Treasure Hunt

Treasure Hunt

http://poj.org/problem?id=1066

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8233		Accepted: 3402

Description

Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid of Key-Ops. Using state-of-the-art technology they are able to determine that the lower floor of the pyramid is constructed from a series of straightline walls, which intersect to form numerous enclosed chambers. Currently, no doors exist to allow access to any chamber. This state-of-the-art technology has also pinpointed the location of the treasure room. What these dedicated (and greedy) archeologists want to do is blast doors through the walls to get to the treasure room. However, to minimize the damage to the artwork in the intervening chambers (and stay under their government grant for dynamite) they want to blast through the minimum number of doors. For structural integrity purposes, doors should only be blasted at the midpoint of the wall of the room being entered. You are to write a program which determines this minimum number of doors. 
An example is shown below: 

Input

The input will consist of one case. The first line will be an integer n (0 <= n <= 30) specifying number of interior walls, followed by n lines containing integer endpoints of each wall x1 y1 x2 y2 . The 4 enclosing walls of the pyramid have fixed endpoints at (0,0); (0,100); (100,100) and (100,0) and are not included in the list of walls. The interior walls always span from one exterior wall to another exterior wall and are arranged such that no more than two walls intersect at any point. You may assume that no two given walls coincide. After the listing of the interior walls there will be one final line containing the floating point coordinates of the treasure in the treasure room (guaranteed not to lie on a wall).

Output

Print a single line listing the minimum number of doors which need to be created, in the format shown below.

Sample Input

7 
20 0 37 100 
40 0 76 100 
85 0 0 75 
100 90 0 90 
0 71 100 61 
0 14 100 38 
100 47 47 100 
54.5 55.4 

Sample Output

Number of doors = 2 

Source

East Central North America 1999

 

枚举边界上的点，要注意，枚举的步长为1会wa,步长要0.5= =

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
const double eps=1e-8;
const double INF=1e20;
const double PI=acos(-1.0);
const int maxp=1010;
int sgn(double x){
    if(fabs(x)<eps) return 0;
    if(x<0) return -1;
    else return 1;
}

double hypot(double x,double y){
    return sqrt(x*x+y*y);
}

inline double sqr(double x){return x*x;}
struct Point{
    double x,y;
    Point(){}
    Point(double _x,double _y){
        x=_x;
        y=_y;
    }
    void input(){
        scanf("%lf %lf",&x,&y);
    }
    void output(){
        printf("%.2f %.2f\n",x,y);
    }
    bool operator == (const Point &b)const{
        return sgn(x-b.x) == 0 && sgn(y-b.y)== 0;
    }
    bool operator < (const Point &b)const{
        return sgn(x-b.x)==0?sgn(y-b.y)<0:x<b.x;
    }
    Point operator - (const Point &b)const{
        return Point(x-b.x,y-b.y);
    }
    //叉积
    double operator ^ (const Point &b)const{
        return x*b.y-y*b.x;
    }
    //点积
    double operator * (const Point &b)const{
        return x*b.x+y*b.y;
    }
    //返回长度
    double len(){
        return hypot(x,y);
    }
    //返回长度的平方
    double len2(){
        return x*x+y*y;
    }
    //返回两点的距离
    double distance(Point p){
        return hypot(x-p.x,y-p.y);
    }
    Point operator + (const Point &b)const{
        return Point(x+b.x,y+b.y);
    }
    Point operator * (const double &k)const{
        return Point(x*k,y*k);
    }
    Point operator / (const double &k)const{
        return Point(x/k,y/k);
    }

    //计算pa和pb的夹角
    //就是求这个点看a,b所成的夹角
    ///LightOJ1202
    double rad(Point a,Point b){
        Point p=*this;
        return fabs(atan2(fabs((a-p)^(b-p)),(a-p)*(b-p)));
    }
    //化为长度为r的向量
    Point trunc(double r){
        double l=len();
        if(!sgn(l)) return *this;
        r/=l;
        return Point(x*r,y*r);
    }
    //逆时针转90度
    Point rotleft(){
        return Point(-y,x);
    }
    //顺时针转90度
    Point rotright(){
        return Point(y,-x);
    }
    //绕着p点逆时针旋转angle
    Point rotate(Point p,double angle){
        Point v=(*this) -p;
        double c=cos(angle),s=sin(angle);
        return Point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
    }
};

struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){
        s=_s;
        e=_e;
    }
    bool operator==(Line v){
        return (s==v.s)&&(e==v.e);
    }
    //根据一个点和倾斜角angle确定直线，0<=angle<pi
    Line(Point p,double angle){
        s=p;
        if(sgn(angle-PI/2)==0){
            e=(s+Point(0,1));
        }
        else{
            e=(s+Point(1,tan(angle)));
        }
    }
    //ax+by+c=0;
    Line(double a,double b,double c){
        if(sgn(a)==0){
            s=Point(0,-c/b);
            e=Point(1,-c/b);
        }
        else if(sgn(b)==0){
            s=Point(-c/a,0);
            e=Point(-c/a,1);
        }
        else{
            s=Point(0,-c/b);
            e=Point(1,(-c-a)/b);
        }
    }
    void input(){
        s.input();
        e.input();
    }
    void adjust(){
        if(e<s) swap(s,e);
    }
    //求线段长度
    double length(){
        return s.distance(e);
    }
    //返回直线倾斜角 0<=angle<pi
    double angle(){
        double k=atan2(e.y-s.y,e.x-s.x);
        if(sgn(k)<0) k+=PI;
        if(sgn(k-PI)==0) k-=PI;
        return k;
    }
    //点和直线的关系
    //1 在左侧
    //2 在右侧
    //3 在直线上
    int relation(Point p){
        int c=sgn((p-s)^(e-s));
        if(c<0) return 1;
        else if(c>0) return 2;
        else return 3;
    }
    //点在线段上的判断
    bool pointonseg(Point p){
        return sgn((p-s)^(e-s))==0&&sgn((p-s)*(p-e))<=0;
    }
    //两向量平行(对应直线平行或重合)
    bool parallel(Line v){
        return sgn((e-s)^(v.e-v.s))==0;
    }
    //两线段相交判断
    //2 规范相交
    //1 非规范相交
    //0 不相交
    int segcrossseg(Line v){
        int d1=sgn((e-s)^(v.s-s));
        int d2=sgn((e-s)^(v.e-s));
        int d3=sgn((v.e-v.s)^(s-v.s));
        int d4=sgn((v.e-v.s)^(e-v.s));
        if((d1^d2)==-2&&(d3^d4)==-2) return 2;
        return (d1==0&&sgn((v.s-s)*(v.s-e))<=0||
                d2==0&&sgn((v.e-s)*(v.e-e))<=0||
                d3==0&&sgn((s-v.s)*(s-v.e))<=0||
                d4==0&&sgn((e-v.s)*(e-v.e))<=0);
    }
    //直线和线段相交判断
    //-*this line -v seg
    //2 规范相交
    //1 非规范相交
    //0 不相交
    int linecrossseg(Line v){
        int d1=sgn((e-s)^(v.s-s));
        int d2=sgn((e-s)^(v.e-s));
        if((d1^d2)==-2) return 2;
        return (d1==0||d2==0);
    }
    //两直线关系
    //0 平行
    //1 重合
    //2 相交
    int linecrossline(Line v){
        if((*this).parallel(v))
            return v.relation(s)==3;
        return 2;
    }
    //求两直线的交点
    //要保证两直线不平行或重合
    Point crosspoint(Line v){
        double a1=(v.e-v.s)^(s-v.s);
        double a2=(v.e-v.s)^(e-v.s);
        return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
    }
    //点到直线的距离
    double dispointtoline(Point p){
        return fabs((p-s)^(e-s))/length();
    }
    //点到线段的距离
    double dispointtoseg(Point p){
        if(sgn((p-s)*(e-s))<0||sgn((p-e)*(s-e))<0)
            return min(p.distance(s),p.distance(e));
        return dispointtoline(p);
    }
    //返回线段到线段的距离
    //前提是两线段不相交，相交距离就是0了
    double dissegtoseg(Line v){
        return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
    }
    //返回点P在直线上的投影
    Point lineprog(Point p){
        return s+(((e-s)*((e-s)*(p-s)))/((e-s).len2()));
    }
    //返回点P关于直线的对称点
    Point symmetrypoint(Point p){
        Point q=lineprog(p);
        return Point(2*q.x-p.x,2*q.y-p.y);
    }
};

Line L[1005];
int n;

bool Check(Line a,Line b){
    if(sgn((a.s-a.e)^(b.s-a.e))*sgn((a.s-a.e)^(b.e-a.e))>0) return false;
    if(sgn((b.s-b.e)^(a.s-b.e))*sgn((b.s-b.e)^(a.e-b.e))>0) return false;
    if(sgn(max(a.s.x,a.e.x)-min(b.s.x,b.e.x))>=0&&sgn(max(b.s.x,b.e.x)-min(a.s.x,a.e.x))>=0
     &&sgn(max(a.s.y,a.e.y)-min(b.s.y,b.e.y))>=0&&sgn(max(b.s.y,b.e.y)-min(a.s.y,a.e.y))>=0)
        return true;
    else return false;
}


double mp[115][115];

int co;

int panduan(Line a){
    int sum=0;
    for(int i=1;i<=n;i++){
        if(Check(a,L[i])){
            sum++;
        }
    }
    return sum;
}

int main(){

    while(~scanf("%d",&n)){
        for(int i=1;i<=n;i++){
            scanf("%lf %lf %lf %lf",&L[i].s.x,&L[i].s.y,&L[i].e.x,&L[i].e.y);
        }
        int ans=0x3f3f3f3f;
        Point goal;
        scanf("%lf %lf",&goal.x,&goal.y);
        Line tmp;
        tmp.s=goal;
        for(double i=0;i+eps<=100;i+=0.5){
            tmp.e.x=0,tmp.e.y=i;
            ans=min(ans,panduan(tmp));
            tmp.e.x=i,tmp.e.y=0;
            ans=min(ans,panduan(tmp));
            tmp.e.x=100,tmp.e.y=i;
            ans=min(ans,panduan(tmp));
            tmp.e.x=i,tmp.e.y=100;
            ans=min(ans,panduan(tmp));
        }
        printf("Number of doors = %d\n",ans+1);
    }

    return 0;
}