poj 2318TOYS

poj 2318(链接)

//第一发：莽写（利用ToLefttest）
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

struct Point{
int x,y;
}edge[5005],point[5005];

int a[5005];

bool ToLeft(int a,int b,int c,int d,int e,int f)
{
    if((a*d+b*e+c*f-d*e-b*c-a*f)<0)return true;
    else return false;
}

int main ()
{
    int n,m;
    while(cin>>n,n)
    {
        cin>>m;
        memset(a,0,sizeof(a));
        cin>>edge[0].x>>edge[0].y>>point[0].x>>point[0].y;
        for(int i=1;i<=n;i++)
            cin>>edge[i].x>>edge[i].y;
        for(int i=1;i<=m;i++)
            cin>>point[i].x>>point[i].y;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
            if(ToLeft(edge[i].x,edge[0].y,edge[i].y,point[0].y,point[j].x,point[j].y))
                a[i]++;
        for(int i=0;i<n;i++)
            cout<<i<<':'<<' '<<a[i+1]-a[i]<<endl;
        cout<<n<<':'<<' '<<m-a[n]<<endl<<endl;
    }
    return 0;
}

通过ToLefttest进行判断是否在每个箱子右边界的逆时针方向的左边；

#include<cstring >
#include<iostream>
using namespace std;

//第二发：参考kuangbin（利用二分+叉积）
struct Point {
    int x,y;
    Point (){};
    Point(int _x,int _y)
    {
        x=_x,y=_y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x-b.x,y-b.y);
    }
    int operator *(const Point &b)const
    {
        return x*b.x+y*b.y;
    }
    int operator ^(const Point &b)const
    {
        return x*b.y-y*b.x;
    }
};

struct Line{
    Point s,e;
    Line(){};
    Line(Point _s,Point _e)
    {
        s=_s,e=_e;
    }
};

int xmult(Point p0,Point p1,Point p2)//求P0p1和 p0p2 的叉积
{
    return (p1-p0)^(p2-p0);
}

const int  MAXN=5050;
Line line[MAXN];
int ans[MAXN];

int main ()
{
    int n,m,x1,y1,x2,y2;
    while(cin>>n,n)
    {
        cin>>m>>x1>>y1>>x2>>y2;
        int U,L;
        for(int i=0;i<n;i++)
        {
            cin>>U>>L;
            line[i]=Line(Point(U,y1),Point(L,y2));
        }
        line[n]=Line(Point(x2,y1),Point(x2,y2));
        int x,y;
        Point p;
        memset(ans,0,sizeof(ans));
        while(m--)
        {
            cin>>x>>y;
            p=Point(x,y);
            int l=0,r=n;
            while(l<r)
            {
                int mid=(l+r)/2;
                if(xmult(p,line[mid].e,line[mid].s)>0)
                    r=mid;
                else
                    l=mid+1;
            }
            ans[l]++;
        }
        for(int i=0;i<=n;i++)
            cout<<i<<": "<<ans[i]<<endl;
            cout<<endl;
    }
    return 0;
}

通过叉积进行判断：