HDOJ 5120 Intersection（求两圆相交面积）

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求两个圆环相交面积，直接容斥即可。

圆环面积 = 两个大圆面积交 - 大圆和小圆交 * 2 + 两个小圆交。

#include <iostream>#include <stdlib.h>#include <stdio.h>#include <math.h>using namespace std;#define eps 1e-8#define pi acos(-1.0)int sig(double x){    return (x > eps) - (x < -eps);}typedef struct Point{        double x, y;        Point() {}        Point(double _x, double _y):                x(_x), y(_y) {}        Point operator -(const Point &argu) const        {                return Point(x - argu.x, y - argu.y);        }        bool operator <(const Point &argu) const        {                if(sig(x - argu.x) == 0)                        return y < argu.y;                return x < argu.x;        }        double dis(const Point &argu) const        {                return sqrt((x - argu.x) * (x - argu.x) + (y - argu.y) * (y - argu.y));        }}pp;typedef struct Circle{    Point o;    double r;    Circle() {}    Circle(Point _o, double _r):        o(_o), r(_r) {}    bool operator <(const Circle &argu) const    {        if(sig(r - argu.r) == 0)            return o < argu.o;        else            return r < argu.r;    }    ///参数圆半径更小些    double Intersection_Area(const Circle &argu) const    {        double d = ((*this).o).dis(argu.o);        if(sig(d - r - argu.r) >= 0)            return 0;        if(sig(r - argu.r - d) >= 0 || sig(d) == 0)            return pi * argu.r * argu.r;        double ang1 = acos((argu.r * argu.r + d * d - r * r) / (2 * argu.r * d));        double ang2 = acos((r * r + d * d - argu.r * argu.r) / (2 * r * d));        return ang1 * argu.r * argu.r + ang2 * r * r - argu.r * d * sin(ang1);    }}cc;cc c1[2], c2[2];double calc(){    scanf("%lf %lf", &c1[0].r, &c1[1].r);    c2[0].r = c1[0].r, c2[1].r = c1[1].r;    scanf("%lf%lf", &c1[0].o.x, &c1[0].o.y);    c1[1].o.x = c1[0].o.x, c1[1].o.y = c1[0].o.y;    scanf("%lf%lf", &c2[0].o.x, &c2[0].o.y);    c2[1].o.x = c2[0].o.x, c2[1].o.y = c2[0].o.y;    return c1[1].Intersection_Area(c2[1]) - c1[1].Intersection_Area(c2[0]) - c2[1].Intersection_Area(c1[0]) + c1[0].Intersection_Area(c2[0]);}int main(){//    freopen("5120.in", "r", stdin);    int t;    scanf("%d", &t);    for(int cas = 1; cas <= t; cas++)    {        printf("Case #%d: %.6lf\n", cas, calc());    }return 0;}

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