[HDOJ6354] 18多校05 Everything Has Changed (计算几何)

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链接：http://acm.hdu.edu.cn/showproblem.php?pid=6354

给你一个圆和互不相交的n个小圆，让你计算这个大圆被n个小圆切割后的周长，大圆内部的不计算。

很容易发现小圆和大圆相交以后，周长增加了小圆的弧长-大圆的弧长。用余弦定理计算一下大小圆关于交处的角度，然后用弧长公式算一下即可。

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#include <bits/stdc++.h>
using namespace std;

typedef struct Node {
  double x, y, r;
}Node;
const double pi = acos(-1.0);
const int maxn = 111;
int n;
double r;
Node p[maxn];

double dis(Node a, Node b) {
  return (double)sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

double gao(Node a, Node b) {
  if(b.x < 0) b.x = -b.x;
  if(b.y < 0) b.y = -b.y;
  double d = dis(a, b);
  double theta1 = 2.0 * acos((d*d+a.r*a.r-b.r*b.r)/(2*d*a.r));
  double theta2 = 2.0 * acos((d*d+b.r*b.r-a.r*a.r)/(2*d*b.r));
  double L1 = theta1 * a.r;
  double L2 = theta2 * b.r;
  return L2 - L1;
}

signed main() {
  // freopen("in", "r", stdin);
  int T, _ = 1;
  scanf("%d", &T);
  while(T--) {
    cin >> n >> r;
    for(int i = 1; i <= n; i++) {
      cin >> p[i].x >> p[i].y >> p[i].r;
    }
    double ret = 2 * pi * r;
    Node o = {0, 0, r};
    for(int i = 1; i <= n; i++) {
      double d = dis(o, p[i]);
      double R = o.r, r = p[i].r;
      if(d >= R + r) continue;
      if(d == R - r) {
        ret += p[i].r * pi * 2.0;
      }
      if(R - r < d && d < R + r) {
        ret += gao(o, p[i]);
      }
    }
    printf("%.15f\n", ret);
  }
  return 0;
}