POJ 2352 Stars

Posted 2021-05-22 smile_to_warm

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 63757		Accepted: 26959

Description

Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers want to know the distribution of the levels of the stars.

 

For example, look at the map shown on the figure above. Level of the star number 5 is equal to 3 (it\'s formed by three stars with a numbers 1, 2 and 4). And the levels of the stars numbered by 2 and 4 are 1. At this map there are only one star of the level 0, two stars of the level 1, one star of the level 2, and one star of the level 3.

You are to write a program that will count the amounts of the stars of each level on a given map.

Input

The first line of the input file contains a number of stars N (1<=N<=15000). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, 0<=X,Y<=32000). There can be only one star at one point of the plane. Stars are listed in ascending order of Y coordinate. Stars with equal Y coordinates are listed in ascending order of X coordinate.

Output

The output should contain N lines, one number per line. The first line contains amount of stars of the level 0, the second does amount of stars of the level 1 and so on, the last line contains amount of stars of the level N-1.

Sample Input

5
1 1
5 1
7 1
3 3
5 5

Sample Output

1
2
1
1
0

 

 1  1 package poj;
 2  2 
 3  3 import java.io.BufferedReader;
 4  4 import java.io.IOException;
 5  5 import java.io.InputStreamReader;
 6  6 import java.util.StringTokenizer;
 7  7 
 8  8 /** 
 9  9  * @ClassName：     POJ_2352.java 
10 10  * @Description：   天文学家经常检查星图，在星图上，星星由平面上的点表示，每颗星星都有笛卡尔坐标。
11 11  *                  某个星星的等级等于高度不大于该星星，且不在此星星右边的星星的数量，天文学家想知道星星的能级分布。 
12 12  *                  你要写一个程序来计算给定地图上每一层星的数量。   
13 13  * @Author：        WangFengyan  
14 14  * @Version：       V1.0   
15 15  * @Date：          2021年5月22日 上午8:49:09 
16 16  * 
17 17  *  分析： 由于y坐标是按升序排列，故星星等级等于在此之前的x坐标小于当前x坐标的星星的个数
18 18  *  可使用Indexed　Tree求星星等级
19 19  */
20 20 
21 21 public class POJ_2352 {
22 22 
23 23 
24 24     static int [] srcArr; 
25 25     static int [] idxTree;
26 26     static int N;
27 27     static int fromIdx;
28 28     static int level ;
29 29     static int [] res;
30 30     public static void main(String[] args) throws IOException {        
31 31         BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
32 32         StringTokenizer st = new StringTokenizer(bf.readLine());
33 33         N = Integer.parseInt(st.nextToken()); 
34 34     
35 35         srcArr = new int[N]; //星星x坐标数组
36 36         res = new int[N];
37 37         for(int i=0;i<N;i++) {
38 38             st = new StringTokenizer(bf.readLine());
39 39             int x = Integer.parseInt(st.nextToken());
40 40             srcArr[i] = x;
41 41         }
42 42         
43 43         init();
44 44         for(int i=0;i<N;i++) {
45 45             level = 0;
46 46             level = query(fromIdx,fromIdx+srcArr[i]-1);
47 47             res[level] += 1 ;
48 48             update(fromIdx+srcArr[i]-1,1);        
49 49         }
50 50         for(int i=0;i<N;i++) {
51 51             System.out.println(res[i]);
52 52         }        
53 53     }
54 54 
55 55     public static void init() {
56 56         int k=0;
57 57         while((1<<k)<srcArr.length) {
58 58             k++;
59 59         }
60 60         idxTree = new int[1<<(k+1)];
61 61         fromIdx = 1<<k;
62 62         for(int i=0,j=0;i<idxTree.length && j<srcArr.length;i++) {
63 63             idxTree[i] = 0;
64 64         }
65 65     }
66 66     
67 67     public static void update(int idx, int target) {
68 68         while(idx>0) {
69 69             idxTree[idx] = idxTree[idx] + target;
70 70             idx = idx >> 1;
71 71         }
72 72     }
73 73     
74 74     public static int query(int s,int e) {
75 75         int res=0;
76 76         while(s<=e) {
77 77             if((s&1)==1) {
78 78                 res = res + idxTree[s];
79 79             }
80 80             
81 81             if((e&1)==0) {
82 82                 res = res + idxTree[e];
83 83             }
84 84             
85 85             s = (s+1) >> 1;
86 86             e = (e-1) >> 1;
87 87         }
88 88         return res;
89 89     }
90 90     
91 91 }
92 92 
93 93 
94 94