hdoj.2892 area【计算几何+圆与多边形相交面积】 2015/08/17

area

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 743 Accepted Submission(s): 289

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
输出一个两位小数，表示实际轰炸到的岛屿的面积。

[align=left]Sample Input[/align]

0 0 2000
100 0
100

4
1900 100
2000 100
2000 -100
1900 -100

[align=left]Sample Output[/align]

15707.96

[align=left]Source[/align]
2009 Multi-University Training Contest 10 - Host by NIT

注：圆与多边形相交面积模板题，根据题中信息可以求出导弹落到岛上时的坐标，即爆炸圆心

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>

using namespace std;

const double eps = 1e-8;
const double PI = acos(-1.0);

int dcmp(double x){
if( x > eps ) return 1;
return x < -eps ? -1 : 0;
}

struct Point{
double x,y;
Point(){
x = y = 0;
}
Point(double a,double b){
x = a;y = b;
}
inline void input(){
scanf("%lf%lf",&x,&y);
}
inline Point operator-(const Point &b)const{
return Point(x - b.x,y - b.y);
}
inline Point operator+(const Point &b)const{
return Point(x + b.x,y + b.y);
}
inline Point operator*(const double &b)const{
return Point(x * b,y * b);
}
inline double dot(const Point &b)const{
return x * b.x + y * b.y;
}
inline double cross(const Point &b,const Point &c)const{
return (b.x - x) * (c.y - y) - (c.x - x) * (b.y - y);
}
inline double Dis(const Point &b)const{
return sqrt((*this-b).dot(*this-b));
}
inline bool InLine(const Point &b,const Point &c)const{ //三点共线
return !dcmp(cross(b,c));
}
inline bool OnSeg(const Point &b,const Point &c)const{ //点在线段上，包括端点
return InLine(b,c) && (*this - c).dot(*this - b) < eps;
}
};

inline double min(double a,double b){
return a < b ? a : b;
}
inline double max(double a,double b){
return a > b ? a : b;
}
inline double Sqr(double x){
return x * x;
}
inline double Sqr(const Point &p){
return p.dot(p);
}

Point LineCross(const Point &a,const Point &b,const Point &c,const Point &d){
double u = a.cross(b,c) , v = b.cross(a,d);
return Point((c.x * v + d.x * u) / (u + v) , (c.y * v + d.y * u) / (u + v));
}

double LineCrossCircle(const Point &a,const Point &b,const Point &r,
double R,Point &p1,Point & p2){
Point fp = LineCross(r , Point(r.x+a.y-b.y , r.y+b.x-a.x) , a , b);
double rtol = r.Dis(fp);
double rtos = fp.OnSeg(a , b) ? rtol : min(r.Dis(a) , r.Dis(b));
double atob = a.Dis(b);
double fptoe = sqrt(R * R - rtol * rtol) / atob;
if( rtos > R - eps ) return rtos;
p1 = fp + (a - b) * fptoe;
p2 = fp + (b - a) * fptoe;
return rtos;
}

double SectorArea(const Point &r,const Point &a,const Point &b,double R){ //不大于180度扇形面积，r->a->b逆时针
double A2 = Sqr(r - a) , B2 = Sqr(r - b) , C2 = Sqr(a - b);
return R * R * acos( (A2 + B2 - C2) * 0.5 / sqrt(A2) / sqrt(B2)) * 0.5;
}

double TACIA(const Point &r,const Point &a,const Point &b,double R){
double adis = r.Dis(a) , bdis = r.Dis(b);
if( adis < R + eps && bdis < R + eps )
return r.cross(a , b) * 0.5;
Point ta , tb;
if( r.InLine(a,b) ) return 0.0;
double rtos = LineCrossCircle(a, b, r, R, ta, tb);
if( rtos > R - eps )
return SectorArea(r, a, b, R);
if( adis < R + eps )
return r.cross(a, tb) * 0.5 + SectorArea(r, tb, b, R);
if( bdis < R + eps )
return r.cross(ta, b) * 0.5 + SectorArea(r, a, ta, R);
return r.cross(ta, tb) * 0.5 + SectorArea(r, tb, b, R) + SectorArea(r, a, ta, R);
}

const int MAXN  = 100010;
Point p[MAXN];

double SPICA(int n,Point r,double R){
int i;
double ret = 0 , if_clock_t;
for( i = 0 ; i < n ; ++i ){
if_clock_t = dcmp(r.cross(p[i], p[(i + 1) % n]));
if( if_clock_t < 0 )
ret -= TACIA(r, p[(i + 1) % n], p[i], R);
else ret += TACIA(r, p[i], p[(i + 1) % n], R);
}
return fabs(ret);
}

int main(){
double xa,xb,ya,yb,h,r;
int n,i;
while( ~scanf("%lf%lf%lf",&xa,&ya,&h) ){
scanf("%lf%lf",&xb,&yb);
double t = sqrt(h/5.0);
Point circle = Point( (xa+t*xb) , (ya+t*yb) );
scanf("%lf",&r);
scanf("%d",&n);
for( i = 0 ; i < n ; ++i )
p[i].input();
printf("%.2lf\n",SPICA(n,circle,r));
}
return 0;
}