Stars

Time limit: 0.25 second
Memory limit: 64 MB
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 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 ofY 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 output
5
1 1
5 1
7 1
3 3
5 5
1
2
1
1
0

分析:树状数组;

代码:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=1e5+;
const int dis[][]={{,},{-,},{,-},{,}};
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
int n,m,k,t,a[maxn],ans[maxn];
int get(int x)
{
int sum=;
for(int i=x;i;i-=i&(-i))
sum+=a[i];
return sum;
}
void add(int x,int y)
{
for(int i=x;i<=;i+=i&(-i))
a[i]+=y;
}
int main()
{
int i,j;
scanf("%d",&n);
rep(i,,n)
{
int x,y;
scanf("%d%d",&x,&y);
x++;
ans[get(x)]++;
add(x,);
}
rep(i,,n-)printf("%d\n",ans[i]);
//system("Pause");
return ;
}

ural1028 Stars的更多相关文章

  1. poj 2352 Stars 数星星 详解

    题目: poj 2352 Stars 数星星 题意:已知n个星星的坐标.每个星星都有一个等级,数值等于坐标系内纵坐标和横坐标皆不大于它的星星的个数.星星的坐标按照纵坐标从小到大的顺序给出,纵坐标相同时 ...

  2. POJ 2352 Stars(树状数组)

    Stars Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 30496   Accepted: 13316 Descripti ...

  3. 【POJ-2482】Stars in your window 线段树 + 扫描线

    Stars in Your Window Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11706   Accepted:  ...

  4. Java基础之在窗口中绘图——填充星型(StarApplet 2 filled stars)

    Applet程序. import javax.swing.*; import java.awt.*; import java.awt.geom.GeneralPath; @SuppressWarnin ...

  5. XidianOJ 1177 Counting Stars

    题目描述 "But baby, I've been, I've been praying hard,     Said, no more counting dollars     We'll ...

  6. POJ-2352 Stars 树状数组

    Stars Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 39186 Accepted: 17027 Description A ...

  7. hdu 1541/poj 2352:Stars(树状数组,经典题)

    Stars Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submi ...

  8. POJ 2482 Stars in Your Window 线段树扫描线

    Stars in Your Window   Description Fleeting time does not blur my memory of you. Can it really be 4 ...

  9. 2015ACM/ICPC亚洲区长春站 G hdu 5533 Dancing Stars on Me

    Dancing Stars on Me Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Ot ...

随机推荐

  1. 解决ORA-00904: invalid identifier标识符无效

    方法/步骤 1 大部分情况下,此错误是由于引用了不存在的列名导致的.比如select name from Studtent 当studeng表中无name列时,系统就会报此错误. 2 解决思路是,确定 ...

  2. javascript操作json

    for (var i = 0; i < selectedPartList.length; i++) { if (selectedPartList[i].vpart_code == jsonRow ...

  3. PS:抠图方法1(利用对比度ctrl+l)

    PS:抠图方法1(利用对比度ctrl+l) 工具/原料   Photoshop.美女照片 方法/步骤     小编使用的是Photoshop cs5版本,大家使用其他版本都没有关系,界面略有不同,但操 ...

  4. flex超链接

    <?xml version="1.0" encoding="utf-8"?> <s:Application xmlns:fx="ht ...

  5. nm applet disable

    http://support.qacafe.com/knowledge-base/how-do-i-prevent-network-manager-from-controlling-an-interf ...

  6. POJ 2031 Building a Space Station 最小生成树模板

    题目大意:在三维坐标中给出n个细胞的x,y,z坐标和半径r.如果两个点相交或相切则不用修路,否则修一条路连接两个细胞的表面,求最小生成树. 题目思路:最小生成树树模板过了,没啥说的 #include& ...

  7. GetClientRect

    这个函数好像就是对应于视口的,获取视口的宽高 #include <windows.h> LRESULT CALLBACK WndProc (HWND, UINT, WPARAM, LPAR ...

  8. angular初始用——简易购物车

    <html> <head> <meta charset="utf-8"> <script src="js/angular.js& ...

  9. 学习笔记——代理模式Proxy

    代理模式,主要是逻辑和实现解耦.具体逻辑如何,由代理Proxy自己来设计,我们只需要把逻辑Subject交给代理即可. 主要应用场景,包括创建大开销对象时,使用代理来慢慢创建:远程代理,如网络不确定时 ...

  10. JavaScript(四)---- 函数

    函数主要用来封装具体的功能代码. 函数是由这样的方式进行声明的:关键字 function.函数名.一组参数,以及置于括号中的待执行代码. 格式: function 函数名(形参列表){         ...