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hdu 2892 多边形与园面积相交

2024-07-03 08:26:01   231人阅读

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 2000
100 0
100 

4
1900 100
2000 100
2000 -100
1900 -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 :_rabbit
Created Time :2014/5/7 10:36:31
File 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;
}



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