CTU Open Contest 2019 C. Beer Coasters（计算几何）

题意：

给定一个圆的圆心和一个矩形的左下方坐标和右上方坐标，除圆的半径之外的 666 个数据均
在−1000-1000−1000 到 100010001000 范围内，圆的半径在 111 到 100010001000 范围。求圆与矩形相交的面积。
计算几何，求圆和凸多边形的相交面积模板。

AC代码：

#include <bits/stdc++.h>
#define inf 1000000000000
#define M 100009
#define eps 1e-12
#define PI acos(-1.0)
using namespace std;
struct Point
{double x, y;Point() {}Point(double xx, double yy){x = xx;y = yy;}Point operator-(Point s) { return Point(x - s.x, y - s.y); }Point operator+(Point s) { return Point(x + s.x, y + s.y); }double operator*(Point s) { return x * s.x + y * s.y; }double operator^(Point s) { return x * s.y - y * s.x; }
} p[M];
double max(double a, double b) { return a > b ? a : b; }
double min(double a, double b) { return a < b ? a : b; }
double len(Point a) { return sqrt(a * a); }
double dis(Point a, Point b) { return len(b - a); } //Á½µãÖ®¼äµÄ¾àÀë
double cross(Point a, Point b, Point c)             //²æ³Ë
{return (b - a) ^ (c - a);
}
double dot(Point a, Point b, Point c) //µã³Ë
{return (b - a) * (c - a);
}
int judge(Point a, Point b, Point c) //ÅÐ¶ÏcÊÇ·ñÔÚabÏß¶ÎÉÏ£¨Ç°ÌáÊÇcÔÚÖ±ÏßabÉÏ£©
{if (c.x >= min(a.x, b.x) && c.x <= max(a.x, b.x) && c.y >= min(a.y, b.y) && c.y <= max(a.y, b.y))return 1;return 0;
}
double area(Point b, Point c, double r)
{Point a(0.0, 0.0);if (dis(b, c) < eps)return 0.0;double h = fabs(cross(a, b, c)) / dis(b, c);if (dis(a, b) > r - eps && dis(a, c) > r - eps) //Á½¸ö¶Ëµã¶¼ÔÚÔ²µÄÍâÃæÔò·ÖÎªÁ½ÖÖÇé¿ö{double angle = acos(dot(a, b, c) / dis(a, b) / dis(a, c));if (h > r - eps)return 0.5 * r * r * angle;else if (dot(b, a, c) > 0 && dot(c, a, b) > 0){double angle1 = 2 * acos(h / r);return 0.5 * r * r * fabs(angle - angle1) + 0.5 * r * r * sin(angle1);}elsereturn 0.5 * r * r * angle;}else if (dis(a, b) < r + eps && dis(a, c) < r + eps)return 0.5 * fabs(cross(a, b, c)); //Á½¸ö¶Ëµã¶¼ÔÚÔ²ÄÚµÄÇé¿öelse                                   //Ò»¸ö¶ËµãÔÚÔ²ÉÏÒ»¸ö¶ËµãÔÚÔ²ÄÚµÄÇé¿ö{if (dis(a, b) > dis(a, c))swap(b, c); //Ä¬ÈÏbÔÚÔ²ÄÚif (fabs(dis(a, b)) < eps)return 0.0; //ab¾àÀëÎª0Ö±½Ó·µ»Ø0if (dot(b, a, c) < eps){double angle1 = acos(h / dis(a, b));double angle2 = acos(h / r) - angle1;double angle3 = acos(h / dis(a, c)) - acos(h / r);return 0.5 * dis(a, b) * r * sin(angle2) + 0.5 * r * r * angle3;}else{double angle1 = acos(h / dis(a, b));double angle2 = acos(h / r);double angle3 = acos(h / dis(a, c)) - angle2;return 0.5 * r * dis(a, b) * sin(angle1 + angle2) + 0.5 * r * r * angle3;}}
}
int main()
{int n = 4;double rx, ry, R;scanf("%lf%lf%lf", &rx, &ry, &R);scanf("%lf%lf%lf%lf", &p[1].x, &p[1].y, &p[3].x, &p[3].y);p[2].x = p[1].x;p[2].y = p[3].y;p[4].x = p[3].x;p[4].y = p[1].y;p[5] = p[1];Point O(rx, ry);for (int i = 1; i <= n + 1; i++)p[i] = p[i] - O;O = Point(0, 0);double sum = 0;for (int i = 1; i <= n; i++){int j = i + 1;double s = area(p[i], p[j], R);if (cross(O, p[i], p[j]) > 0)sum += s;elsesum -= s;}printf("%.4lf\n", fabs(sum));
}

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CTU Open Contest 2019 C. Beer Coasters（计算几何）

题意：

AC代码：

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