TOYS

Time Limit: 2000MS	Memory Limit: 65536K
Total Submissions: 9011	Accepted: 4284

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.


題目大意：將一個二維平面取出一個矩形來，劃分成若干塊，如圖所示。然後給你若干個座標，問你每個區域聚集了多少個座標並輸出。


想法：還是很簡單的叉積運算，對於一個劃分區域的兩個點和給定的某個座標進行叉積運算，如果線上的下方叉積小於0，直到出現大於等於0的情況記錄cnt陣列++。優化可以使用二分查詢，時間複雜度O(n*lgm),不過直接查詢也能過。大部分程式碼是模板，cross（）為計算叉積。

#include <cstdio>
#include <cstring>
#include<vector>
#include <algorithm>
#include <iostream>
#include <climits>
#include <numeric>
#include<cmath>
#define foreach(e,x) for(__typeof(x.begin()) e=x.begin();e!=x.end();++e)
#define REP(i,n) for(int i=0;i<n;++i)
using namespace std;

const double EPS = 1e-8;
inline int sign(double a)
{
    return a < -EPS ? -1 : a > EPS;
}

struct Point
{
    double x, y;
    Point()
    {
    }
    Point(double _x, double _y) :
        x(_x), y(_y)
    {
    }
    Point operator+(const Point&p) const
    {
        return Point(x + p.x, y + p.y);
    }
    Point operator-(const Point&p) const
    {
        return Point(x - p.x, y - p.y);
    }
    Point operator*(double d) const
    {
        return Point(x * d, y * d);
    }
    Point operator/(double d) const
    {
        return Point(x / d, y / d);
    }
    bool operator<(const Point&p) const
    {
        int c = sign(x - p.x);
        if (c)
            return c == -1;
        return sign(y - p.y) == -1;
    }
    double dot(const Point&p) const
    {
        return x * p.x + y * p.y;
    }
    double det(const Point&p) const
    {
        return x * p.y - y * p.x;
    }
    double alpha() const
    {
        return atan2(y, x);
    }
    double distTo(const Point&p) const
    {
        double dx = x - p.x, dy = y - p.y;
        return hypot(dx, dy);
    }
    double alphaTo(const Point&p) const
    {
        double dx = x - p.x, dy = y - p.y;
        return atan2(dy, dx);
    }
    void read()
    {
        scanf("%lf%lf", &x, &y);
    }
    double abs()
    {
        return hypot(x, y);
    }
    double abs2()
    {
        return x * x + y * y;
    }
    void write()
    {
        cout << "(" << x << "," << y << ")" << endl;
    }
};
#define cross(p1,p2,p3) ((p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y))

#define crossOp(p1,p2,p3) sign(cross(p1,p2,p3))

Point isSS(Point p1, Point p2, Point q1, Point q2)       //可求p1,p2 直線與q1，q2的焦點。。不過不確定
{
    double a1 = cross(q1,q2,p1), a2 = -cross(q1,q2,p2);
    Point temp;
    temp.x=sign((p1.x*a2+p2.x*a1)/(a1+a2))==0?0:(p1.x*a2+p2.x*a1)/(a1+a2);
    temp.y=sign((p1.y*a2+p2.y*a1)/(a1+a2))==0?0:(p1.y*a2+p2.y*a1)/(a1+a2);
    return temp;
}
struct Border
{
    Point p1, p2;
    long double alpha;
    void setAlpha()
    {
        alpha = atan2(p2.y - p1.y, p2.x - p1.x);
    }
    void read()
    {
        p1.read();
        p2.read();
        setAlpha();
    }
};

int main()
{
    int n, m, x1, y1, x2, y2;
    int t1, t2;
    Point a;
    Border line[5001];
    int cnt[5002];
    while (scanf ("%d", &m) && m)
    {
        scanf ("%d%d%d%d%d", &n, &x1, &y1, &x2, &y2);
        for (int i = 0; i < m; i++)
        {
            scanf ("%d%d", &t1, &t2);
            line[i].p1.x = t1;
            line[i].p1.y = y1;
            line[i].p2.x = t2;
            line[i].p2.y = y2;
        }
        int flag=0;
        memset(cnt, 0, sizeof (cnt));
        for(int i=1; i<=n; i++)
        {
            Point toy;
            toy.read();
             flag=0;
            for(int j=0; j<m; j++)
            {
                if(crossOp(line[j].p1,toy,line[j].p2)>=0)
                {
                    flag=1;
                    cnt[j]++;
                    break;
                }
            } if(flag==0)
            cnt[m]++;
        }
        for (int i=0; i <=m; i++)
            printf ("%d: %d\n",i,cnt[i]);
        printf("\n");
    }
    return 0;
}