bjfu1235 两圆公共面积
给定两个圆，求其覆盖的面积，其实也就是求其公共面积(然后用两圆面积和减去此值即得最后结果)。
我一开始是用计算几何的方法做的，结果始终不过。代码如下：
/* * Author : ben */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #include <iostream> #include <algorithm> #include <queue> #include <set> #include <map> #include <stack> #include <string> #include <vector> #include <deque> #include <list> #include <functional> #include <numeric> #include <cctype> using namespace std; const double pi = acos(-1); typedef struct MyPoint { double x, y; MyPoint(double xx = 0, double yy = 0) { x = xx; y = yy; } } MyPoint; inline double mydistance2(const MyPoint &p1, const MyPoint &p2) { return (p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y); } inline double mydistance(const MyPoint &p1, const MyPoint &p2) { return sqrt(mydistance2(p1, p2)); } MyPoint intersection(MyPoint u1, MyPoint u2, MyPoint v1, MyPoint v2) { MyPoint ret = u1; double t = ((u1.x - v1.x) * (v1.y - v2.y) - (u1.y - v1.y) * (v1.x - v2.x)) / ((u1.x - u2.x) * (v1.y - v2.y) - (u1.y - u2.y) * (v1.x - v2.x)); ret.x += (u2.x - u1.x) * t; ret.y += (u2.y - u1.y) * t; return ret; } void intersection_line_circle(MyPoint c, double r, MyPoint l1, MyPoint l2, MyPoint& p1, MyPoint& p2) { MyPoint p = c; double t; p.x += l1.y - l2.y; p.y += l2.x - l1.x; p = intersection(p, c, l1, l2); t = sqrt(r * r - mydistance(p, c) * mydistance(p, c)) / mydistance(l1, l2); p1.x = p.x + (l2.x - l1.x) * t; p1.y = p.y + (l2.y - l1.y) * t; p2.x = p.x - (l2.x - l1.x) * t; p2.y = p.y - (l2.y - l1.y) * t; } void intersection_circle_circle(MyPoint c1, double r1, MyPoint c2, double r2, MyPoint& p1, MyPoint& p2) { MyPoint u, v; double t; t = (1 + (r1 * r1 - r2 * r2) / mydistance(c1, c2) / mydistance(c1, c2)) / 2; u.x = c1.x + (c2.x - c1.x) * t; u.y = c1.y + (c2.y - c1.y) * t; v.x = u.x + c1.y - c2.y; v.y = u.y - c1.x + c2.x; intersection_line_circle(c1, r1, u, v, p1, p2); } int main() { freopen("data.in", "r", stdin); // freopen("data.out", "w", stdout); int T; double x, y, r1, r2; scanf("%d", &T); double ans; while (T--) { scanf("%lf%lf%lf", &x, &y, &r1); MyPoint c1(x, y); scanf("%lf%lf%lf", &x, &y, &r2); MyPoint c2(x, y); double dis2 = mydistance2(c1, c2); double dis = sqrt(dis2); if (dis >= r1 + r2) { //相离 ans = pi * r1 * r1 + pi * r2 * r2; } else if (dis <= fabs(r1 - r2)) { //包含 double r = r1 > r2 ? r1 : r2; ans = pi * r * r; } else { //相交 MyPoint p1, p2; intersection_circle_circle(c1, r1, c2, r2, p1, p2); double d2 = mydistance(p1, p2) / 2; double angle1 = asin(d2 / r1); double angle2 = asin(d2 / r2); double Sanjiao1 = sqrt(r1 * r1 - d2 * d2) * d2; double Sanjiao2 = sqrt(r2 * r2 - d2 * d2) * d2; double San1 = r1 * r1 * angle1; double San2 = r2 * r2 * angle2; ans = pi * r1 * r1 + pi * r2 * r2; ans -= San1 + San2 - Sanjiao1 - Sanjiao2; } printf("%.6f\n", ans); } return 0; }
根据后来的调试，应该是对如下图b的情况处理不正确。
于是后来上网找了几个中学的解析几何公式，终于a了。
做法是联立两个圆的方程(相减)，得到相交弦所在直线方程，然后用点到直接的距离公式得到h1和h2，接着算出θ1和θ2，然后就能求得三角形的面积和扇形的面积了。一开始我以为需要分类讨论上面图a和图b两种情况，后来发现，直接去掉求距离时的取绝对值运算就可以了，因为距离为负的时候，得到的夹角也是负的，这样求的三角形面积是负的，扇形也是原先的相补的那部分，具体的图我就不画了，很容易想明白的。
AC代码如下：
/* * Author : ben */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #include <iostream> #include <algorithm> #include <queue> #include <set> #include <map> #include <stack> #include <string> #include <vector> #include <deque> #include <list> #include <functional> #include <numeric> #include <cctype> using namespace std; const double pi = acos(-1); typedef struct MyPoint { double x, y; MyPoint(double xx = 0, double yy = 0) { x = xx; y = yy; } } MyPoint; inline double mydistance2(const MyPoint &p1, const MyPoint &p2) { return (p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y); } int main() { // freopen("data.in", "r", stdin); int T; scanf("%d", &T); double ans, r1, r2; MyPoint c1, c2; while (T--) { scanf("%lf%lf%lf", &c1.x, &c1.y, &r1); scanf("%lf%lf%lf", &c2.x, &c2.y, &r2); double dis2 = mydistance2(c1, c2); double dis = sqrt(dis2); if (dis >= r1 + r2) { //相离 ans = pi * r1 * r1 + pi * r2 * r2; } else if (dis <= fabs(r1 - r2)) { //包含 double r = r1 > r2 ? r1 : r2; ans = pi * r * r; } else { //相交 //h1和h2可能为负 double h1 = (dis2 + r1 * r1 - r2 * r2) / dis / 2.0; double h2 = dis - h1; double angle1 = acos(h1 / r1); double angle2 = acos(h2 / r2); double Sanjiao = sqrt(r1 * r1 - h1 * h1) * dis; double Sanxin1 = r1 * r1 * angle1; double Sanxin2 = r2 * r2 * angle2; ans = pi * r1 * r1 + pi * r2 * r2; ans -= Sanxin1 + Sanxin2 - Sanjiao; } printf("%.6f\n", ans); } return 0; }
