Problem 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

翻译题意

天文学家把天上的每颗星星都画在一个平面上，并给定每颗的坐标。同时把星星左下方(包括横坐标或者纵坐标一样的)的星星数目定义为该星星的等级。天文学家想知道星星等级的分布。
例如上图中，5号星星的等级为3(左下方的星星为1，2，4)，2号与4号星星等级为1。上图中等级为0的星星有1颗，等级为1的星星有2颗，等级为2的星星有1颗，等级为3的星星有1颗。
请写一个程序计算屏幕上每一个等级的星星数。

仔细审题之后，发现是树状数组，因为输入的xy的值是有序的，所以只需要建立c[x]表示横坐标小于等于x的符合要求的点有几个，运用到树状数组的基础知识，统计出即可。

下面给出代码

 1 //树状数组基本框架的搭建（维护和查询都是O(lgn)的复杂度）
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstdlib>
 5 #include<cstring>
 6 #include<string>
 7 using namespace std;
 8 int n,a,b;
 9 int c[40000],ans[40000];//注意数组的大小
10
11 int lowbit(int k)
12 {
13      return k&(-k);
14
15 }
16 int add(int k,int num)
17 {
18     while(k<=32010)//此处的32010应为n的最大值+2
19     {
20          c[k]+=num;
21          k+=lowbit(k);
22     }
23 }
24 int sum(int k)
25 {
26     int Sum=0;
27     while(k>0)
28     {
29         Sum+=c[k];
30         k-=lowbit(k);
31     }
32     return Sum;
33 }
34
35
36 int main()
37 {
38      cin>>n;
39      for(int i=1;i<=n;i++)
40      {
41         cin>>a>>b;
42         a=a+1;//树状数组的下标需要从1开始，而数据从0开始，故此+1;
43         ans[sum(a)]++;
44         add(a,1);
45      }
46      for(int i=0;i<n;i++)
47        cout<<ans[i]<<endl;
48     return 0;
49 }