POJ 2352 Stars

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29602		Accepted: 12924

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

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

Source

Ural Collegiate Programming Contest 1999

 

树状数组计数问题，因为这里全部星星都是按从Y值从小到大给出的，所以只要使用一维树状数组便可

因为题目中的数据范围是从0开始的，为了把0去掉，要将全部数据的X值加1

按输入数据的顺序，每组在树状树组中先求c[1]+…+c[x+1]，以后再在c[x+1]上加1

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #define MAX 35000
 5 
 6 using namespace std;
 7 
 8 int c[MAX];
 9 int total[MAX];
10 
11 int lowbit(int x)
12 {
13     return x&(-x);
14 }
15 
16 int sum(int x)
17 {
18     int ret=0;
19 
20     while(x>0)
21     {
22         ret+=c[x];
23         x-=lowbit(x);
24     }
25 
26     return ret;
27 }
28 
29 void add(int x,int t)
30 {
31     while(x<=MAX)
32     {
33         c[x]+=t;
34         x+=lowbit(x);
35     }
36 }
37 
38 int main()
39 {
40     int n;
41 
42     while(scanf("%d",&n)==1)
43     {
44         memset(c,0,sizeof(c));
45         memset(total,0,sizeof(total));
46 
47         int x,y;
48         for(int i=1;i<=n;i++)
49         {
50             scanf("%d %d",&x,&y);
51             total[sum(x+1)]++;
52             add(x+1,1);
53         }
54 
55         for(int i=0;i<n;i++)
56             printf("%d\n",total[i]);
57     }
58 
59     return 0;
60 }

[C++]