# poj2398

/***************************************************************\
*Author:Hu Wenbiao
*Created Time: Tue 27 Jul 2010 07:10:34 PM CST
*File Name: main.cpp
*Description:几何题目。对poj2318的扩展
\***************************************************************/

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
/*----------------------*Global Variable*----------------------*/
struct Cardboard{//隔板
int u,l;
}card[1010];
int toynum[1010],ans[1010],n,m,x1,y1,x2,y2,X,Y,_max;//toynum各分区的数目
//*=======================*Main Program*=======================*//
using namespace std;

bool cmp(Cardboard a,Cardboard b){//返回a是否在b的左边，以此排序
int s=b.u-b.l,t=y1-y2,p=a.u-b.l,q=y1-y2;
return s*q-t*p>0;
}
bool left(int m){//判断toy在第m个隔板左边
int a=card[m].u-card[m].l,b=y1-y2,c=X-card[m].l,d=Y-y2;
return a*d-b*c>0;
}

void location(){//二分法，将toy所在的区间数目加1
int s=0,t=n;
int m=(s+t)/2;
while(s<t){
if(left(m)){
t=m;
m=(s+t)/2;
}
else{
s=m+1;
m=(s+t)/2;
}
}
toynum[t]++;
}
int main(){
//freopen("input","r",stdin);
while(scanf("%d%d%d%d%d%d",&n,&m,&x1,&y1,&x2,&y2)!=EOF&&n){
for(int i=0;i<n;i++)
scanf("%d%d",&card[i].u,&card[i].l);
sort(card,card+n,cmp);//隔板从左到右的顺序排序
card[n].u=card[n].l=x2;//用box右边界作第n个区间的‘隔板’
memset(toynum,0,sizeof(toynum));
while(m--){
scanf("%d%d",&X,&Y);
location();
}
memset(ans,0,sizeof(ans));
for(int i=0;i<=n;i++){
_max=_max>toynum[i]?_max:toynum[i];
ans[toynum[i]]++;
}
printf("Box\n");
for(int i=1;i<=_max;i++)
if(ans[i])
printf("%d: %d\n",i,ans[i]);
}
}


posted @ 2010-07-28 08:49  open source  阅读(194)  评论(0编辑  收藏