# POJ2318 TOYS 判断点与直线位置关系 【计算几何】

Calculate the number of toys that land in each bin of a partitioned toy box.

Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John is rebellious and obeys his parents by simply throwing his toys into the box. All the toys get mixed up, and it is impossible for John to find his favorite toys. 妈妈和爸爸有一个问题，他们的孩子约翰从来没有在玩完玩具后把玩具都放好，他们有了约翰一个矩形的箱子用来放他的玩具，但是约翰很叛逆，他服从了他父母，不过只是简单地把玩具扔进了箱子里。所有的玩具都搞混了，并且约翰不可能找到他最喜欢的玩具。

John's parents came up with the following idea. They put cardboard partitions into the box. Even if John keeps throwing his toys into the box, at least toys that get thrown into different bins stay separated. The following diagram shows a top view of an example toy box.

For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box.

Input

The input file contains one or more problems. The first line of a problem consists of six integers, n m x1 y1 x2 y2. The number of cardboard partitions is n (0 < n <= 5000) and the number of toys is m (0 < m <= 5000). The coordinates of the upper-left corner and the lower-right corner of the box are (x1,y1) and (x2,y2), respectively. The following n lines contain two integers per line, Ui Li, indicating that the ends of the i-th cardboard partition is at the coordinates (Ui,y1) and (Li,y2). You may assume that the cardboard partitions do not intersect each other and that they are specified in sorted order from left to right. The next m lines contain two integers per line, Xj Yj specifying where the j-th toy has landed in the box. The order of the toy locations is random. You may assume that no toy will land exactly on a cardboard partition or outside the boundary of the box. The input is terminated by a line consisting of a single 0.

Output

The output for each problem will be one line for each separate bin in the toy box. For each bin, print its bin number, followed by a colon and one space, followed by the number of toys thrown into that bin. Bins are numbered from 0 (the leftmost bin) to n (the rightmost bin). Separate the output of different problems by a single blank line.

Sample Input 5 6 0 10 60 0 3 1 4 3 6 8 10 10 15 30 1 5 2 1 2 8 5 5 40 10 7 9 4 10 0 10 100 0 20 20 40 40 60 60 80 80 5 10 15 10 25 10 35 10 45 10 55 10 65 10 75 10 85 10 95 10 0

Sample Output

0: 2 1: 1 2: 1 3: 1 4: 0 5: 1

0: 2 1: 2 2: 2 3: 2 4: 2

Hint

As the example illustrates, toys that fall on the boundary of the box are "in" the box. 如图所示，落在盒子边界上的玩具在盒子里

1. k1=k2,三点共线
2. k1>k2,点在直线的右侧
3. k1<k2,点在直线的左侧

#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
struct point{
int x,y;
};//记录坐标
struct Node{
point a,b;
}A[5010];//隔板上下的点
int pos[5010];
//判断点(x,y)与线段A[mid]的关系
bool judge(int xx,int yy,int mid){
int ans=(A[mid].a.x-xx)*(A[mid].b.y-yy)-(A[mid].a.y-yy)*(A[mid].b.x-xx);
if(ans<0)return false;//点在线段左边
return true;//点在线段右边
}
//二分查找点(x,y)最近的线段
void search(int xx,int yy,int n){
int left=0,right=n-1;
while(left<=right){
int mid=(left+right)>>1;
if(judge(xx,yy,mid)){//点在mid所在那条线段的右侧
left=mid+1;
}
else { //点在mid所在那条线段的左侧
right=mid-1;
}
}
pos[left]++;//记录不同箱子所得到的玩具的数量
}
int main()
{
int n,m,i,j,x1,x2,y1,y2;
while(scanf("%d",&n),n){
scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
for(i=0;i<n;++i){
int xd,xu;
scanf("%d%d",&xu,&xd);
A[i].a.x=xu;
A[i].a.y=y1;
A[i].b.x=xd;
A[i].b.y=y2;
}//存储隔板的位置
memset(pos,0,sizeof(pos));
for(i=0;i<m;++i){
int xx,yy;
scanf("%d%d",&xx,&yy);
search(xx,yy,n);
}//寻找这个玩具在那个区间

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

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