poj 2138-TOYS解题报告

链接：http://poj.org/problem?id=2318

这道题是比较简单的计算几何问题，只要找每个点应该在第几块内部就可以了，用叉积判断方向，二分查找

#include<stdio.h> 
#include<string.h>
struct point
{
    double x,y;
    };
struct
point a[10005];
int num[5005];
double mul(point p1,point p2,point p3)//叉积
{
    return (p2.x-p1.x)*(p3.y-p2.y)-(p3.x-p2.x)*(p2.y-p1.y);
    }
int main()
{
    int n,m;
    point t1,t2;
    int flage=0;
    while(scanf("%d",&n)&&n)
    {
        if(flage)
        printf("\n");
        flage=1;
        memset(num,0,sizeof(num));
        scanf("%d%lf%lf%lf%lf",&m,&t1.x,&t1.y,&t2.x,&t2.y);
        a[0].x=t1.x;
        a[0].y=t1.y;
        a[1].x=t1.x;
        a[1].y=t2.y;
        for(int i=2;i<=2*n;i+=2)
        {
            double x1,x2;
            scanf("%lf%lf",&x1,&x2);
            a[i].x=x1;
            a[i+1].x=x2;
            a[i].y=t1.y;
            a[i+1].y=t2.y;
            }
        a[2*n+2].x=t2.x;
        a[2*n+2].y=t1.y;
        a[2*n+3].x=t2.x;
        a[2*n+3].y=t2.y;
        point p1;
        for(int j=0;j<m;j++)
        {
            scanf("%lf%lf",&p1.x,&p1.y);
            int low=0;
            int high=n;
            int mid=(low+high)/2;
            while(mul(p1,a[2*mid+1],a[2*mid])*mul(p1,a[2*mid+3],a[2*mid+2])>0)//二分查找
            {
                if(mul(p1,a[2*mid+1],a[2*mid])>0)
                {
                    high=mid-1;
                    mid=(low+high)/2;
                    }
                else
                {
                    low=mid+1;
                    mid=(low+high)/2;
                    }
                }
            num[mid]++;
            }
        for(int i=0;i<=n;i++)
        {
            printf("%d: %d\n",i,num[i]);
            }
        }
        return 0;
    }