P2742 [USACO5.1]圈奶牛Fencing the Cows /【模板】二维凸包

凸包模板题

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5+10;
const double eps = 1e-10;//精度
//atan2(y,x)求极角的弧度
struct Point{
    double x,y;
    Point(double x = 0, double y = 0):x(x),y(y){}
};//点
typedef Point Vector;//向量
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)//判断位置，先按照x坐标排序然后按照y坐标排序
{
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}

int dcmp(double x)//判断x的符号
{
    if(fabs(x) < eps)
        return 0;
    else
        return x < 0 ? -1 : 1;
}
bool operator == (const Point &a,const Point &b)//判断两个点是否重合
{
    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}
double Dot(Vector A,Vector B)//两个向量的点积
{
    return A.x*B.x + A.y*B.y;
}
double Length(Vector A)//向量的长度
{
    return sqrt(Dot(A,A));
}
double Angle(Vector A,Vector B)//求两个向量夹角(弧度)
{
    return acos(Dot(A,B)/Length(A)/Length(B));
}
//A×B,B在A的左边是正，在右边是负
double Cross(Vector A,Vector B)//两个向量的叉积
{
    return A.x*B.y - A.y*B.x;
}
double Area2(Point A,Point B,Point C)//ABC三角形有向面积的两倍
{
    return Cross(B-A,C-A);
}
Vector Rotate(Vector A,double rad)//向量逆时针旋转rad
{
    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
//把向量向左旋转90°,得到法向量,得保证A向量不是零向量
Vector Normal(Vector A)
{
    double L = Length(A);
    return Vector(-A.y/L,A.x/L);
}
//两条直线的交点,向量式--参数方程.前提是v,w向量不平行,Cross(v,w)!=0
Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)
{
    Vector u = P - Q;
    double t = Cross(w,u) / Cross(v,w);//P,v直线的参数t
    return P + v*t;
}
//P到直线A-B距离
double DistanceToLine(Point P, Point A, Point B)
{
    Vector v1=B-A, v2=P-A;
    return fabs(Cross(v1,v2) / Length(v1));
}
//P到线段A-B距离
//设投影点为Q，如果Q在线段AB上，则所求距离就是P点直线AB的距离，如果Q在射线BA上，则所求距离为PA距离；否则为PB距离。
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);//PA距离
    else if(dcmp(Dot(v1,v3)) > 0) return Length(v3);//PB距离
    else return fabs(Cross(v1,v2)) / Length(v1);//PQ距离
}
//求P到直线A-B垂点
Point GetLineProjection(Point P, Point A, Point B)
{
    Vector v=B-A;
    return A+v*(Dot(v,P-A)) / Dot(v,v);
}
//两线是否规范相交
bool SegmentProPerIntersection(Point a1, Point a2, Point b1,Point b2)
{
    double c1=Cross(a2-a1,b1-a1), c2=Cross(a2-a1,b2-a1),
            c3=Cross(b2-b1,a1-b1), c4=Cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}
//p点是否在a1,a2线段上(不在a1,a2点上)
bool OnSegment(Point p, Point a1, Point a2)
{
    return dcmp(Cross(a1-p,a2-p))==0 && dcmp(Dot(a1-p,a2-p)) < 0;
}
//判断多边形的面积(凸，非凸都行)
double PolygomArea(Point *p, int n)
{
    double area = 0;
    for(int i=1;i<n;i++)
    {
        area += Cross(p[i]-p[0],p[i+1]-p[0]);
    }
    return area/2;
}
//Andrew算法求凸包
int ConvexHull(Point *p,int n,Point *ch)
{
    sort(p,p+n);
    int m = 0;
    for(int i=0;i<n;i++)
    {
        while(m > 1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]) <= 0)
            m--;
        ch[m++] = p[i];
    }
    int k = m;
    for(int i=n-2;i>=0;i--)
    {
        while(m > k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]) <= 0)
            m--;
        ch[m++] = p[i];
    }
    if(m > 1)
        m--;
    return m;
}
Point p[N],ch[N];
void solve()
{
    int n;
    cin >> n;
    for(int i=0;i<n;i++)
    {
        cin >> p[i].x >> p[i].y;
    }
    int m = ConvexHull(p,n,ch);
    double ans = 0;
    for(int i=1;i<=m;i++)
    {
        ans += Length(ch[i]-ch[i-1]);
    }
    printf("%0.2lf
",ans);
}
int main()
{
    int T = 1;
    while(T--)
    {
        solve();
    }
    return 0;
}