new: CSES Geometry problems

Author: omar-besbes

solutions/CSES/Geometry/2195 - Convex Hull.cpp

@@ -0,0 +1,69 @@
+#include <bits/stdc++.h>
+using namespace std;
+typedef long long ll;
+
+struct Po {
+   typedef Po P;
+   ll x, y, idx;
+   Po(ll x = 0, ll y = 0) : x(x), y(y) {}
+   bool operator==(P t) const { return x == t.x && y == t.y; }
+   P operator-(P p) const { return P(x - p.x, y - p.y); }
+   ll cross(P p) const { return x * p.y - y * p.x; }
+   ll cross(P a, P b) const { return (a - *this).cross(b - *this); }
+   friend ostream &operator<<(ostream &os, P p) {
+      return os << p.x << " " << p.y << "\n";
+   }
+};
+
+void solve() {
+   int n;
+   cin >> n;
+   vector<Po> v(n);
+   for (int i = 0; i < n; i++) {
+      cin >> v[i].x >> v[i].y;
+      v[i].idx = i;
+   }
+
+   vector<Po> convex_hull;
+
+   auto getlast2 = [&]() {
+      auto x = *(convex_hull.end() - 2), y = *(convex_hull.end() - 1);
+      return make_tuple(x, y);
+   };
+
+   auto check = [&](Po z) {
+      auto [x, y] = getlast2();
+      return x.cross(y, z) <= 0;
+   };
+
+   // construct upper hull
+   sort(v.begin(), v.end(),
+        [](Po a, Po b) { return a.x == b.x ? a.y > b.y : a.x < b.x; });
+   for (auto cur : v) {
+      while (convex_hull.size() >= 2 && !check(cur)) convex_hull.pop_back();
+      convex_hull.push_back(cur);
+   }
+
+   // construct lower hull
+   reverse(v.begin(), v.end());
+   convex_hull.pop_back();
+   int s = convex_hull.size();
+   for (auto cur : v) {
+      while (convex_hull.size() - s >= 2 && !check(cur)) convex_hull.pop_back();
+      convex_hull.push_back(cur);
+   }
+   convex_hull.pop_back();
+
+   cout << convex_hull.size() << "\n";
+   for (auto i : convex_hull) cout << i;
+}
+
+int main() {
+   ios::sync_with_stdio(false);
+   cin.tie(0);
+   cout.tie(0);
+
+   int t = 1;
+   // cin >> t;
+   while (t--) solve();
+}

solutions/CSES/Graph Algorithms/1193 - Labyrinth.cpp

@@ -85,7 +85,7 @@ int main() {
       //					char x = -s[i][j];
       //					cout << x << " ";
       //				} else cout << (s[i][j] > n * m ?
-      //INT_MAX - s[i][j] : s[i][j]) << " ";
+      // INT_MAX - s[i][j] : s[i][j]) << " ";
       //			}
       //			cout << "\n";
       //		}