rand tput art cat stay not in inpu lib part

TOYS

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17974		Accepted: 8539

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.

Source

Rocky Mountain 2003

1. 題意:給n條邊，劃分成n+1個區域，再給定m個點坐標，點不會落在邊界上和區域外，問每個區域中各自存在多少個點
2. 代碼如下

    1 #include<iostream>
    2 #include<cstring> 
    3 #include<string>
    4 #include<algorithm> 
    5 #include<vector> 
    6 #include<stack> 
    7 #include<bitset>
    8 #include<cstdio> 
    9 #include<cstdlib> 
   10 #include<cmath> 
   11 #include<set> 
   12 #include<list> 
   13 #include<deque> 
   14 #include<map> 
   15 #include<queue>
   16 using namespace std;
   17 const int MAX = 5005;
   18 
   19 typedef struct point {
   20     int x;
   21     int y;
   22 }point;
   23 typedef struct value {
   24     point start;
   25     point end;
   26 }v;
   27 v edge[MAX];
   28 int sum[MAX];
   29 int n, m, x1, y11, x2, y2, flag = 1;
   30 point tp;
   31 int Xj, Yj;
   32 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
   33     return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);
   34 }
   35 void inset(point p) {
   36     int low = 0, high = n;
   37     while (low <= high) {
   38         int mid = (high + low) / 2;
   39         if (multi(p, edge[mid].start, edge[mid].end) < 0)    /*點p1在邊的左側*/
   40             high = mid - 1;
   41         else    //點p在邊的右側
   42             low = mid + 1;
   43     }
   44     if (multi(p, edge[low-1].start, edge[low-1].end) < 0 )
   45         sum[low-1]++;
   46     else
   47         sum[low]++;
   48 }
   49 int main() {
   50     while (~scanf("%d", &n) && n) {
   51         memset(sum, 0, sizeof(sum));
   52         if (flag == 1)flag++;
   53         else printf("\n");
   54         scanf("%d%d%d%d%d", &m, &x1, &y11, &x2, &y2);
   55         int Ui, Li;
   56         for (int i = 0; i < n; i++) {
   57             scanf("%d%d", &Ui, &Li);
   58             edge[i].start.x = Ui;
   59             edge[i].start.y = y11;
   60             edge[i].end.x = Li;
   61             edge[i].end.y = y2;
   62         }
   63         edge[n].start.x = x2;
   64         edge[n].start.y = y11;
   65         edge[n].end.x = x2;
   66         edge[n].end.y = y2;
   67         for (int j = 0; j < m; j++) {
   68             scanf("%d%d", &Xj, &Yj);
   69             tp.x = Xj;
   70             tp.y = Yj;
   71             inset(tp);
   72         }
   73         for (int i = 0; i <= n; i++)
   74             printf("%d: %d\n", i, sum[i]);
   75     }
   76     return 0;
   77 }

   View Code

3. Experience:　前面點的構造寫成

   

   1 edge[i].start = { Ui,y11 };
   2 edge[i].end = { Li,y2 };

   View Code當發現這個錯誤的時候，我自己都被自己蠢哭了，Wa了2頁，一直以為是叉積方向搞錯了，原來不是ORZ
4. 這個是我真正意義上第一道計算幾何,mark一下。

POJ 2318--TOYS(二分找點，叉積判斷方向)