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-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;

}

View Code