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;
}