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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

• 題意:給定n條無順序的邊，將一個矩形劃分成n+1個區域，再給定m個點，求每個區域各有多少個點，輸出將按區域內存在的點的數目進行升序排序。
  和POJ2318類似，但是這裏的點需要排序，還有輸出結果不同
• 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) {  //判斷p1p0和p2p0的關系，<0,p1p0在p2p0的逆時針方向
  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)    /*點p1在邊的左側*/
  37             high = mid - 1;
  38         else    //點p在邊的右側
  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
• 再熟悉一下叉積函數

  1 再熟悉一下叉積函數
  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  //判斷p1p0和p2p0的關系
  6  //結果<0, p1p0在p2p0的逆時針方向，即點p1在p2p0的左側
  7  //結果>0, p1p0在p2p0的順時針方向，即點p1在p2p0的右側

  View Code

POJ 2398--Toy Storage(叉積判斷，二分找點，點排序)