POJ 1066

由于到达点时不能绕墙，因为这是无意义的，所以，两点间的最小墙依然是按照直线所穿过的墙计算。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const double eps=0.000000001;
struct point{
	double x,y;
}Point[100],des;
struct edge{
	point start,end;
}Edge[100],Tmp;

int ne,np;

double multi(point p1,point p2, point p0){
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}

bool cross(edge v1, edge v2){
    if(max(v1.start.x,v1.end.x)>=min(v2.start.x,v2.end.x)&&
    max(v2.start.x,v2.end.x)>=min(v1.start.x,v1.end.x)&&
    max(v1.start.y,v1.end.y)>=min(v2.start.y,v2.end.y)&&
    max(v2.start.y,v2.end.y)>=min(v1.start.y,v1.end.y)&&
    multi(v2.start,v1.end,v1.start)*multi(v1.end,v2.end,v1.start)>eps&&
    multi(v1.start,v2.end,v2.start)*multi(v2.end,v1.end,v2.start)>eps)
    return true;
    return false;
}

int main(){
	while(scanf("%d",&ne)!=EOF){
		np=0;
		for(int i=0;i<ne;i++){
			scanf("%lf%lf",&Point[np].x,&Point[np].y);
			np++;
			scanf("%lf%lf",&Point[np].x,&Point[np].y);
			np++;
			Edge[i].start=Point[np-2]; Edge[i].end=Point[np-1];
		}
		Point[np].x=0; Point[np].y=0;
		np++;
		Point[np].x=0; Point[np].y=100;
		np++;
		Point[np].x=100; Point[np].y=0;
		np++;
		Point[np].x=100; Point[np].y=100;
		np++;
		scanf("%lf%lf",&des.x,&des.y);
		int ans=1000,tmp;
		for(int i=0;i<np;i++){
			Tmp.start=des; Tmp.end=Point[i]; tmp=0;
			for(int i=0;i<ne;i++){
				if(cross(Tmp,Edge[i]))
				tmp++;
			}
			ans=min(ans,tmp);
		}
		printf("Number of doors = %d\n",ans+1);
	}
}

POJ 1066