POJ 2318 TOYS

TOYS

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2318

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-thcardboard 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.

题意：给出矩形的左上和右下的坐标，在矩形中有n个木棒，木棒之间不会相交，然后给出木棒上下端点的横坐标，这些木棒将矩形分成多个区域，接着有m个玩具，给出玩具的坐标。输出在每个区域中玩具的个数。
思路：叉积判断点是否在四边形区域内。
向量的叉积：
       向量的叉积性质可以用来判断点在直线的某侧。进而可以解决点是否在三角形内，两个矩形是否重叠等问题。向量的叉积的模表示这两个向量围成的平行四边形的面积。 
设矢量P = ( x1, y1 )，Q = ( x2, y2 )，则矢量叉积定义为由(0,0)、p1、p2和p1+p2所组成的平行四边形的带符号的面积，即：P×Q = x1*y2 - x2*y1，其结果是一个伪矢量。 
显然有性质 P × Q = - ( Q × P ) 和 P × ( - Q ) = - ( P × Q )。 
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。 
       叉积的方向与进行叉积的两个向量都垂直，所以叉积向量即为这两个向量构成平面的法向量。
       如果向量叉积为零向量，那么这两个向量是平行关系。因为向量叉积是这两个向量平面的法向量，如果两个向量平行无法形成一个平面，其对应也没有平面法向量。所以，两个向量平行时，其向量叉积为零。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxn = 5010;
struct point
{
    double x , y;
};
double Xmult(point a , point b , point c)
{
    return (b.x-a.x)*(c.y-a.y) - (b.y-a.y)*(c.x-a.x);
}
point P[maxn] , Pu[maxn] , Pd[maxn];
int  sum[maxn];
int n , m , k;
void Bin(point X)
{
    int l , r , ans;
    double tmp;
    l = 0;
    r = n+1;
    while(l <= r)
    {
        int mid = (l+r)/2;
        tmp = Xmult(X , Pu[mid] , Pd[mid]);
        if(tmp < 0)
        {
            ans = mid;
            r = mid-1;
        }
        else l = mid+1;
    }
    sum[ans-1] ++;
}
int main()
{
    double x1 , y1 , x2 , y2;
    int i , j;
    while(~scanf("%d",&n))
    {
        if(!n) break;
        scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);
        for(i = 1 ; i <= n ; i ++)
        {
            scanf("%lf%lf",&Pu[i].x,&Pd[i].x);
            Pu[i].y = y1;
            Pd[i].y = y2;
        }
        Pu[0].x = x1;
        Pu[0].y=y1;
        Pd[0].x = x1;
        Pd[0].y=y2;
        Pu[n+1].x = x2;
        Pu[n+1].y=y1;
        Pd[n+1].x = x2;
        Pd[n+1].y=y2;
        for(i = 0 ; i < m ; i ++) scanf("%lf%lf",&P[i].x,&P[i].y);
        k  = 0;
        memset(sum , 0 , sizeof(sum));
        for(i = 0 ; i < m ; i ++) Bin(P[i]);
        for(i = 0 ; i <= n ; i ++) printf("%d: %d
",i,sum[i]);
        printf("
");
    }
}