POJ 2318 toys 叉积的简单运用

POJ 2318 toys 叉积的简单运用

题目意思很简单，把一个盒子分成很多部分，往盒子里扔小球，小球的落点当然要告诉你拉，让你统计每个盒子里最后拥有的小球数。
我的做法是叉积+二分。

#include < iostream >
#include < cmath >
#include < algorithm >
using   namespace  std;

struct  Point 
{              // 二维点或矢量 
    double x, y;  
    Point() {} 
    Point(double x0, double y0): x(x0), y(y0) {} 
} ; 

struct  Line
  {               // 二维的直线或线段 
    Point p1, p2; 
    Line() {} 
    Line(Point p10, Point p20): p1(p10), p2(p20) {} 
} ; 


double  multiply(Point sp,Point ep,Point op)
{
    return((sp.x-op.x)*(ep.y-op.y) - (ep.x-op.x)*(sp.y-op.y));
}



Line myline[ 5100 ];
int  record[ 5100 ];


int  main()
{

    int n,m,i;
    Point left;
    Point right;
    while(scanf("%d",&n))
    {

        if(n==0)
            break;
        memset(record,0,sizeof(record));
        scanf("%d%lf%lf%lf%lf",&m,&left.x,&left.y,&right.x,&right.y);
        myline[0].p1.x=left.x;
        myline[0].p1.y=right.y;
        myline[0].p2.x=left.x;
        myline[0].p2.y=left.y;
        for(i=1;i<=n;i++)
        {
            scanf("%lf%lf",&myline[i].p2.x,&myline[i].p1.x);
            myline[i].p1.y=right.y;
            myline[i].p2.y=left.y;

        }
        myline[n+1].p1.x=right.x;
        myline[n+1].p1.y=right.y;
        myline[n+1].p2.x=right.x;
        myline[n+1].p2.y=left.y;
        
        Point toy;
        for(i=1;i<=m;i++)
        {
            scanf("%lf%lf",&toy.x,&toy.y);
            int front=0;
            int rear=n+1;
            while(front<=rear)
            {

                int mid=(front+rear)>>1;
                if(multiply(toy,myline[front].p2,myline[front].p1)>0&&multiply(toy,myline[mid].p2,myline[mid].p1)<0)
                {

                    if(mid==front+1)
                    {
                        record[front]++;
                        break;
                    }
                    rear=mid;
                    continue;
                }
                else
                {

                    if(mid+1==rear)
                    {
                        record[mid]++;
                        break;
                    }
                    front=mid;
                    continue;
                }

            }
        }

        for(i=0;i<=n;i++)
        {
            printf("%d: %d\n",i,record[i]);
        }
        printf("\n");
    }
    return 0;
}

恭喜此题成为我收集计算几何模板的第一题 呵呵~