ZOJ 3058 Circle and Ring【圆与环相交面积】【圆与圆相交面积模板】-白红宇

ZOJ 3058 Circle and Ring【圆与环相交面积】【圆与圆相交面积模板】

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发布时间：2019-06-25

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Circle and Ring

Time Limit: 1 Second

Memory Limit: 32768 KB

Given a circle and a ring, your task is to calculate the area of their intersection.

Input

This problem contains multiple test cases, process to the end of file.

For each case, there are two lines. The first line contains three real numbers x', y' and r' (0 <= r' <= 1024) representing the circle. The second line contains four real numbers x, y, r and R (0 <= r <= R <= 1024) representing the ring.

Output

For each case, output the area with the accuracy of three digits after decimal point in a signal line.

Never output "-0.000"!

Sample Input

10 0 20-10 0 10 2020 30 1540 30 0 30

Sample Output

351.041608.366

Author:

WU, Zejun

Source:

ZOJ Monthly, November 2008

题意：给出一个圆的圆心坐标和半径，以及一个圆环的坐标和内外圆的半径，求这个圆和圆环的相交面积。

经分析可得，一共有如下几种情况：

通过计算可得出：

圆和圆环的相交面积 = 圆和圆环中大圆的相交面积 - 圆和圆环中小圆的相交面积。

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          using namespace std;const int maxn = 100010;const double pi = acos(-1.0);struct point { double x, y;};double x1, y11, r1, x2, y2, r2, R2;double D(point a, point b) //两点间距离{ return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));}double calc(point a, double r1, point b, double r2) //计算面积{ double d = D(a, b); double cosaa = (r1*r1 + d*d - r2*r2) / (2 * r1*d); double cosa = acos(cosaa); double l = r1*cosa * 2; double s = l*r1 / 2; double s2 = r1*r1*sin(cosa)*cos(cosa); return s - s2;}bool ainb(point a, double r1, point b, double r2) //判断a是否在b中{ double d = D(a, b); if (d <= r2 - r1) return 1; return 0;}bool ainstb(point a, double r1, point b, double r2) //判断a和b是否有相交的区域{ double d = D(a, b); if (d >= fabs(r1 - r2) && d <= r1 + r2) return 1; return 0;}double S(point a, double r1, point b, double r2) //计算相交面积{ if (ainb(a, r1, b, r2)) return pi*r1*r1; if (ainb(b, r2, a, r1)) return pi*r2*r2; if (!ainstb(a, r1, b, r2)) return 0; return calc(a, r1, b, r2) + calc(b, r2, a, r1);}int main(){ point a, b; double r1, r2, r3; while (~scanf("%lf%lf%lf", &a.x, &a.y, &r1)) { scanf("%lf%lf%lf%lf", &b.x, &b.y, &r2, &r3); printf("%.3f\n", S(a, r1, b, r3) - S(a, r1, b, r2)); } return 0;}