POJ 1066 Treasure Hunt(线段相交判断)

Treasure Hunt

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4797		Accepted: 1998

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

题目是穿过墙的中点出来。其实枚举起点和终点，看中间和多少条线段相交。

/************************************************************
 * Author        : kuangbin
 * Email         : [e-mail]
 * Last modified : 2013-07-14 21:59
 * Filename      : POJ1066TreasureHunt.cpp
 * Description   :
 * *********************************************************/

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>

using namespace std;
const double eps = 1e-;
int sgn(double x)
{
    if(fabs(x) < eps)return ;
    if(x < )return -;
    else return ;
}
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);
    }
    //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;
    }
    //绕原点旋转角度B（弧度值），后x,y的变化
    void transXY(double B)
    {
        double tx = x,ty = y;
        x = tx*cos(B) - ty*sin(B);
        y = tx*sin(B) + ty*cos(B);
    }
};
struct Line
{
    Point s,e;
    double k;
    Line(){}
    Line(Point _s,Point _e)
    {
        s = _s;e = _e;
        k = atan2(e.y - s.y,e.x - s.x);
    }
    //两条直线求交点，
    //第一个值为0表示直线重合，为1表示平行，为0表示相交,为2是相交
    //只有第一个值为2时，交点才有意义
    pair<int,Point> operator &(const Line &b)const
    {
        Point res = s;
        if(sgn((s-e)^(b.s-b.e)) == )
        {
            if(sgn((s-b.e)^(b.s-b.e)) == )
                return make_pair(,res);//重合
            else return make_pair(,res);//平行
        }
        double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
        res.x += (e.x-s.x)*t;
        res.y += (e.y-s.y)*t;
        return make_pair(,res);
    }
};
//两点间距离
double dist(Point a,Point b)
{
    return sqrt((a-b)*(a-b));
}

bool inter(Line l1,Line l2)
{
    return
        max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
        max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
        max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
        max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
        sgn((l2.s-l1.s)^(l1.e-l1.s))*sgn((l2.e-l1.s)^(l1.e-l1.s)) <=  &&
        sgn((l1.s-l2.s)^(l2.e-l1.s))*sgn((l1.e-l2.s)^(l2.e-l2.s)) <= ;
}

const int MAXN = ;
Line line[MAXN];

Point s;

Point p[MAXN];

int main()
{
    int n;
    while(scanf("%d",&n)==)
    {
        double x1,y1,x2,y2;
        for(int i = ;i <= n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            line[i] = Line(Point(x1,y1),Point(x2,y2));
            p[*i-] = Point(x1,y1);
            p[*i] = Point(x2,y2);
        }
        scanf("%lf%lf",&x1,&y1);
        s = Point(x1,y1);
        int ans = ;
        for(int i = ;i <= *n;i++)
        {
            int tmp = ;
            Line line1 = Line(s,p[i]);
            for(int j = ;j <= n;j++)
                if(inter(line1,line[j]))
                    tmp++;
            ans = min(ans,tmp);
        }
        Line line1;
        line1 = Line(s,Point(,));
        int tmp = ;
        for(int i = ;i <= n;i++)
            if(inter(line1,line[i]))
                tmp++;
        ans = min(ans,tmp+);
        line1 = Line(s,Point(,));
        tmp = ;
        for(int i = ;i <= n;i++)
            if(inter(line1,line[i]))
                tmp++;
        ans = min(ans,tmp+);
        line1 = Line(s,Point(,));
        tmp = ;
        for(int i = ;i <= n;i++)
            if(inter(line1,line[i]))
                tmp++;
        ans = min(ans,tmp+);
        line1 = Line(s,Point(,));
        tmp = ;
        for(int i = ;i <= n;i++)
            if(inter(line1,line[i]))
                tmp++;
        ans = min(ans,tmp+);
        printf("Number of doors = %d\n",ans);
    }

    return ;
}