqrt( (A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y) );  }
Point pa[maxn];
int a[maxn];
int b[maxn];
bool up(Point a,Point b)
{
    if(a.x!=b.x) return a.x<b.x;
    return a.y<b.y;
}
vector<Point> vs;
vector<int> vp;
vector<int> vc;
double work(){
    Point pa[60];
    Point pt[60];
    int n=vs.size();
    for(int i=0;i<n;i++) pa[i]=vs[i];
    sort(pa,pa+n,up);
    int m=0;
    for(int i=0;i<n;i++)
    {
        while(m>1 && Cross(pt[m-1]-pt[m-2],pa[i]-pt[m-2])<=0) m--;
        pt[m++]=pa[i];
    }
    int k=m;
    for(int i=n-2;i>=0;i--)
    {
        while(m>k && Cross(pt[m-1]-pt[m-2],pa[i]-pt[m-2])<=0) m--;
        pt[m++]=pa[i];
    }
    if(n>1) m--;
    double ans=0;
    for(int i=1;i<m;i++) ans+=dis(pt[i],pt[i+1]); ans+=dis(pt[m],pt[1]);
    return ans;
}
int main(){
   int n; int tot=0;
   while(scanf("%d",&n) && n!=0){
     for(int i=1;i<=n;i++){
        scanf("%lf %lf",&pa[i].x,&pa[i].y);
          scanf("%d %d",&a[i],&b[i]);
     }
     vp.clear();
     int ans1=0;
     int ans2=0;
     double ans3=0;
     for(int i=0;i<(1<<n);i++){
        vs.clear();
        vc.clear();
        int value=0;
        int tree=0;
        int num=0;
        for(int j=1;j<=n;j++){
            if(i&(1<<(j-1))){  // 砍掉
                tree+=b[j];   vc.push_back(j);
            }else {   // 剩余
                value+=a[j]; vs.push_back(pa[j]); num++;
            }
        }
        double k=work();  ;
        if(tree>=k){
            if(value>ans1){
                ans1=value; ans2=num;  ans3=tree-k; vp.clear();
                for(int w=0;w<vc.size();w++) vp.push_back(vc[w]);
            }
            else if(value==ans1){
               if(num>=ans2){
                 ans1=value; ans2=num;  ans3=tree-k; vp.clear();
                for(int w=0;w<vc.size();w++) vp.push_back(vc[w]);
               }
            }
        }
     }
     if(tot!=0) printf("
");
     printf("Forest %d
",++tot);
     printf("Cut these trees:"); for(int i=0;i<vp.size();i++){ printf(" %d",vp[i]); } printf("
");
     printf("Extra wood: %.2f
",ans3);

   }
}
View Code

poj 2007 凸包

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>
#define ll long long
using namespace std;
const int maxn=1e5+10;
struct Point  { double x,y;   Point(double a=0,double b=0) { x=a;y=b; };   };
typedef Point Vector;
Vector operator - (Point  A, Point  B){ return Vector(B.x-A.x, B.y-A.y); } // 向量生成
double Cross(Vector A, Vector B) { return A.x*B.y-A.y*B.x;   }  // 叉积
Point pa[maxn];
bool cmp1(Point a,Point b){ return atan2(a.y,a.x)<atan2(b.y,b.x); }
bool cmp2(Point a,Point b){ Point c; return Cross(c-a,c-b)>0;     }
int main(){
    Point s; scanf("%lf %lf",&s.x,&s.y);
    int tot=1;
    while(~scanf("%lf %lf",&pa[tot].x,&pa[tot].y) && pa[tot].x!=1100) {++tot; }
    //cout<<tot<<endl;
    sort(pa+1,pa+tot,cmp2);
    printf("(%.0f,%.0f)
",s.x,s.y);
    for(int i=1;i<tot;i++){
        printf("(%.0f,%.0f)
",pa[i].x,pa[i].y);
    }
}
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原文地址：https://www.cnblogs.com/Andromeda-Galaxy/p/11152067.html