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计算几何专项:UVa 11072

2013-04-26 21:51 976 查看
一道纯的模板,就是先求一个凸包,然后再判断点是否在凸包内。需要注意给定点与凸包顶点重合的情况(样例里就有这样的情况,恩,样例还是给了提示的)。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
#define M 100010
struct point
{
int x,y;
point(int x=0,int y=0):x(x),y(y){}
};
point operator-(point a,point b){return point(a.x-b.x,a.y-b.y);}
point operator+(point a,point b){return point(a.x+b.x,a.y+b.y);}
bool operator<(const point& a,const point& b)
{
return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
int cross(point a,point b){return a.x*b.y-a.y*b.x;}
int dot(point a,point b){return a.x*b.x+a.y*b.y;}
bool onseg(point p,point a1,point a2)
{
return cross(a1-p,a2-p)==0&&dot(a1-p,a2-p)<0;
}
vector<point> p,ch;
int P,R;
vector<point> convex()
{
int n=p.size();
vector<point> CH(n+1);
sort(p.begin(),p.end());
int m=0;
for(int i=0;i<n;i++)
{
while(m>1&&cross(CH[m-1]-CH[m-2],p[i]-CH[m-2])<=0) m--;
CH[m++]=p[i];
}
int k=m;
for(int i=n-2;i>=0;i--)
{
while(m>k&&cross(CH[m-1]-CH[m-2],p[i]-CH[m-2])<=0) m--;
CH[m++]=p[i];
}
if(n>1) m--;
CH.resize(m);
return CH;
}
int isin(point p0)
{
int wn=0;
int n=ch.size();
for(int i=0;i<n;i++)
{
if(onseg(p0,ch[i],ch[(i+1)%n])) return 1;
if(p0.x==ch[i].x&&p0.y==ch[i].y) return 1;
int k=cross(ch[(i+1)%n]-ch[i],p0-ch[i]);
int d1=ch[i].y-p0.y;
int d2=ch[(i+1)%n].y-p0.y;
if(k>0&&d1<=0&&d2>0) wn++;
if(k<0&&d2<=0&&d1>0) wn--;
}
if(wn!=0) return 1;
return 0;
}
int main()
{
freopen("in.txt","r",stdin);
while(cin>>P)
{
p.clear();ch.clear();
int x,y;
for(int i=0;i<P;i++)
{
cin>>x>>y;
p.push_back(point(x,y));
}
ch=convex();
cin>>R;
while(R--)
{
cin>>x>>y;
if(isin(point(x,y))) cout<<"inside"<<endl;
else cout<<"outside"<<endl;
}
}
return 0;
}
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