hdu 2892 Area

http://acm.hdu.edu.cn/showproblem.php?pid=2892

解题思路：

求多边形与圆的相交的面积是多少。

以圆心为顶点，将多边形划分为n个三角形。

接下来就求出每个三角形与圆相交的面积。

因为三角形的一个点是圆心，所以三角形的另外两个点与圆的情况有以下几种：

（1）两点都在圆里，三角形与圆相交的面积=三角形的面积。

（2）一个点在圆外，一个点在圆里，三角形与圆相交的面积=小三角形的面积+扇形面积

（3）两点都在圆外，又分为几种情况：

　　1、两点构成的线段与圆相交的点数0或1个时，三角形与圆相交的面积=扇形的面积

　　2.两点构成的线段与圆相交的点数2个时，三角形与圆相交的面积=大扇形面积+小三角形面积-小扇形的面积

#include<cmath>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;

#define MAXN 100000+10
#define PI acos(-1.0)
#define EPS 0.00000001

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) {}
};

struct Circle{
    Point c;
    double r;
    Circle(Point c = Point(0, 0), double r = 0): c(c), r(r) {}
};

typedef Point Vector;

Vector operator + (Vector A, Vector B){
    return Vector(A.x + B.x, A.y + B.y);
}
Vector operator - (Point A, Point 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);
}

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));
}

double cross(Vector A, Vector B){
    return A.x * B.y - A.y * B.x;
}

Circle bomb;//炸弹爆炸的坐标及半径
Point p[MAXN];//岛屿的点
int n;//岛屿点数

double point_line_distance(Point P, Point A, Point B){//点到直线的距离
    Vector AP = P - A, AB = B - A;
    return fabs(cross(AP, AB) / length(AB));
}

Point point_line_projection(Point P, Point A, Point B){//点在直线上的映射
    Vector v = B - A;
    return A + v * (dot(v, P - A) / dot(v, v));
}

int circle_line_intersect(Circle C, Point A, Point B, vector<Point> &v){
    double dist = point_line_distance(C.c, A, B);
    int d = dcmp(dist - C.r);
    if(d > 0){
        return 0;
    }
    Point pro = point_line_projection(C.c, A, B);
    if(d == 0){
        v.push_back(pro);
        return 1;
    }
    double len = sqrt(C.r * C.r - dist * dist);//勾股定理
    Vector AB = B - A;
    Vector l = AB / length(AB) * len;
    v.push_back(pro + l);
    v.push_back(pro - l);
    return 2;
}

bool point_on_segment(Point P, Point A, Point B){//判断点在线段上
    Vector PA = A - P, PB = B - P;
    return dcmp(cross(PA, PB)) == 0 && dcmp(dot(PA, PB)) <= 0;
}

double circle_delta_intersect_area(Circle C, Point A, Point B){
    Vector CA = A - C.c, CB = B - C.c;
    double da = length(CA), db = length(CB);

    da = dcmp(da - C.r), db = dcmp(db - C.r);

    if(da <= 0 && db <= 0){//三角形在圆里面
        return fabs(cross(CA, CB)) * 0.5;
    }

    vector<Point> v;
    int num = circle_line_intersect(C, A, B, v);//圆和直线的关系
    double carea = C.r * C.r * PI;
    Point t;
    if(da <= 0 && db > 0){//左边的点在圆里 右边的点在圆外
        t = point_on_segment(v[0], A, B) ? v[0] : v[1];

        double area = fabs(cross(CA, t - C.c)) * 0.5, an = angle(CB, t - C.c);
        return area + carea * an / PI / 2;
    }
    if(da > 0 && db <= 0){//左边点在圆外 右边点在圆里
        t = point_on_segment(v[0], A, B) ? v[0] : v[1];

        double area = fabs(cross(CB, t - C.c)) * 0.5, an = angle(CA, t - C.c);
        return area + carea * an / PI / 2;
    }
    //两个点都在圆外
    if(num == 2){
        double bigarea = carea * angle(CA, CB) / PI / 2,
            smallarea = carea * angle(v[0] - C.c, v[1] - C.c) / PI / 2,
            deltaarea = fabs(cross(v[0] - C.c, v[1] - C.c)) * 0.5;
        return bigarea + deltaarea - smallarea;
    }
    return carea * angle(CA, CB) / PI / 2;//两点都在圆外 直线AB与圆交点1个或两个
}

double circle_polygon_intersect_area(){//源于多边形相交面积
    p[n] = p[0];
    double ans = 0;
    for(int i = 0; i < n; i++ ){
        double area = circle_delta_intersect_area( bomb, p[i], p[i + 1] );
        if(cross(p[i] - bomb.c, p[i + 1] - bomb.c) < 0){
            area = -area;
        }
        ans += area;
    }
    return ans > 0 ? ans : -ans;
}

void solve(){
    scanf("%d", &n );
    for(int i = 0; i < n; i++ ){
        scanf("%lf%lf", &p[i].x, &p[i].y );
    }
    printf("%.2lf\n", circle_polygon_intersect_area() );
}

int main(){
    //freopen("data.in", "r", stdin );
    double x, y, h, x1, y1, r;
    while(~scanf("%lf%lf%lf", &x, &y, &h )){
        scanf("%lf%lf%lf", &x1, &y1, &r  );

        double t = sqrt(0.2 * h);//h = 0.5 * G * t^2 重力加速度公式

        bomb = Circle( Point(x1 * t + x, y1 * t + y), r );

        solve();
    }
    return 0;
}