poj 1066 Treasure Hunt

题目的意思就是从边界某个点出发到目标点问最少要和多少条线段相交。

枚举边界上的点判断就行了。

注意n=0时ans=1

CODE：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<vector>using namespace std;#define FOR(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)#define DOR(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)#define oo 1e6#define eps 1e-8#define nMax 1000#define pb push_back#define bug puts("OOOOh.....");#define zero(x) (((x)>0?(x):-(x))<eps)int dcmp(double x){    if(fabs(x)<eps) return 0;    return x>0?1:-1;}struct point {    double x,y;    point(double x=0,double y=0): x(x),y(y) {}    void make(double _x,double _y) {x=_x;y=_y;}    void read(){ scanf("%lf%lf",&x,&y); }    double len(){ return sqrt(x*x+y*y); }    friend point operator -(point const& u,point const& v) {        return point(u.x-v.x,u.y-v.y);    }    friend point operator +(point const& u,point const& v) {        return point(u.x+v.x,u.y+v.y);    }    friend double operator *(point const& u,point const& v) {        return u.x*v.y-u.y*v.x;    }    friend double operator ^(point const& u,point const& v) {        return u.x*v.x+u.y*v.y;    }    friend point operator *(double const& k,point const& v) {        return point(k*v.x,k*v.y);    }    friend point operator /(point const& u,double const& k){        return point(u.x/k,u.y/k);    }    friend bool operator < (point const& u,point const& v) {        if(dcmp(u.x-v.x) == 0) return dcmp(u.y-v.y)<0;        return dcmp(u.x-v.x)<0;    }    friend bool operator == (point const& u,point const& v) {        return (dcmp(u.x-v.x) == 0) && dcmp(u.y-v.y)==0;    }    friend int dots_online(point,point,point);};int dots_online(point a,point b,point c){ return dcmp((a-c)*(b-c))==0; }typedef struct line{    point a,b;    line() {}    line(point a,point b): a(a),b(b) {}    void make(point _a,point _b) {a=_a;b=_b;}    void read() { a.read(),b.read(); }    friend int intersection(line,line);} segment;int dot_in_line(point p,line l){    return dcmp((l.a-p)*(p-l.b))==0 && dcmp((l.a-p)^(p-l.b))>=0;}int sameside(point a,point b,line l){    return dcmp((l.a-a)*(a-l.b)) * dcmp((l.a-b)*(b-l.b)) > 0;}int intersection(line u,line v){    if(dots_online(u.a,u.b,v.a) && dots_online(u.a,u.b,v.b))        return dot_in_line(u.a,v) || dot_in_line(u.b,v) || dot_in_line(v.a,u) || dot_in_line(v.a,u);    else        return !sameside(u.a,u.b,v) && !sameside(v.a,v.b,u);}int n,cnt;point p,p1,p2;line l[nMax];vector<point> s[4];int g[nMax][nMax];int main(){#ifndef ONLINE_JUDGE    freopen("input.txt","r",stdin);#endif    scanf("%d",&n);    for(int i=0;i<n;i++) {        p1.read(),p2.read();        l[i].make(p1,p2);    }    p.read();    int ans = oo,f;    if(n==0) ans=0;    for(int i=0;i<n;i++) {        segment s(p,l[i].a);        f=0;        for(int k=0;k<n;k++) if(!dot_in_line(s.b,l[k]) && intersection(s,l[k])) f++;        if(ans>f) ans =f;        s.b=l[i].b;        f=0;        for(int k=0;k<n;k++) if(!dot_in_line(s.b,l[k]) && intersection(s,l[k])) f++;        ans = min(ans,f);    }    printf("Number of doors = %d\n",ans+1);    return 0;}