[POJ 2318]TOYS

Description

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.

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.

题目大意

给出矩形的左上、右下坐标，按顺序给出n条线段在矩形上两条横边上的截距。这n条直线将矩形分成n+1块。给出m个点，问每个小块中有几个点。

题解

还记得之前的某篇博客讲过叉积的性质吗？

叉积的一个非常重要性质是可以通过它的符号判断两矢量相互之间的顺逆时针关系：

• 若 $P × Q > 0$ , 则$P$在$Q$的顺时针方向。
• 若 $P × Q < 0$ , 则$P$在$Q$的逆时针方向。
• 若 $P × Q = 0$ , 则$P$与$Q$共线，但可能同向也可能反向。

这里同理，我们只需要扫过每条直线判断点在线的左端还是右端。

当然可以二分实现：我们设判断的点为$(x,y)$，直线的上下截距分别为$u$，$d$，矩形上边纵坐标$ya$，下边$yb$。

对于二分出的$mid$值

if (Point(x-d[mid],y-yb)*Point(u[mid]-d[mid],ya-yb)>0) l=mid+1;
else r=mid-1,ans=mid;

这里直线$0～n-1$编号，所以$ans$初值应该赋为$n$。

#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<queue>
#include<stack>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define LL long long
#define RE register
#define IL inline
using namespace std;
const int N=5000;

int n,m,xa,xb,ya,yb;
int d[N+5],u[N+5];
struct Point
{
    int x,y;
    Point (){};
    Point (int _x,int _y) {x=_x,y=_y;}
    int operator *(const Point &a)
    const{
        return x*a.y-y*a.x;
    }
};
int cnt[N+5],x,y;

IL int Dev(int x,int y);

int main()
{
    scanf("%d",&n);
    while (n)
    {
        memset(cnt,0,sizeof(cnt));
        scanf("%d%d%d%d%d",&m,&xa,&ya,&xb,&yb);
        for (RE int i=0;i<n;i++) scanf("%d%d",&u[i],&d[i]);
        for (RE int i=1;i<=m;i++)
        {
            scanf("%d%d",&x,&y);
            cnt[Dev(x,y)]++;
        }
        for (RE int i=0;i<=n;i++) printf("%d: %d\n",i,cnt[i]);
        printf("\n");
        scanf("%d",&n);
    }
    return 0;
}

IL int Dev(int x,int y)
{
    int l=0,r=n-1,ans=n,mid;
    while (l<=r)
    {
        mid=(l+r)>>1;
        if (Point(x-d[mid],y-yb)*Point(u[mid]-d[mid],ya-yb)>0) l=mid+1;
        else r=mid-1,ans=mid;
    }
    return ans;
}