hihoCoder#1040 矩形判断

原题地址

一开始写了一个求交点然后算角度的，特别复杂不说，还很容易出错，WA了很久也不知道错在哪

后来想想，应该用向量做

四条直线能围成一个矩形 <=> 四条直线两两之间有4个平行关系，8个垂直关系

注意，直线重合并不算平行

代码清晰明了，一遍AC，这酸爽。

代码：

#include <iostream>
#include <vector>

using namespace std;

struct Line {
  int x[2];
  int y[2];
  Line(int x1, int y1, int x2, int y2) {
    x[0] = x1;
    x[1] = x2;
    y[0] = y1;
    y[1] = y2;
  }
  int vx() {return x[0] - x[1];}
  int vy() {return y[0] - y[1];}
};

bool parallelp(Line &a, Line &b) {
  return a.vx() * b.vy() - b.vx() * a.vy() == 0;
}

bool samep(Line &a, Line &b) {
  Line c(a.x[0], a.y[0], b.x[0], b.y[0]);
  return parallelp(a, c);
}

bool verticalp(Line &a, Line &b) {
  return a.vx() * b.vx() + a.vy() * b.vy() == 0;
}

int parallel_count(vector<Line> &lines) {
  int res = 0;

  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      if (i == j)
        continue;
      if (parallelp(lines[i], lines[j]) && !samep(lines[i], lines[j]))
        res++;
    }
  }

  return res;
}

int vertical_count(vector<Line> &lines) {
  int res = 0;

  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      if (i == j)
        continue;
      if (verticalp(lines[i], lines[j]))
        res++;
    }
  }

  return res;
}

bool rectanglep(vector<Line> &lines) {
  int p_count = parallel_count(lines);
  int v_count = vertical_count(lines);
  return p_count == 4 && v_count == 8;
}

int main() {
  int n;
  double x1, y1, x2, y2;

  cin >> n;
  while (n--) {
    vector<Line> lines;
    for (int i = 0; i < 4; i++) {
      cin >> x1 >> y1 >> x2 >> y2;
      lines.push_back(Line(x1, y1, x2, y2));
    }

    if (rectanglep(lines))
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}