POJ 2398--Toy Storage(叉积判断，二分找点，点排序)

Toy Storage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6534		Accepted: 3905

Description

Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza 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 Reza to find his favorite toys anymore.
Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top:

We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.

Input

The input consists of a number of cases. The first line consists of six integers n, m, x1, y1, x2, y2. The number of cardboards to form the partitions is n (0 < n <= 1000) and the number of toys is given in m (0 < m <= 1000). 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 each consists of two integers Ui Li, indicating that the ends of the ith cardboard is at the coordinates (Ui, y1) and (Li, y2). You may assume that the cardboards do not intersect with each other. The next m lines each consists of two integers Xi Yi specifying where the ith toy has landed in the box. You may assume that no toy will land on a cardboard.

A line consisting of a single 0 terminates the input.

Output

For each box, first provide a header stating "Box" on a line of its own. After that, there will be one line of output per count (t > 0) of toys in a partition. The value t will be followed by a colon and a space, followed the number of partitions containing t toys. Output will be sorted in ascending order of t for each box.

Sample Input

4 10 0 10 100 0
20 20
80 80
60 60
40 40
5 10
15 10
95 10
25 10
65 10
75 10
35 10
45 10
55 10
85 10
5 6 0 10 60 0
4 3
15 30
3 1
6 8
10 10
2 1
2 8
1 5
5 5
40 10
7 9
0

Sample Output

Box
2: 5
Box
1: 4
2: 1

Source

Tehran 2003 Preliminary

题意:给定n条无顺序的边，将一个矩形划分成n+1个区域，再给定m个点，求每个区域各有多少个点，输出将按区域内存在的点的数目进行升序排序。
和POJ2318类似，但是这里的点需要排序，还有输出结果不同

code:

 1 #include<iostream>
 2 #include<cstring> 
 3 #include<string>
 4 #include<algorithm> 
 5 #include<cstdio> 
 6 #include<cstdlib> 
 7 #include<cmath> 
 8 using namespace std;
 9 const int MAX = 5005;
10 typedef struct point {
11     int x;
12     int y;
13 }point;
14 typedef struct value {
15     point start;
16     point end;
17 }v;
18 v edge[MAX];
19 int sum[MAX], ans[MAX];
20 int n, m, x1, y11, x2, y2, flag = 1,Ui, Li;
21 point tp;
22 int Xj, Yj;
23 bool com(const v t1, const v t2) {
24     return t1.start.x < t2.start.x;
25 }
26 bool com2(const int a, const int b) {
27     return a < b;
28 }
29 int multi(point p1, point p2, point p0) {  //判断p1p0和p2p0的关系，<0,p1p0在p2p0的逆时针方向
30     return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);
31 }
32 void inset(point p) {
33     int low = 0, high = n;
34     while (low <= high) {
35         int mid = (high + low) / 2;
36         if (multi(p, edge[mid].start, edge[mid].end) < 0)    /*点p1在边的左侧*/
37             high = mid - 1;
38         else    //点p在边的右侧
39             low = mid + 1;
40     }
41     if (multi(p, edge[low-1].start, edge[low-1].end) < 0 )
42         sum[low-1]++;
43     else
44         sum[low]++;
45 }
46 int main() {
47     while (cin>>n && n) {
48         memset(sum, 0, sizeof(sum));
49         memset(ans, 0, sizeof(ans));
50         cin >> m >> x1 >> y11 >> x2 >> y2;
51         for (int i = 0; i < n; i++) {
52             cin >> Ui >> Li;
53             edge[i].start.x = Ui;
54             edge[i].start.y = y11;
55             edge[i].end.x = Li;
56             edge[i].end.y = y2;
57         }
58         edge[n].start.x = x2;
59         edge[n].start.y = y11;
60         edge[n].end.x = x2;
61         edge[n].end.y = y2;
62         sort(edge, edge + n + 1, com);
63         for (int j = 0; j < m; j++) {
64             cin >> Xj >> Yj;
65             tp.x = Xj;
66             tp.y = Yj;
67             inset(tp);
68         }
69         for (int i = 0; i <= n; i++)
70         {
71             if (sum[i] != 0)
72                 ans[sum[i]]++;
73         }
74         cout << "Box" << endl;
75         for (int i = 0; i <= n; i++)
76         {
77             if (ans[i] != 0)
78                 cout << i << ": " << ans[i] << endl;
79         }    
80     }
81     return 0;
82 }

View Code

再熟悉一下叉积函数

1 再熟悉一下叉积函数
2 int multi(point p1, point p2, point p0) { 
3     return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);
4 }
5  //判断p1p0和p2p0的关系
6  //结果<0, p1p0在p2p0的逆时针方向，即点p1在p2p0的左侧
7  //结果>0, p1p0在p2p0的顺时针方向，即点p1在p2p0的右侧