POJ 2318 TOYS （向量叉积的右手法则）

TOYS

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8292		Accepted: 3929

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

思路：题意是判断n个线段分成的n+1个区域，给出m个玩具的坐标，问各个区域里玩具的个数。
向量的叉积：a = (x1,y1),b = (x2,y2), 则a x b = x1*y2 - x2*y1.
右手法则：若b在a的顺时针转动方向（夹角<90度），则a x b < 0,否则若b在a的逆时针方向,则a x b >0, 若a,b共线，a x b = 0.
这道题对于输入的每次的玩具坐标x,y,那么设线段上端的坐标为x1,y1,下端的坐标为x2,y2,则向量a=(x1-x,y1,x)和向量b=(x2-x,y2-x),如果玩具在线段左端，
则有a x b <= 0(包括玩具在线段上)，由于输入的每条线段的横坐标是按升序输入的，根据玩具的输入依次枚举每条线段，记录答案即可。

View Code

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 #define MAX 5005
 6 
 7 using namespace std;
 8 
 9 struct node
10 {
11     int x,y;
12 };
13 
14 node up[MAX],down[MAX];
15 int ans[MAX];
16 int n,m,xx1,yy1,xx2,yy2;
17 
18 bool check(node tar,node u,node d)
19 {
20     node temp1,temp2;
21     temp1.x = u.x - tar.x;
22     temp1.y = u.y - tar.y;
23     temp2.x = d.x - tar.x;
24     temp2.y = d.y - tar.y;
25     if(temp1.x*temp2.y - temp2.x*temp1.y <= 0)
26         return true;
27     return false;
28 }
29 
30 int main()
31 {
32     while(~scanf("%d",&n) && n)
33     {
34         scanf("%d%d%d%d%d",&m,&xx1,&yy1,&xx2,&yy2);
35         memset(ans,0,sizeof(ans));
36         for(int i=0; i)
37         {
38             scanf("%d%d",&up[i].x,&down[i].x);
39             up[i].y = yy1; down[i].y = yy2;
40         }
41         up[n].x = xx2; down[n].x = xx2;
42         up[n].y = yy1; down[n].y = yy2;
43         for(int i=0; i)
44         {
45             node tar;
46             scanf("%d%d",&tar.x,&tar.y);
47             for(int i=0; i<=n; i++)
48             {
49                 if(check(tar,up[i],down[i]))
50                 {
51                     ans[i] ++;
52                     break;
53                 }
54             }
55         }
56         for(int i=0; i<=n; i++)
57         {
58             printf("%d: %d\n",i,ans[i]);
59         }
60         printf("\n");
61     }
62     return 0;
63 }