枚举+判断线段相交
#include<cstdio> #include<cmath> #include<cstring> #include<cmath> #include<algorithm> #include<map> #include<vector> using namespace std; const int INF=0x7FFFFFFF; const int maxn=30+10; #define EPS 1e-8 int n,ans,num; double Px,Py; struct Line { double Sx,Sy; double Ex,Ey; } L[maxn]; struct Point { double x; double y; Point(double a,double b) { x=a; y=b; } }; double cross_pro(Point p0, Point p1, Point p2) { return (p1.x-p0.x)*(p2.y-p0.y) - (p2.x-p0.x)*(p1.y-p0.y); } int dblcmp(double d) { if (fabs(d) < EPS) return 0; return (d > 0) ? 1 : -1; } bool is_intersect(Point p1, Point p2, Point p3, Point p4) { return dblcmp(cross_pro(p3, p4, p1)) * dblcmp(cross_pro(p3, p4, p2)) == -1; } int main() { while(~scanf("%d",&n)) { ans=INF; for(int i=1; i<=n; i++) scanf("%lf%lf%lf%lf",&L[i].Sx,&L[i].Sy,&L[i].Ex,&L[i].Ey); scanf("%lf%lf",&Px,&Py); for(int i=1; i<=n; i++) { Point A(L[i].Sx,L[i].Sy); Point B(Px,Py); num=0; for(int j=1; j<=n; j++) { Point C(L[j].Sx,L[j].Sy); Point D(L[j].Ex,L[j].Ey); if(is_intersect(A,B,C,D)) num++; } ans=min(ans,num); Point AA(L[i].Ex,L[i].Ey); Point BB(Px,Py); num=0; for(int j=1; j<=n; j++) { Point C(L[j].Sx,L[j].Sy); Point D(L[j].Ex,L[j].Ey); if(is_intersect(AA,BB,C,D)) num++; } ans=min(ans,num); } if(n==0) printf("Number of doors = 1\n"); else printf("Number of doors = %d\n",ans+1); } return 0; }
时间: 2024-10-27 00:45:42