Little changes on Colombian Programming Contest solutions.
[and.git] / 10078 - The Art Gallery / p10078.cpp
blob254cc3f5c03e62ffde808aba110ed490c2428b1d
1 #include <iostream>
2 #include <algorithm>
3 #include <vector>
5 using namespace std;
7 struct Point{
8 int x, y;
9 bool operator < (const Point &p) const {
10 return (x < p.x || (x == p.x && y < p.y));
14 // 2D cross product.
15 // Return a positive value, if OAB makes a counter-clockwise turn,
16 // negative for clockwise turn, and zero if the points are collinear.
17 int cross(const Point &O, const Point &A, const Point &B)
19 return (A.x - O.x) * (B.y - O.y) - (A.y - O.y) * (B.x - O.x);
22 // Returns a list of points on the convex hull in counter-clockwise order.
23 // Note: the last point in the returned list is the same as the first one.
24 vector<Point> convexHull(vector<Point> P)
26 int n = P.size(), k = 0;
27 vector<Point> H(2*n);
29 // Sort Points lexicographically
30 sort(P.begin(), P.end());
32 // Build lower hull
33 for (int i = 0; i < n; i++) {
34 while (k >= 2 && cross(H[k-2], H[k-1], P[i]) <= 0) k--;
35 H[k++] = P[i];
38 // Build upper hull
39 for (int i = n-2, t = k+1; i >= 0; i--) {
40 while (k >= t && cross(H[k-2], H[k-1], P[i]) <= 0) k--;
41 H[k++] = P[i];
44 H.resize(k);
45 return H;
48 void printPnt(const Point &p){
49 cout << "(" << p.x << "," << p.y << ")";
52 int main(){
53 int n;
54 cin >> n;
55 while (n){
56 //procesar caso
57 vector<Point> g(n);
58 for (int i=0; i<n; ++i){
59 int x, y;
60 cin >> x >> y;
61 g[i].x = x;
62 g[i].y = y;
64 /*cout << "g.size() es: " << g.size() << endl;
65 for (int i=0; i<n; ++i){
66 printPnt(g[i]);
67 cout << " ";
69 cout << endl; */
70 vector<Point> chull = convexHull(g);
71 /*cout << "chull.size() es: " << chull.size() << endl;
72 for (int i=0; i<chull.size(); ++i){
73 printPnt(chull[i]);
74 cout << " ";
76 cout << endl;*/
77 chull.pop_back();
78 if (chull.size() == n){
79 cout << "No\n";
80 }else{
81 cout << "Yes\n";
83 cin >> n;
85 return 0;