hdu 2892 多边形与园面积相交

area

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 623    Accepted Submission(s): 233

Problem Description

小白最近被空军特招为飞行员，参与一项实战演习。演习的内容是轰炸某个岛屿。。。
作为一名优秀的飞行员，任务是必须要完成的，当然，凭借小白出色的操作，顺利地将炸弹投到了岛上某个位置，可是长官更关心的是，小白投掷的炸弹到底摧毁了岛上多大的区域？
岛是一个不规则的多边形，而炸弹的爆炸半径为R。
小白只知道自己在（x，y，h）的空间坐标处以（x1,y1,0）的速度水平飞行时投下的炸弹，请你计算出小白所摧毁的岛屿的面积有多大. 重力加速度G = 10.

 

Input

首先输入三个数代表小白投弹的坐标（x，y，h）；
然后输入两个数代表飞机当前的速度（x1, y1）;
接着输入炸弹的爆炸半径R；
再输入一个数n，代表岛屿由n个点组成；
最后输入n行，每行输入一个（x'，y'）坐标，代表岛屿的顶点（按顺势针或者逆时针给出）。(3<= n < 100000)

 

Output

输出一个两位小数，表示实际轰炸到的岛屿的面积。

 

Sample Input

0 0 2000100 0100 41900 1002000 1002000 -1001900 -100

 

Sample Output

15707.96

多边形与园面积相交主要思想就是把多边形拆成一个一个三角形，计算三角形与园面积相交的结果，然后累加。

三角形与圆面积的计算分了四种情况，具体看这里：http://www.cnblogs.com/lxglbk/archive/2012/08/12/2634192.html

具体代码：

/* ***********************************************Author :_rabbitCreated Time :2014/5/7 10:36:31File Name :F.cpp************************************************ */#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <iostream>#include <algorithm>#include <sstream>#include <stdlib.h>#include <string.h>#include <limits.h>#include <string>#include <time.h>#include <math.h>#include <queue>#include <stack>#include <set>#include <map>using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8#define pi acos(-1.0)typedef long long ll;int dcmp(double x){if(fabs(x)<eps)return 0;return x>0?1:-1;}struct Point{double x,y;Point(double _x=0,double _y=0){x=_x;y=_y;}};Point operator + (Point a,Point b){return Point(a.x+b.x,a.y+b.y);}Point operator - (Point a,Point &b){return Point(a.x-b.x,a.y-b.y);}Point operator * (Point a,double p){return Point(a.x*p,a.y*p);}Point operator / (Point a,double p){return Point(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);}bool operator == (const Point &a,const Point &b){return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;}double Dot(Point a,Point b){return a.x*b.x+a.y*b.y;}double Length(Point a){return sqrt(Dot(a,a));}double Angle(Point a,Point b){return acos(Dot(a,b)/Length(a)/Length(b));}double angle(Point a){return atan2(a.y,a.x);}double Cross(Point a,Point b){return a.x*b.y-a.y*b.x;}Point vecunit(Point a){return a/Length(a);}Point Normal(Point a){return Point(-a.y,a.x)/Length(a);}Point Rotate(Point a,double rad){return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));}bool OnSegment(Point p,Point a1,Point a2){return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<=0;}Point GetLineIntersection(Point p,Point v,Point q,Point w){Point u=p-q;double t=Cross(w,u)/Cross(v,w);return p+v*t;}struct Line{Point p,v;double ang;Line(){}Line(Point _p,Point _v){p=_p;v=_v;ang=atan2(v.y,v.x);}bool operator < (const Line &L) const {return ang<L.ang;}Point point(double a){return p+(v*a);}};Point GetLineIntersection(Line a,Line b){return GetLineIntersection(a.p,a.v,b.p,b.v);}struct Circle  {      Point c;      double r;      Circle(){}      Circle(Point c, double r):c(c), r(r){}      Point point(double a) //根据圆心角求点坐标      {          return Point(c.x+cos(a)*r, c.y+sin(a)*r);      }  }; bool InCircle(Point x,Circle c){return dcmp(c.r-Length(c.c-x))>=0;}bool OnCircle(Point x,Circle c){return dcmp(c.r-Length(c.c-x))==0;}int getSegCircleIntersection(Line L,Circle C,Point *sol){Point nor=Normal(L.v);Line p1=Line(C.c,nor);Point ip=GetLineIntersection(p1,L);double dis=Length(ip-C.c);if(dcmp(dis-C.r)>0)return 0;Point dxy=vecunit(L.v)*sqrt(C.r*C.r-dis*dis);int ret=0;sol[ret]=ip+dxy;if(OnSegment(sol[ret],L.p,L.point(1)))ret++;sol[ret]=ip-dxy;if(OnSegment(sol[ret],L.p,L.point(1)))ret++;return ret;}double SegCircleArea(Circle C,Point a,Point b){double a1=angle(a-C.c);double a2=angle(b-C.c);double da=fabs(a1-a2);if(da>pi)da=pi*2-da;return dcmp(Cross(b-C.c,a-C.c))*da*C.r*C.r/2.0;}double PolyCircleArea(Circle C,Point *p,int n){double ret=0;Point sol[2];p[n]=p[0];for(int i=0;i<n;i++){double t1,t2;int cnt=getSegCircleIntersection(Line(p[i],p[i+1]-p[i]),C,sol);if(cnt==0){if(!InCircle(p[i],C)||!InCircle(p[i+1],C))ret+=SegCircleArea(C,p[i],p[i+1]);else ret+=Cross(p[i+1]-C.c,p[i]-C.c)/2;}if(cnt==1){if(InCircle(p[i],C)&&!InCircle(p[i+1],C))ret+=Cross(sol[0]-C.c,p[i]-C.c)/2,ret+=SegCircleArea(C,sol[0],p[i+1]);else ret+=SegCircleArea(C,p[i],sol[0]),ret+=Cross(p[i+1]-C.c,sol[0]-C.c)/2;}if(cnt==2){if((p[i]<p[i+1])^(sol[0]<sol[1]))swap(sol[0],sol[1]);ret+=SegCircleArea(C,p[i],sol[0]);ret+=Cross(sol[1]-C.c,sol[0]-C.c)/2;ret+=SegCircleArea(C,sol[1],p[i+1]);}}return fabs(ret);}Point p[200000];int main(){     //freopen("data.in","r",stdin);     //freopen("data.out","w",stdout);     Point a,b; double h,R; int n; while(~scanf("%lf%lf%lf",&a.x,&a.y,&h)){ scanf("%lf%lf%lf%d",&b.x,&b.y,&R,&n); for(int i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y); double t=sqrt(h/5); Point g=Point(a.x+b.x*t,a.y+b.y*t); Circle h=Circle(g,R); double ans=PolyCircleArea(h,p,n); printf("%.2lf\n",ans); }     return 0;}