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);
}
}
原文:http://www.cnblogs.com/jie-dcai/p/3872805.html