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POJ 1066 Treasure Hunt --几何,线段相交

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-8using namespace std;#define N 100017struct 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 segmentstruct Seg{    Point P[2];}seg[45];//data endsint 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;}
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POJ 1066 Treasure Hunt --几何,线段相交