POJ1375 Intervals

嘟嘟嘟


题意简述：给出一个光源\((x_0, y_0)\)，和一些圆，求投影区间。

这道题其实就是求经过\((x_0, y_0)\))的圆的切线。 刚开始我想到了一个用向量旋转的方法，但是写起来特别麻烦，于是在网上找到了一个特别巨的大佬的题解，主程序代码不过\(30\)行，这里分享给大家。 这是原文地址

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 505;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n;
struct Point
{
  int x, y;
}P, C;
struct Node
{
  db L, R;
  bool operator < (const Node& oth)const
  {
    return L < oth.L;
  }
}ans[maxn];

db dis(Point A, Point B)
{
  return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}

int main()
{
  while(scanf("%d", &n) && n)
    {
      P.x = read(); P.y = read();
      for(int i = 1; i <= n; ++i)
    {
      C.x = read(); C.y = read(); db r = read();
      db d = dis(P, C);
      db a = asin(r / d), b = asin((P.x - C.x) / d);
      ans[i].L = P.x - P.y * tan(a + b);
      ans[i].R = P.x - P.y * tan(b - a);
    }
      sort(ans + 1, ans + n + 1);
      db l = ans[1].L, r = ans[1].R;
      for(int i = 2; i <= n; ++i)
    {
      if(ans[i].L > r)
        {
          printf("%.2f %.2f\n", l, r);
          l = ans[i].L; r = ans[i].R;
        }
      else r = max(r, ans[i].R);
    }
      printf("%.2f %.2f\n", l, r);
      enter;
    }
  return 0;
}