前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >poj------2352 Stars(树状数组)

poj------2352 Stars(树状数组)

作者头像
Gxjun
发布2018-03-22 13:02:02
8580
发布2018-03-22 13:02:02
举报
文章被收录于专栏:ml

Stars

Time Limit: 1000MS

Memory Limit: 65536K

Total Submissions: 30268

Accepted: 13202

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

代码语言:javascript
复制
5
1 1
5 1
7 1
3 3
5 5

Sample Output

代码语言:javascript
复制
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

代码:

代码语言:javascript
复制
 1 #include<stdio.h>
 2 #include<string.h>
 3 #define maxn 32005
 4 #define lowbit(x) ((x)&(-x))
 5 int aa[maxn];
 6 int bb[15005];
 7  void ope(int x)
 8 {
 9     while(x<maxn)
10     {
11      aa[x]++;
12      x+=lowbit(x);
13     }
14 }
15 int getsum(int x)
16 {
17     int ans=0;
18     while(x>0)
19     {
20         ans+=aa[x];
21         x-=lowbit(x);
22     }
23     return ans;
24 }
25 int main()
26 {
27     int nn,i,a;
28     while(scanf("%d",&nn)!=EOF)
29     {
30         memset(aa,0,sizeof(aa));
31         memset(bb,0,sizeof(int)*(nn));
32         for(i=0;i<nn;i++)
33         {
34             scanf("%d%*d",&a);
35             a++;
36             bb[getsum(a)]++;
37             ope(a);
38         }
39          for(i=0;i<nn;i++)
40              printf("%d\n",bb[i]);
41     }
42     return 0;
43 }
本文参与 腾讯云自媒体同步曝光计划,分享自作者个人站点/博客。
原始发表:2014-04-19 ,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 作者个人站点/博客 前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体同步曝光计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档