HDU 5130 Signal Interference（圆与多边形交的面积）

题意：

A国与B国是敌国，A国有一个信号塔，B国有一个信号塔，A国领土是一个凸多边形，当一个位置距B国信号塔的距离是距A国信号塔距离k倍以内的时候，消息将被干扰。
求A国领土上消息被干扰的面积。

思路：

写出方程发现是个圆方程，所以这就是个圆与凸多边形面积交的裸题。。测试下模板。

代码：

#include<bits/stdc++.h>using namespace std;const double eps = 1e-12 ;const double PI = acos( -1.0 ) ;inline double sqr( double x ){ return x * x ; }inline int sgn( double x ){    if ( fabs(x) < eps ) return 0 ;    return x > 0? 1 : -1 ;}struct Point{    double x , y ;    Point(){}    Point( double _x , double _y ): x(_x) , y(_y) {}     void input() { scanf( "%lf%lf" ,&x ,&y ); }    double norm() { return sqrt( sqr(x) + sqr(y) ); }    friend Point operator + ( const Point &a , const Point &b ) { return Point( a.x + b.x , a.y + b.y ) ; }    friend Point operator - ( const Point &a , const Point &b ) { return Point( a.x - b.x , a.y - b.y ) ; }    friend Point operator * ( const Point &a , const double &b ) { return Point( a.x * b , a.y * b ) ; }    friend Point operator * ( const double &a , const Point &b ) { return Point( b.x * a , b.y * a ) ; }    friend Point operator / ( const Point &a , const double &b ) { return Point( a.x / b , a.y / b ) ; }    friend bool operator == ( const Point &a , const Point &b ) { return sgn( a.x - b.x ) == 0 && sgn( a.y - b.y ) == 0 ; }    bool operator < ( const Point &a )const{        return ( sgn( x - a.x ) < 0 ) || ( sgn( x - a.x ) == 0 && sgn( y - a.y ) < 0 ) ;    }};double dot( Point a , Point b ) { return a.x * b.x + a.y * b.y ; }double det( Point a , Point b ) { return a.x * b.y - a.y * b.x ; }double dist( Point a , Point b ) { return ( a - b ).norm() ; }int n ;double k ;Point A,B ;Point p[505] ;Point o ;double r ;int CircleInterLine( Point a, Point b, Point o, double r, Point *p ){    Point p1 = a - o ;    Point d = b - a ;    double A = dot( d, d ) ;    double B = 2 * dot( d, p1 ) ;    double C = dot( p1, p1 ) - sqr(r) ;    double delta = sqr(B) - 4*A*C ;    if ( sgn(delta) < 0 ) return 0 ;//相离    if ( sgn(delta) == 0 ) { //相切        double t = -B / (2*A) ; // 0 <= t <= 1说明交点在线段上        if ( sgn( t - 1 ) <= 0 && sgn( t ) >= 0 ) {            p[0] = a + t * d ;             return 1 ;        }    }    if ( sgn(delta) > 0 ) { //相交        double t1 = ( -B - sqrt(delta) ) / (2*A) ;        double t2 = ( -B + sqrt(delta) ) / (2*A) ; //0 <= t1, t2 <= 1说明交点在线段上        int k = 0 ;        if ( sgn( t1 - 1 ) <= 0 && sgn( t1 ) >= 0 )             p[k++] = a + t1 * d ;         if ( sgn( t2 - 1 ) <= 0 && sgn( t2 ) >= 0 )             p[k++] = a + t2 * d ;        return k ;    }}double Triangle_area( Point a, Point b ){    return fabs( det( a , b ) ) / 2.0  ;}double Sector_area( Point a, Point b ){    double ang = atan2( a.y , a.x ) - atan2( b.y, b.x  ) ;    while ( ang <= 0 ) ang += 2 * PI ;    while ( ang > 2 * PI ) ang -= 2 * PI ;    ang = min( ang, 2*PI - ang ) ;    return sqr(r) * ang/2 ;}double calc( Point a , Point b , double r ){    Point pi[2] ;    if ( sgn( a.norm() - r ) < 0 ) {        if ( sgn( b.norm() - r ) < 0 ) {            return Triangle_area( a, b ) ;        }        else {            CircleInterLine( a, b, Point(0,0), r, pi) ;            return Sector_area( b, pi[0] ) + Triangle_area( a, pi[0] ) ;        }    }    else {        int cnt = CircleInterLine( a, b, Point(0,0), r, pi ) ;        if ( sgn( b.norm() - r ) < 0 ) {            return Sector_area( a, pi[0] ) + Triangle_area( b, pi[0] ) ;        }        else {            if ( cnt == 2 )                return Sector_area( a, pi[0] ) + Sector_area( b, pi[1] ) + Triangle_area( pi[0], pi[1] ) ;            else                return Sector_area( a, b ) ;        }    }}double area_CircleAndPolygon( Point *p , int n , Point o , double r ){    double res = 0 ;    p[n] = p[0] ;    for ( int i = 0 ; i < n ; i++ ) {        int tmp = sgn( det( p[i] - o , p[i+1] - o ) ) ;        if ( tmp )              res += tmp * calc( p[i] - o , p[i+1] - o , r ) ;    }    return fabs( res ) ;}void init() {    double tmp = 1.0 - sqr(k) ;    o.x = ( B.x - sqr(k)*A.x ) / tmp ;    o.y = ( B.y - sqr(k)*A.y ) / tmp ;    r = sqr(B.x) - sqr(k*A.x) + sqr(B.y) - sqr(k*A.y);    r /= tmp ;    r = sqr(o.x) + sqr(o.y) - r ;    r = sqrt( r ) ;}int main(){    int kase = 1 ;    while ( cin >> n >> k ) {        for ( int i = 0 ; i < n ; i++ )            p[i].input() ;        A.input() ; B.input() ;        init() ;        printf( "Case %d: %.12f\n", kase++, area_CircleAndPolygon( p, n, o, r )  ) ;    }    return 0;}