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(-);
typedef struct MyPoint {
    double x, y;
    MyPoint(double xx = , double yy = ) {
        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 = ( + (r1 * r1 - r2 * r2) / mydistance(c1, c2) / mydistance(c1, c2)) / ;
    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) / ;
            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 ;
}

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(-);
typedef struct MyPoint {
    double x, y;
    MyPoint(double xx = , double yy = ) {
        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 ;
}