🧠 题解思路

✅ 目标简化

我们从球出发做一圈 $360^\circ$ 的全角扫描，求出哪些方向可以投进球门（含反弹一次），统计这些角度区间总长度，占比 $360^\circ$。

✅ 关键思路：角度扫描 + 可行方向判断

将每个球门看作一个矩形区域；

从球的位置看球门，计算该球门在球眼中的角度可视区间；

还需考虑球反弹一次后的视角（对称镜像）：

左右边界的镜像球门（沿 $x$ 轴对称）；

上下边界的镜像球门（沿 $y$ 轴对称）；

将所有有效角度段合并，求总角度和；

最后输出：

结果

( 可得分角度总和 2 𝜋 ) × 100 % 结果=( 2π 可得分角度总和 ​ )×100%

✅ 实现细节建议

使用 atan2() 来计算角度，范围 $[-\pi, \pi]$；

每个矩形门用四个顶点转为两个角度段；

所有角度段统一映射为 $[0, 2\pi)$，便于合并；

合并所有有效区间后统计总长度；

输出格式注意百分号与保留两位小数。

代码实现

#include <math.h>
#include <stdio.h>
#define eps (1e-8)
#define pi 3.1415926536
struct rectangle {
	double x1, x2, y1, y2;
} b, g[800];
double x, y, l[1005], r[1005];
void adjust(rectangle & r) {
	double t;
	if (r.x1 - r.x2 > eps) {
		t = r.x1;
		r.x1 = r.x2;
		r.x2 = t;
	}
	if (r.y1 - r.y2 > eps) {
		t = r.y1;
		r.y1 = r.y2;
		r.y2 = t;
	}
}
int pos(rectangle a) {
	if ((a.x1 - x <= eps)  && (x - a.x2 <= eps) && (a.y1 - y <= eps) && (y - a.y2 <= eps)) {
		return 0;
	}
	if (x - a.x1 < eps) {
		if (y > a.y2) {
			return 8;
		} else if (y - a.y1 < eps) {
			return 6;
		} else {
			return 7;
		}
	} else if (x - a.x2 > eps) {
		if (y > a.y2) {
			return 2;
		} else if (y - a.y1 < eps) {
			return 4;
		} else {
			return 3;
		}
	} else {
		if (y - a.y2 > eps) {
			return 1;
		} else {
			return 5;
		}
	}
}
int main() {
	int T, n;
	scanf("%d", &T);
	while (T--) {
		scanf("%lf%lf%lf%lf", &b.x1, &b.y1, &b.x2, &b.y2);
		adjust(b);
		scanf("%lf%lf%d", &x, &y, &n);
		for (int i = 1; i <= n; i++) {
			scanf("%lf%lf%lf%lf", &g[i].x1, &g[i].y1, &g[i].x2, &g[i].y2);
			adjust(g[i]);
			g[i + n].x1 = g[i].x1;
			g[i + n].y1 = 2 * b.y1 - g[i].y1;
			g[i + n].x2 = g[i].x2;
			g[i + n].y2 = 2 * b.y1 - g[i].y2;
			adjust(g[i + n]);
			g[i + 2 * n].x1 = g[i].x1;
			g[i + 2 * n].y1 = 2 * b.y2 - g[i].y1;
			g[i + 2 * n].x2 = g[i].x2;
			g[i + 2 * n].y2 = 2 * b.y2 - g[i].y2;
			adjust(g[i + 2 * n]);
			g[i + 3 * n].x1 = 2 * b.x1 - g[i].x1;
			g[i + 3 * n].y1 = g[i].y1;
			g[i + 3 * n].x2 = 2 * b.x1 - g[i].x2;
			g[i + 3 * n].y2 = g[i].y2;
			adjust(g[i + 3 * n]);
			g[i + 4 * n].x1 = 2 * b.x2 - g[i].x1;
			g[i + 4 * n].y1 = g[i].y1;
			g[i + 4 * n].x2 = 2 * b.x2 - g[i].x2;
			g[i + 4 * n].y2 = g[i].y2;
			adjust(g[i + 4 * n]);
		}
		n *= 5;
		int m = 0;
		bool flag = false;
		double a1, a2;
		for (int i = 1; i <= n; i++) {
			int tmp = pos(g[i]);
			if (tmp == 0) {
				flag = true;
				break;
			} else if (tmp == 1) {
				m++;
				a1 = atan2(y - g[i].y2, x - g[i].x1);
				a2 = atan2(y - g[i].y2, g[i].x2 - x);
				l[m] = pi + a1;
				r[m] = 2 * pi - a2;
			} else if (tmp == 2) {
				m++;
				a1 = atan2(y - g[i].y2, x - g[i].x1);
				a2 = atan2(x - g[i].x2, y - g[i].y1);
				l[m] = pi + a1;
				r[m] = 1.5 * pi - a2;
			} else if (tmp == 3) {
				m++;
				a1 = atan2(x - g[i].x2, g[i].y2 - y);
				a2 = atan2(x - g[i].x2, y - g[i].y1);
				l[m] = 0.5 * pi + a1;
				r[m] = 1.5 * pi - a2;
			} else if (tmp == 4) {
				m++;
				a1 = atan2(x - g[i].x2, g[i].y2 - y);
				a2 = atan2(g[i].y1 - y, x - g[i].x1);
				l[m] = 0.5 * pi + a1;
				r[m] = pi - a2;
			} else if (tmp == 5) {
				m++;
				a1 = atan2(g[i].y1 - y, g[i].x2 - x);
				a2 = atan2(g[i].y1 - y, x - g[i].x1);
				l[m] = a1;
				r[m] = pi - a2;
			} else if (tmp == 6) {
				m++;
				a1 = atan2(g[i].y1 - y, g[i].x2 - x);
				a2 = atan2(g[i].x1 - x, g[i].y2 - y);
				l[m] = a1;
				r[m] = 0.5 * pi - a2;
			} else if (tmp == 7) {
				m++;
				a1 = atan2(g[i].y2 - y, g[i].x1 - x);
				l[m] = 0;
				r[m] = a1;
				m++;
				a2 = atan2(y - g[i].y1, g[i].x1 - x);
				l[m] = 2 * pi - a2;
				r[m] = 2 * pi;
			} else {
				m++;
				a1 = atan2(g[i].x1 - x, y - g[i].y1);
				a2 = atan2(y - g[i].y2, g[i].x2 - x);
				l[m] = 1.5 * pi + a1;
				r[m] = 2 * pi - a2;
			}
		}
		if (flag) {
			printf("100.00%%\n");
		} else {
			double tot = 0, t, ll, rr;
			for (int i = 1; i < n; i++) {
				for (int j = i + 1; j <= n; j++) {
					if (l[i] - l[j] > eps) {
						t = l[i];
						l[i] = l[j];
						l[j] = t;
						t = r[i];
						r[i] = r[j];
						r[j] = t;
					}
				}
			}
			ll = l[1];
			rr = r[1];
			m++;
			l[m] = 361;
			r[m] = 461;
			for (int i = 2; i <= m; i++) {
				if (l[i] - rr <= eps) {
					if (rr - r[i] < eps) {
						rr = r[i];
					}
				} else {
					tot += (rr - ll);
					ll = l[i];
					rr = r[i];
				}
			}
			printf("%.2lf%%\n", tot * 50 / pi);
		}
	}
	return 0;
}