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//计算几何第一题
//学习了叉积和利用叉积判断左右位置关系
//叉积+二分
//延续了这两天比赛的PE黑手
#include<iostream>
using namespace std;
#define MAXN 5000+1
#define eps 1e-8
#define _sign(x) ((x)>eps?1:((x)<-eps?2:0))

struct point{int x,y;};
struct line{point a,b;};

int ans[MAXN];
point toys[MAXN];
line board[MAXN];

double xmult(point p0,point p1,point p2){//叉积,囧
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}

int main()
{
    
int n,m,x1,y1,x2,y2,u1,u2,i;
    
bool flag=false;
    
while(scanf("%d",&n)!=EOF && n )
    
{
        
if(flag) printf("\n");
        memset(ans,
0,sizeof(ans));
        scanf(
"%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
        
for(i=0;i<n;i++)
        
{
            scanf(
"%d%d",&u1,&u2);
            board[i].a.x
=u1;
            board[i].a.y
=y1;
            board[i].b.x
=u2;
            board[i].b.y
=y2;
        }

        board[n].a.x
=x2;
        board[n].a.y
=y1;
        board[n].b.x
=x2;
        board[n].b.y
=y2;
        
for(i=0;i<m;i++)
        
{
            scanf(
"%d%d",&(toys[i].x),&(toys[i].y));
            
int low=0,hig=n,mid;
            
while(low+1<hig)
            
{
                mid
=(low+hig)>>1;
                
double res=xmult(board[mid].a,board[mid].b,toys[i]);
                
//if(res>0)
                if(res>0) low=mid;
                
else hig=mid;
            }

            
if(xmult(board[low].a,board[low].b,toys[i])<0) ans[low]++;
            
else  ans[low+1]++;
        }

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

        flag
=true;
    }

    
return 0;
}
posted on 2009-10-04 12:14 wyiu 阅读(138) 评论(0)  编辑 收藏 引用 所属分类: POJ

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