嘟嘟嘟


題意看題中的圖就行：問你從給定的點出發最少需要穿過幾條線段才能從正方形中出去（邊界也算）。


因為\(n\)很小，可以考慮比較暴力的做法。列舉在邊界中的哪一個點離開的。也就是列舉四周的點\((x, y)\)，並和起點\((x_0, y_0)\)連成線段，求和多少條線段相交。
但是因為點可以是實數，所以不知道怎麼列舉。不過想想就知道，同一個區間中的點是等價的。因此我們只要列舉線段的端點即可。
至於判斷線段相交，用叉積實現：對於線段\(AB\)和\(CD\)，如果\((\overrightarrow{AB} \times \overrightarrow{AC}) * (\overrightarrow{AB} \times \overrightarrow{AD}) < 0\)

#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 = 50;
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 Vec
{
  db x, y;
  db operator * (const Vec& oth)const
  {
    return x * oth.y - oth.x * y;
  }
};
struct Point
{
  db x, y;
  Vec operator - (const Point& oth)const
  {
    return (Vec){x - oth.x, y - oth.y};
  }
}a[maxn], b[maxn], P;

int solve(Point A, Point B)
{
  Vec AB = B - A;
  int ret = 0;
  for(int i = 1; i <= n; ++i)
    {
      Vec AC = a[i] - A, AD = b[i] - A;
      Vec CD = b[i] - a[i], CB = B - a[i];
      if((AB * AC) * (AB * AD) < -eps && (CD * AC) * (CD * CB) > eps) ret++;
    }
  return ret;
}

int ans = INF;

int main()
{
  n = read();
  for(int i = 1; i <= n; ++i)
    a[i].x = read(), a[i].y = read(), b[i].x = read(), b[i].y = read();
  scanf("%lf%lf", &P.x, &P.y);
  for(int i = 1; i <= n; ++i)
    {
      ans = min(ans, solve(a[i], P));
      ans = min(ans, solve(b[i], P));
    }
  if(!n) ans = 0;
  printf("Number of doors = ");
  write(ans + 1), enter;
  return 0;
}