POJ 1066Treasure Hunt 计算几何

F - Treasure Hunt

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1066

Appoint description: System Crawler  (2016-05-06)

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

n条线段把一个正方形的空间分割成若干个空间

问从外界到达正方形内部点（x，y）最少要穿个多少面墙

思路枚举每个线段的端点和要达到的点组成的线段

判断这条线段与其他线段规范相交的个数

个数最小的那条线段的相交个数就是答案

ACcode：

#include <iostream>#include <cstdio>#include <cmath>#define eps 1e-8using namespace std;int sgn(double x){    if(fabs(x)<eps)return 0;    if(x<0)return -1;    return 1;}struct Point{    double x,y;    Point(){}    Point(double _x,double _y){        x=_x;        y=_y;    }    Point operator -(const Point &b)const{        return Point(x-b.x,y-b.y);    }    double operator ^(const Point &b)const{        return x*b.y-y*b.x;    }    double operator *(const Point &b)const{        return x*b.x+y*b.y;    }};struct Line{    Point s,e;    Line(){}    Line(Point _s,Point _e){        s=_s;        e=_e;    }    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);    }}my[40];int n;int ju(Point t,Point m){    Line tmp=Line(t,m);    int res=0;    for(int i=1;i<=n;++i)        if(tmp.segcrossseg(my[i])==2)        res++;    return res;}int main(){    double x1,x2,y1,y2,x,y;    while(~scanf("%d",&n)){        for(int i=1;i<=n;++i){            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);            my[i]=Line(Point(x1,y1),Point(x2,y2));        }        my[++n]=Line(Point(0,100),Point(100,100));        my[++n]=Line(Point(0,0),Point(0,100));        my[++n]=Line(Point(0,0),Point(100,0));        my[++n]=Line(Point(100,0),Point(100,100));        scanf("%lf%lf",&x,&y);        Point tmp=Point(x,y);        int ans=1e8;        for(int i=1;i<=n;++i){            ans=min(ans,ju(my[i].s,tmp));            ans=min(ans,ju(my[i].e,tmp));        }        printf("Number of doors = %d\n",ans+1);    }    return 0;}