HDU 5120 Intersection——两圆面积并

HDU 5120 Intersection——两圆面积并

题意：求两个圆环的面积并

思路：设两个圆环为S1 S2，S1 由 两个圆 c1（大） c2（小）组成，S2由两个圆c3（大） c4（小）组成，那么结果就是c1与c3的面积并减去c1与c4的面积并减去c3与c2的面积并加上c2与c4的面积并

#include #include #include #include #include using namespace std;const double PI = acos(-1.0);double cal(double r1, double x1, double y1, double r2, double x2, double y2) { double d_2 = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2); double d = sqrt(d_2); if (1e-6 + d - r1 - r2 > 0) return 0.0; if (d < fabs(r1 - r2) + 1e-6) { double r = min(r1, r2); return PI * r * r; } double d2 = (d_2 - r1 * r1 + r2 * r2) / (2.0 * d); double d1 = d - d2; double angle1 = 2.0 * acos(d1 / r1); double angle2 = 2.0 * acos(d2 / r2); double temp = sqrt(r1 * r1 - d1 * d1); return angle1 * r1 * r1 / 2.0 - temp * d1 + angle2 * r2 * r2 / 2.0 - temp * d2;}int main() { int T; scanf("%d", &T); for (int kase = 1; kase <= T; kase++) { double r, R, x1, y1, x2, y2; scanf("%lf%lf%lf%lf%lf%lf", &r, &R, &x1, &y1, &x2, &y2); printf("Case #%d: ", kase); double ans1 = cal(R, x1, y1, R, x2, y2); double ans2 = cal(R, x1, y1, r, x2, y2); double ans3 = cal(r, x1, y1, R, x2, y2); double ans4 = cal(r, x1, y1, r, x2, y2); //cout << ans1 << " " << ans2 << " " << ans3 << " " << ans4 << endl; printf("%.6lf\n", ans1 - ans2 - ans3 + ans4); } return 0;}

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