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leoanjos

Have you driven a fjord lately?

Mar 22nd, 2023
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C++ 2.85 KB | None | 0 0
  1. #include <bits/stdc++.h>
  2.  
  3. using namespace std;
  4.  
  5. #define llong long long int
  6. #define ldouble long double
  7. #define Vector Point
  8.  
  9. const ldouble EPS = 1e-6;
  10. const ldouble INF = 1e18 + 5;
  11.  
  12. int sign(ldouble x) {
  13.     return fabsl(x) < EPS ? 0 : (x < 0 ? -1 : 1);
  14. }
  15.  
  16. template<typename T>
  17. struct Point {
  18.     T x, y;
  19.  
  20.     Point(): x(0), y(0) {}
  21.     Point(T x, T y): x(x), y(y) {}
  22.  
  23.     Vector<T> operator -(Point<T> other) {
  24.         return Vector<T>(x - other.x, y - other.y);
  25.     }
  26.  
  27.     T operator *(Vector<T> other) {
  28.         return x * other.x + y * other.y;
  29.     }
  30. };
  31.  
  32. template<typename T>
  33. ldouble abs(Vector<T> v) {
  34.     return sqrtl(v * v);
  35. }
  36.  
  37. template<typename T>
  38. ldouble angle(Vector<T> u, Vector<T> v) {
  39.     return acosl(u * v / abs(u) / abs(v));
  40. }
  41.  
  42. int n, m;
  43. vector<Point<ldouble>> points;
  44. vector<vector<ldouble>> memo;
  45.  
  46. ldouble savings(ldouble a, ldouble c, ldouble theta, ldouble mx) {
  47.     ldouble left = a * cosl(theta);
  48.     ldouble right = sqrtl(a * a * cosl(theta) * cosl(theta) - a * a + c * c);
  49.     cout << a << " " << c << " " << theta << " " << (left + right) << " " << mx << " " << (left + right - mx) << " " << (fabsl(left + right - mx) < EPS) << " " << (left - right) << "\n";
  50.     return sign(left + right - mx) <= 0 ? left + right : (sign(left - right) >= 0 ? left - right : -INF);
  51. }
  52.  
  53. ldouble savings(int fjord, int bridge) {
  54.     Vector<ldouble> u = points[fjord] - points[fjord + 1];
  55.     Vector<ldouble> v = points[fjord + 2] - points[fjord + 1];
  56.  
  57.     ldouble l = 0.0L, r = abs(u);
  58.     ldouble d = abs(v), theta = angle(u, v);
  59.  
  60.     for (int i = 0; i < 50; i++) {
  61.         ldouble m1 = l + (r - l) / 3.0L;
  62.         ldouble m2 = r - (r - l) / 3.0L;
  63.  
  64.         ldouble a1 = m1 + savings(m1, bridge, theta, d);
  65.         ldouble a2 = m2 + savings(m2, bridge, theta, d);
  66.  
  67.         if (sign(a1 - a2) < 0) l = m1;
  68.         else r = m2;
  69.     }
  70.  
  71.     ldouble ans = l + savings(l, bridge, theta, d);
  72.     cout << fjord << " " << bridge << " " << l << " " << ans << "\n";
  73.  
  74.     return l + savings(l, bridge, theta, d);
  75. }
  76.  
  77. ldouble total_savings(int i = 0, int bridge = 0) {
  78.     if (i >= 2 * n) return 0.0L;
  79.  
  80.     ldouble &ans = memo[i][bridge];
  81.     if (sign(ans) >= 0) return ans;
  82.  
  83.     ans = total_savings(i + 2, bridge);
  84.     for (int j = 1; j <= m - bridge; j++)
  85.         ans = max(ans, total_savings(i + 2, bridge + j) + savings(i, j) - j);
  86.  
  87.     return ans;
  88. }
  89.  
  90. int main() {
  91.     ios_base::sync_with_stdio(false);
  92.     cin.tie(NULL);
  93.  
  94.     int t = 1;
  95.     while (cin >> n >> m, n || m) {
  96.         points.resize(2 * n + 1);
  97.         for (int i = 0; i <= 2 * n; i++) {
  98.             int x, y;
  99.             cin >> x >> y;
  100.             points[i] = Point<ldouble>(x, y);
  101.         }
  102.  
  103.         memo.assign(n + 5, vector<ldouble>(m + 5, -1.0L));
  104.  
  105.         ldouble ans = total_savings();
  106.         cout << "Case " << t++ << ": " << ans << "\n";
  107.     }
  108. }
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