凸包 及 多邊形面積

標籤：凸包   演算法   動態規劃   線段樹   typedef   

首先求多邊形面積，這個比較簡單，用的就是把一個多邊形劃分為多個三角形，然後求三角形面積。

代碼：

double Cross(Vector A,Vector B) { return (A.x*B.y-A.y*B.x); }double ConvexPolygonArea(Point* p,int n)//多邊形面積,，點按順序{    double area=0;    for(int i=1;i<n-1;i++)        area+=Cross(p[i]-p[0],p[i+1]-p[0]);    return area/2;}

求凸包的演算法有很多，這裡給出Andrew演算法。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <queue>#include <vector>#include <map>#include <set>#include <string>using namespace std;#define Del(a,b) memset(a,b,sizeof(a))const int N = 1010;const double esp = 1e-10;struct Point{    double x,y;    Point(double x=0,double y=0):x(x),y(y) {}};typedef Point Vector;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){    return a.x<b.x || (a.x==b.x && a.y<b.y);}int dcmp(double x)  //{    if(fabs(x)<esp) return 0;    else return x<0?-1:1;}bool operator == (const Point& a,const Point& b){    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y)==0;}///計算點積，及向量長度，及向量夾角double Dot(Vector A,Vector B) { return A.x*B.x+A.y*B.y; }double Length(Vector A) { return sqrt(Dot(A,A)); }double Angle(Vector A,Vector B) { return acos(Dot(A,B))/Length(A)/Length(B); }//計算叉積，向量逆時針旋轉double Cross(Vector A,Vector B) { return (A.x*B.y-A.y*B.x); }double Area2(Vector A,Vector B,Vector C)  { return Cross(B-A,C-A); }Vector Rotate(Vector A,double rad){    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}int cmp(Point a,Point b){    if(a.x!=b.x)        return a.x<b.x;    if(a.y!=b.y)        return a.y<b.y;}double ConvexPolygonArea(Point* p,int n)//多邊形面積,，點按順序{    double area=0;    for(int i=1;i<n-1;i++)        area+=Cross(p[i]-p[0],p[i+1]-p[0]);    return area/2;}int ConvexHull(Point *p,Point *ch,int n)//求凸包{    sort(p,p+n);    int i,m=0,k;    for(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];    }    k=m;    for(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(n>1)m--;    return m;}int main(){    int n,k;    Point a[N],ch[N];    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)            scanf("%lf%lf",&a[i].x,&a[i].y);        int num=ConvexHull(a,ch,n);        double ans=ConvexPolygonArea(ch,num);        ans/=50.0;        printf("%d\n",(int)ans);    }    return 0;}