[找规律] SRM599 Hard FindPolygons

$Description$

找到一个边数最小的格点多边形使得周长等于$L$，若有多解，要找到最长边和最短边相差最短的。

$Solution$

• 所有边为整数。
• $L=2$无解。
• $L$为奇数无解：
  
  • 对于其中的一条边连接了两个点$(x_1,y_1)$和$(x_2,y_2)$。
  • $c^2=(x_1-x_2)^2+(y_1-y_2)^2$ 。
  • 若$c$为奇数，$x$或$y$的奇偶性会有一个发生改变。
  • 若$c$为偶数，$x$和$y$的奇偶性会同时发生改变或同时不改变。
  • $L$为奇数，说明边长为奇数的边有奇数条，而多边形是从某一点出发回到这点，最终$x$和$y$的奇偶性与最初不可能相同。与$x_n=x_0$且$y_n=y_0$矛盾。
  • $L \equiv0\pmod 4$可以组成一个边长为$L\over4$的正方形
  • $L \equiv2\pmod 4$可以组成一个边长分别为${{L+2\over4}}$，${{L-2\over4}}$的矩形。
  • 关于是否存在整点的三角形，只需要枚举第一象限内合法的点。

  tc的评测机是真的快~~

  // BEGIN CUT HERE
  
  // END CUT HERE
  #line 5 "FindPolygons.cpp"
  #include <bits/stdc++.h>
  using namespace std;
  
  const int INF = 1 << 30;
  const int N = 5050;
  typedef pair<int, int> Pairs;
  
      short s[N * N];
      int ans;
      vector<Pairs> p;
  
  class FindPolygons {
  
  public:
  
      inline int Abs(int x) {
          return x < 0 ? -x : x;
      }
      double minimumPolygon(int L) {  
      //$CARETPOSITION$
          if (L & 1) return -1;
          if (L == 2) return -1;
          for (int i = 1; i <= L; i++)
              s[i * i] = i;
          p.clear();
          for (int i = 0; i <= L / 2; i++)
              for (int j = 0; j * j + i * i <= L * L / 4; j++) 
                  if ((i || j) && s[i * i + j * j]) {
                      p.push_back(Pairs(i, j));
                  }
          ans = INF;
          for (int i = 0; i < p.size(); i++) {
              Pairs u = p[i]; int v, x = u.first, y = u.second, a, b, c;
              a = s[x * x + y * y];
              for (int j = i + 1; j < p.size(); j++) {
                  Pairs v = p[j];
                  if (u == v) continue;
                  int xx = v.first, yy = v.second;
                  b = s[xx * xx + yy * yy];
                  c = (x - xx) * (x - xx) + (y - yy) * (y - yy);
                  if (c > L * L) continue; c = s[c];
                  if (!c) continue; if (a + b + c != L) continue;
                  if (x * yy == xx * y) continue;
                  ans = min(ans, max(Abs(a - b), max(Abs(b - c), Abs(c - a))));
              }
          }
          if (ans == INF) ans = (L >> 1) & 1;
          return ans;
      }
  
  // BEGIN CUT HERE
      public:
      void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); }
      private:
      template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
      void verify_case(int Case, const double &Expected, const double &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
      void test_case_0() { int Arg0 = 4; double Arg1 = 0.0; verify_case(0, Arg1, minimumPolygon(Arg0)); }
      void test_case_1() { int Arg0 = 5; double Arg1 = -1.0; verify_case(1, Arg1, minimumPolygon(Arg0)); }
      void test_case_2() { int Arg0 = 12; double Arg1 = 2.0; verify_case(2, Arg1, minimumPolygon(Arg0)); }
      void test_case_3() { int Arg0 = 4984; double Arg1 = 819.0; verify_case(3, Arg1, minimumPolygon(Arg0)); }
  
  // END CUT HERE
  
  
  };
  
  // BEGIN CUT HERE
  
  
  
  int main(void) {
      FindPolygons ___test;
      ___test.minimumPolygon(16);
      ___test.run_test(-1);
      system("pause");
  }
  // END CUT HERE