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.题意是有m个玩具（看成一个点），有n个板子把一个矩形区域隔成n+1个区域，问每个区域内有多少个玩具。
利用叉积判断点在一条线段的左边还是右边，可以直接暴力循环找在哪个区域，也可以二分。
叉积：为x1*y2 - x2*y1
叉积的结果也是一个向量，是垂直于向量a，b所形成的平面，如果看成三维坐标的话是在 z 轴上，上面结果是它的模。
方向判定：右手定则，（右手半握，大拇指垂直向上，四指右向量a握向b，大拇指的方向就是叉积的方向）
若 a x b > 0表示a在b的顺时针方向上
若 a x b < 0表示a在b的逆时针方向上
若 a x b == 0表示a在b共线，但不确定方向是否相同
如果要判断一个点在线段的左边还是右边，只用计算一下这个点指向线段上面那个点的向量与这个点指向线段下面那个点的向量的叉积，如果是负这点就在线段的左边，如果是正就在右边。
#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct Point
{
int x,y;
Point(){}
Point(int _x,int _y)
{
x = _x;y = _y;
}
Point operator -(const Point &b)const
{
return Point(x - b.x,y - b.y);
}
int operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
int operator ^(const Point &b)const
{
return x*b.y - y*b.x;
}
};
struct Line
{
Point p1,p2;
Line(){}
Line(Point _p1,Point _p2)
{
p1 = _p1;
p2 = _p2;
}
};
int xmult(Point p0,Point p1,Point p2) //p0p1 X p0p2
{
return (p1-p0)^(p2-p0);
}
int n;
Line a[5500];
int ans[5500];
int binarysearch(Point p)
{
int l = 0,r = n;
int ans = r;
while(l <= r)
{
int mid = (l+r)/2;
if(xmult(p,a[mid].p1,a[mid].p2) < 0)
{
ans = mid;
r = mid - 1;
}
else
l = mid + 1;
}
return ans;
}
int main(void)
{
int m,x,y,xx,yy,i,j;
int first = 1;
while(scanf("%d",&n)==1)
{
if(n == 0)
break;
if(first == 1)
first = 0;
else
printf("\n");
scanf("%d %d%d%d%d",&m,&x,&y,&xx,&yy);
for(i=0;i<n;i++)
{
int u,d;
scanf("%d%d",&u,&d);
a[i] = Line(Point(u,y),Point(d,yy));
}
a[n] = Line(Point(xx,y),Point(xx,yy));
memset(ans,0,sizeof(ans));
for(i=1;i<=m;i++)
{
int x,y;
scanf("%d%d",&x,&y);
Point p(x,y);
if(xmult(p,a[0].p1,a[0].p2) < 0)
{
ans[0]++;
continue;
}
for(j=1;j<=n;j++)
{
if(xmult(p,a[j-1].p1,a[j-1].p2) > 0 && xmult(p,a[j].p1,a[j].p2) < 0)
{
ans[j]++;
break;
}
}
/*
int t = binarysearch(p); //二分找区间
ans[t]++;
*/
}
for(i=0;i<=n;i++)
printf("%d: %d\n",i,ans[i]);
}
return 0;
}