题意: 正方形的房子,给一些墙,墙在区域内是封闭的,给你人的坐标,每穿过一道墙需要一把钥匙,问走出正方形需要多少把钥匙。
解法: 因为墙是封闭的,所以绕路也不会减少通过的墙的个数,还不如不绕路走直线,所以枚举角度,得出直线,求出与正方形内的所有墙交点最少的值,最后加1(正方形边界)。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> #define Mod 1000000007 #define pi acos(-1.0) #define eps 1e-8 using namespace std; #define N 100017 struct Point{ double x,y; Point(double x=0, double y=0):x(x),y(y) {} void input() { scanf("%lf%lf",&x,&y); } }; typedef Point Vector; struct Circle{ Point c; double r; Circle(){} Circle(Point c,double r):c(c),r(r) {} Point point(double a) { return Point(c.x + cos(a)*r, c.y + sin(a)*r); } void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); } }; struct Line{ Point p; Vector v; double ang; Line(){} Line(Point p, Vector v):p(p),v(v) { ang = atan2(v.y,v.x); } Point point(double t) { return Point(p.x + t*v.x, p.y + t*v.y); } bool operator < (const Line &L)const { return ang < L.ang; } }; int dcmp(double x) { if(x < -eps) return -1; if(x > eps) return 1; return 0; } template <class T> T sqr(T x) { return x * x;} Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); } Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); } Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); } Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); } bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); } bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; } bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; } bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; } double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } double Length(Vector A) { return sqrt(Dot(A, A)); } double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } Vector VectorUnit(Vector x){ return x / Length(x);} Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);} double angle(Vector v) { return atan2(v.y, v.x); } bool SegmentIntersection(Point A,Point B,Point C,Point D) { if(dcmp(Cross(C-A,B-A)*Cross(D-A,B-A)) <= 0 && dcmp(Cross(A-C,D-C)*Cross(B-C,D-C)) <= 0) return true; return false; } //data segment struct Seg{ Point P[2]; }seg[45]; //data ends int main() { int n,m,i,j; scanf("%d",&n); for(i=1;i<=n;i++) seg[i].P[0].input(), seg[i].P[1].input(); Point C,D; C.input(); int Mini = Mod; double delta = 2*pi*0.001; for(i=1;i<=1000;i++) { double ang = delta*i; D.x = 10000.0*cos(ang) + C.x; D.y = 10000.0*sin(ang) + C.y; int cnt = 0; for(j=1;j<=n;j++) if(SegmentIntersection(seg[j].P[0],seg[j].P[1],C,D)) cnt++; Mini = min(Mini,cnt); } printf("Number of doors = %d\n",Mini+1); return 0; }