计算几何模板

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邝斌的计算几何模板:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+10,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=2147493647;
const double eps = 1e-8;
const double PI = acos(-1.0);
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}
struct Point
{
    double x,y;
    Point() {}
    Point(double _x,double _y)
    {
        x = _x;
        y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x,y - b.y);
    }
    //叉积
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
//点积
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
//绕原点旋转角度B(弧度值),后x,y的变化
    void transXY(double B)
    {
        double tx = x,ty = y;
        x= tx*cos(B) - ty*sin(B);
        y= tx*sin(B) + ty*cos(B);
    }
};
struct Line
{
    Point s,e;
    Line() {}
    Line(Point _s,Point _e)
    {
        s = _s;
        e = _e;
    }
//两直线相交求交点 //第一个值为0表示直线重合,为1表示平行,为0表示相交,为2是相交 //只有第一个值为2时,交点才有意义
    pair<int,Point> operator &(const Line &b)const
    {
        Point res = s;
        if(sgn((s-e)^(b.s-b.e)) == 0)
        {
            if(sgn((s-b.e)^(b.s-b.e)) == 0) return make_pair(0,res);//重合
            else return make_pair(1,res);//平行
        }
        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 make_pair(2,res);
    }
};
//*两点间距离
double dist(Point a,Point b)
{
    return sqrt((a-b)*(a-b));
}
// *判断线段相交
bool inter(Line l1,Line l2)
{
    return
        max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
        max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
        max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
        max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
        sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0 &&
        sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e)) <= 0;
}
//判断直线l1和线段l2是否相交
bool Seg_inter_line(Line  l1,Line l2)
{
    return sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0;
}
//点到直线距离 //返回为result,是点到直线最近的点
Point PointToLine(Point P,Line L)
{
    Point result;
    double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
    result.x = L.s.x + (L.e.x-L.s.x)*t;
    result.y = L.s.y + (L.e.y-L.s.y)*t;
    return result;
}
//点到线段的距离
//返回点到线段最近的点
Point NearestPointToLineSeg(Point P,Line L)
{
    Point result;
    double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
    if(t >= 0 && t <= 1)
    {
        result.x = L.s.x + (L.e.x - L.s.x)*t;
        result.y = L.s.y + (L.e.y - L.s.y)*t;
    }
    else
    {
        if(dist(P,L.s) < dist(P,L.e)) result = L.s;
        else result = L.e;
    }
    return result;
}
//计算多边形面积
//点的编号从0~n-1
double CalcArea(Point p[],int n)
{
    double res = 0;
    for(int i = 0; i < n; i++)
        res += (p[i]^p[(i+1)%n])/2;
    return fabs(res);
}
//*判断点在线段上
bool OnSeg(Point P,Line L)
{
    return
        sgn((L.s-P)^(L.e-P)) == 0 &&
        sgn((P.x - L.s.x) * (P.x - L.e.x)) <= 0 && sgn((P.y - L.s.y) * (P.y - L.e.y)) <= 0;
}
//*判断点在凸多边形内
//点形成一个凸包,而且按逆时针排序(如果是顺时针把里面的<0改为>0) //点的编号:0~n-1
//返回值:
//-1:点在凸多边形外
//0:点在凸多边形边界上
//1:点在凸多边形内
int inConvexPoly(Point a,Point p[],int n)
{
    for(int i = 0; i < n; i++)
    {
        if(sgn((p[i]-a)^(p[(i+1)%n]-a)) < 0)return -1;
        else if(OnSeg(a,Line(p[i],p[(i+1)%n])))return 0;
    }
    return 1;
}
//*判断点在任意多边形内
//射线法,poly[]的顶点数要大于等于3,点的编号0~n-1 //返回值
//-1:点在凸多边形外
//0:点在凸多边形边界上
//1:点在凸多边形内
int inPoly(Point p,Point poly[],int n)
{
    int cnt;
    Line ray,side;
    cnt = 0;
    ray.s = p;
    ray.e.y = p.y;
    ray.e.x = -100000000000.0;//-INF,注意取值防止越界
    for(int i = 0; i < n; i++)
    {
        side.s = poly[i];
        side.e = poly[(i+1)%n];
        if(OnSeg(p,side))return 0;
//如果平行轴则不考虑
        if(sgn(side.s.y - side.e.y) == 0)
            continue;
        if(OnSeg(side.s,ray))
        {
            if(sgn(side.s.y - side.e.y) > 0)cnt++;
        }
        else if(OnSeg(side.e,ray))
        {
            if(sgn(side.e.y - side.s.y) > 0)cnt++;
        }
        else if(inter(ray,side))
            cnt++;
    }
    if(cnt % 2 == 1)return 1;
    else return -1;
}
//判断凸多边形 //允许共线边
//点可以是顺时针给出也可以是逆时针给出
//点的编号1~n-1
bool isconvex(Point poly[],int n)
{
    bool s[3];
    memset(s,false,sizeof(s));
    for(int i = 0; i < n; i++)
    {
        s[sgn( (poly[(i+1)%n]-poly[i])^(poly[(i+2)%n]-poly[i]) )+1] = true;
        if(s[0] && s[2])return false;
    }
    return true;
}
/*
0求凸包,Graham算法
0点的编号0~n-1
0返回凸包结果Stack[0~top-1]为凸包的编号
 */
const int MAXN = 1010;
Point listt[MAXN];
int Stack[MAXN],top; //相对于listt[0]的极角排序
bool _cmp(Point p1,Point p2)
{
    double tmp = (p1-listt[0])^(p2-listt[0]);
    if(sgn(tmp) > 0)return true;
    else if(sgn(tmp) == 0 && sgn(dist(p1,listt[0]) - dist(p2,listt[0])) <= 0) return true;
    else return false;
}
void Graham(int n)
{
    Point p0;
    int k = 0;
    p0 = listt[0]; //找最下边的一个点
    for(int i = 1; i < n; i++)
    {
        if( (p0.y > listt[i].y) || (p0.y == listt[i].y && p0.x > listt[i].x) )
        {
            p0 = listt[i];
            k = i;
        }
    }
    swap(listt[k],listt[0]);
    sort(listt+1,listt+n,_cmp);
    if(n == 1)
    {
        top = 1;
        Stack[0] = 0;
        return;
    }
    if(n == 2)
    {
        top = 2;
        Stack[0] = 0;
        Stack[1] = 1;
        return ;
    }
    Stack[0] = 0;
    Stack[1] = 1;
    top = 2;
    for(int i = 2; i < n; i++)
    {
        while(top > 1 && sgn((listt[Stack[top-1]]-listt[Stack[top-2]])^(listt[i]-listt[Stack[top-2]])) <= 0)
            top--;
        Stack[top++] = i;
    }
}

void change(double &x0,double &y0,double a,double b,double p)//x0,y0绕a,b旋转P度
{
double x = a + (x0 - a) * cos(p) - (y0 - b) * sin(p);
double y = b + (x0 - a) * sin(p) + (y0 - b) * cos(p);
x0=x;
y0=y;
}

int main()
{

    return 0;
}
原文地址:https://www.cnblogs.com/jhz033/p/6688067.html