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vjudge_UVA-1571_How_I_Mathematician_Wonder_What_You_Are.cpp
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vjudge_UVA-1571_How_I_Mathematician_Wonder_What_You_Are.cpp
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/**vjudge - UVA-1571 - How I Mathematician Wonder What You Are!
* Observe that this problem is basically just art gallery defendable by 1 guard. Use the
* cutpolygon method.
*
* Time: O(n), Space: O(n)
*/
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
#define fast_cin() \
ios_base::sync_with_stdio(false); \
cin.tie(NULL); \
cout.tie(NULL);
typedef long double ld;
const ld EPS = 1e-9;
struct point {
ld x, y;
point() { x = y = 0; }
point(ld _x, ld _y) : x(_x), y(_y) {}
// Compare x-coordinate, if equal compare y-coordinate
bool operator<(const point &p) const {
if (fabs(x - p.x) > EPS) return x < p.x;
return y < p.y;
}
// Compare both x and y
bool operator==(const point &p) const { return (fabs(x - p.x) < EPS && (fabs(y - p.y) < EPS)); }
// Arithmetic Operations (Translation and Scaling)
point operator+(const point &p) const { return point(x + p.x, y + p.y); }
point operator-(const point &p) const { return point(x - p.x, y - p.y); }
point operator*(const ld &r) const { return point(x * r, y * r); }
point operator/(const ld &r) const { return point(x / r, y / r); }
};
// Output Representation of a point
ostream &operator<<(ostream &os, const point &p) { return os << "(" << p.x << "," << p.y << ")"; }
// Euclidean Distance
ld dist(point p1, point p2) { return hypot(p1.x - p2.x, p1.y - p2.y); }
// Vector Struct
struct vec {
ld x, y;
vec(ld _x, ld _y) : x(_x), y(_y) {}
vec(point p1, point p2) : x(p2.x - p1.x), y(p2.y - p1.y) {}
// Vector Operations
vec operator+(const vec &v) const { return vec(x + v.x, y + v.y); }
vec operator-(const vec &v) const { return vec(x - v.x, y - v.y); }
vec operator*(const ld &r) const { return vec(x * r, y * r); }
vec operator/(const ld &r) const { return vec(x / r, y / r); }
// Length
ld length() { return sqrt(x * x + y * y); }
// Length Square
ld length_sq() { return x * x + y * y; }
};
double dot(vec a, vec b) { return (a.x * b.x + a.y * b.y); }
double angle(point a, point o, point b) { // returns angle aob in rad
vec oa = vec(o, a), ob = vec(o, b);
return acos((dot(oa, ob)) / (oa.length() * ob.length()));
}
double cross(vec a, vec b) { return a.x * b.y - a.y * b.x; }
// note: to accept collinear points, we have to change the `> 0'
// returns true if point r is on the left side of line pq
bool ccw(point p, point q, point r) { return cross(vec(p, q), vec(p, r)) > 0; }
// compute the intersection point between line segment p-q and line A-B
point lineIntersectSeg(point p, point q, point A, point B) {
double a = B.y-A.y, b = A.x-B.x, c = B.x*A.y - A.x*B.y;
double u = fabs(a*p.x + b*p.y + c);
double v = fabs(a*q.x + b*q.y + c);
return point((p.x*v + q.x*u) / (u+v), (p.y*v + q.y*u) / (u+v));
}
// cuts polygon Q along the line formed by point A->point B (order matters)
// (note: the last point must be the same as the first point)
vector<point> cutPolygon(point A, point B, const vector<point> &Q) {
vector<point> P;
for (int i = 0; i < (int)Q.size(); ++i) {
double left1 = cross(vec(A, B), vec(A, Q[i])), left2 = 0;
if (i != (int)Q.size()-1) left2 = cross(vec(A, B), vec(A, Q[i+1]));
if (left1 > -EPS) P.push_back(Q[i]); // Q[i] is on the left
if (left1*left2 < -EPS) // crosses line AB
P.push_back(lineIntersectSeg(Q[i], Q[i+1], A, B));
}
if (!P.empty() && !(P.back() == P.front()))
P.push_back(P.front()); // wrap around
return P;
}
int main() {
vector<point> P;
for (int floor=1; ;floor++){
int n;
cin >> n;
if (n == 0)break;
P.clear();
for (int i=0; i<n;i++){
int x, y;
cin >> x >> y;
P.push_back(point(x, y));
}
P.push_back(P[0]);
vector<point> space = P;
for (int i=1; i<(int)P.size(); i++){
point B = P[i], A = P[i-1];
space = cutPolygon(A, B, space);
}
cout << ((int) space.size() != 0 ? "1" : "0") << endl;
}
return 0;
}