poj 1066(枚举+线段相交)

///判断直线与线段相交
///做法：枚举每两个端点,要是存在一条直线经过这两个端点并且和所有线段相交就OK，但是不能为重合点.
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int N = 50;
const double eps = 1e-8;
struct Point
{
    double x,y;
};
struct Line
{
    Point a,b;
} line[N];
int n;
double cross(Point a,Point b,Point c){
    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
}
bool isCross(Point a, Point b, Point c, Point d)
{
    if (cross(c, b, a)*cross(b, d, a)<0)return false;
    if (cross(a, d, c)*cross(d, b, c)<0)return false;
    return true;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf%lf%lf",&line[i].a.x,&line[i].a.y,&line[i].b.x,&line[i].b.y);
        }
        Point e;
        scanf("%lf%lf",&e.x,&e.y);
        int mi = 99999999;
        int cnt;
        for(int j=1; j<=4; j++)
        {
            for(int i=0; i<=100; i++)
            {
                Point s;
                if(j==1) s.x=i,s.y=0;
                if(j==2) s.x=0,s.y=i;
                if(j==3) s.x=100,s.y = i;
                if(j==4) s.x=i,s.y=100;
                cnt=0;
                for(int k=0; k<n; k++){
                    if(fabs(s.x-line[k].a.x)<eps&&fabs(s.y-line[k].a.y)<eps) continue;
                    if(fabs(s.x-line[k].b.x)<eps&&fabs(s.y-line[k].b.y)<eps) continue;
                    if(isCross(s,e,line[k].a,line[k].b)){
                        cnt++;
                    }
                }
                //printf("%d\n",cnt);
                if(mi>cnt) mi = cnt;
            }
        }
        if(n==0) printf("Number of doors = 1\n");
        else printf("Number of doors = %d\n",mi+1);
    }

    return 0;
}