艰难入门ing
感觉我这样的去写计算几何光调bug就得5小时
参考资料(很多):
POJ1127 Jack Straws
先用向量判交,然后传递闭包
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <vector> using namespace std; typedef double db; namespace Geo { const db eps = 1e-8; const db Pi = acos(-1); inline int sgn(db x) { if (x < -eps) return -1; else if (x > eps) return 1; else return 0; } struct Vector { db x, y; Vector(db x = 0, db y = 0) : x(x), y(y) {} inline friend Vector operator + (Vector u, Vector v) { return Vector(u.x + v.x, u.y + v.y); } inline friend Vector operator - (Vector u, Vector v) { return Vector(u.x - v.x, u.y - v.y); } inline friend Vector operator * (Vector u, Vector v) { return Vector(u.x * v.x, u.y * v.y); } inline friend Vector operator / (Vector u, Vector v) { return Vector(u.x / v.x, u.y / v.y); } inline friend bool operator == (Vector u, Vector v) { return sgn(u.x - v.x) == 0 && sgn(u.y - v.y) == 0; } }; typedef Vector Point; typedef vector <Point> Polygon; inline db dot(Vector u, Vector v) { return u.x * v.x + u.y * v.y; } inline db crs(Vector u, Vector v) { return u.x * v.y - u.y * v.x; } inline db length(Vector u) { return max(0.0, sqrt(dot(u, u))); } inline db angle(Vector u, Vector v) { return acos(dot(u, v) / length(u) / length(v)); } inline Vector rotate(Vector u, db rad) { return Vector(u.x * cos(rad) - u.y * sin(rad), u.x * sin(rad) + u.y * cos(rad)); } struct Line { Point p; Vector v; Line(Point p = Point(), Vector v = Vector()) : p(p), v(v) {} }; struct Ray : public Line { Ray(Point p = Point(), Vector v = Vector()) : Line(p, v) {} }; struct Segment { Point a, b; Segment(Point a = Point(0, 0), Point b = Point(0, 0)) : a(a), b(b) {} }; inline bool isCrs(Segment u, Segment v) { Point p1 = u.a, p2 = u.b, q1 = v.a, q2 = v.b; if (min(p1.x, p2.x) <= max(q1.x, q2.x) && min(q1.x, q2.x) <= max(p1.x, p2.x) && min(p1.y, p2.y) <= max(q1.y, q2.y) && min(q1.y, q2.y) <= max(p1.y, p2.y)) { //puts("in"); if (sgn(crs(q1 - p1, p2 - p1) * crs(q2 - p1, p2 - p1)) > 0) return 0; //puts("1"); if (sgn(crs(p1 - q1, q2 - q1) * crs(p2 - q1, q2 - q1)) > 0) return 0; //puts("2"); return 1; } else return 0; } } using namespace Geo; const int N = 20; int n; Segment s[N]; bool f[N][N]; int main() { #ifndef ONLINE_JUDGE freopen("sample.in", "r", stdin); #endif for (; scanf("%d", &n); getchar()) { if (n == 0) break; for (int i = 1; i <= n; i++) { int x1, y1, x2, y2; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); // printf("%f %f %f %f ", Point(x1, y1).x, Point(x1, y1).y, Point(x2, y2).x, Point(x2, y2).y); s[i] = Segment(Point(x1, y1), Point(x2, y2)); } for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) if (isCrs(s[i], s[j])) f[i][j] = 1; else f[i][j] = 0; for (int k = 1; k <= n; k++) for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) f[i][j] |= f[i][k] & f[k][j]; /* for (int i = 1; i <= n; i++) printf("%f %f %f %f ", s[i].a.x, s[i].a.y, s[i].b.x, s[i].b.y); */ for (int u, v; ; ) { scanf("%d%d", &u, &v); if (u == 0 && v == 0) break; if (f[u][v]) puts("CONNECTED"); else puts("NOT CONNECTED"); } } return 0; }
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