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.
 
求一个star 左下角有多少星星  
这个和求逆序对也差不多
几乎没有不同
 
 #include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define pi acos(-1.0)
#define eps 1e-6
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define bug printf("******\n")
#define mem(a,b) memset(a,b,sizeof(a))
#define fuck(x) cout<<"["<<x<<"]"<<endl
#define f(a) a*a
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define FIN freopen("DATA.txt","r",stdin)
#define lowbit(x) x&-x
#pragma comment (linker,"/STACK:102400000,102400000") using namespace std;
typedef long long LL ;
const int maxn = 4e4 + ;
const int limit = ;
int n, c[maxn], lev[maxn];
void update(int x) {
while(x < ) {
c[x] += ;
x += lowbit(x);
}
}
int sum(int x) {
int ret = ;
while(x > ) {
ret += c[x];
x -= lowbit(x);
}
return ret;
}
int main() {
scanf("%d", &n);
mem(c, );
mem(lev, );
for (int i = ; i <= n ; i++) {
int x, y;
scanf("%d%d", &x, &y);
x++;
lev[sum(x)]++;
update(x);
}
for (int i = ; i < n ; i++)
printf("%d\n", lev[i]);
return ;
}

Stars POJ - 2352的更多相关文章

  1. (线段树 -星星等级)Stars POJ - 2352

    题意: 给出n个星星的坐标 x,y ,当存在其他星星的坐标x1,y1满足x>=x1&&y>=y1时 这个星星的等级就加1. 注意: 题中给的数据是有规律的 ,y是逐渐增加的 ...

  2. 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 ...

  3. POJ 2352 Stars(线段树)

    题目地址:id=2352">POJ 2352 今天的周赛被虐了. . TAT..线段树太渣了..得好好补补了(尽管是从昨天才開始学的..不能算补...) 这题还是非常easy的..维护 ...

  4. POJ 2352 Stars(树状数组)

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

  5. POJ 2352 &amp;&amp; HDU 1541 Stars (树状数组)

    一開始想,总感觉是DP,但是最后什么都没想到.还暴力的交了一发. 然后開始写线段树,结果超时.感觉自己线段树的写法有问题.改天再写.先把树状数组的写法贴出来吧. ~~~~~~~~~~~~~~~~~~~ ...

  6. 【树状数组】POJ 2352 Stars

    /** * @author johnsondu * @time 2015-8-22 * @type Binary Index Tree * ignore the coordinate of y and ...

  7. 【POJ 2352】 Stars

    [题目链接] http://poj.org/problem?id=2352 [算法] 树状数组 注意x坐标为0的情况 [代码] #include <algorithm> #include ...

  8. 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 ...

  9. POJ 2352 Stars(HDU 1541 Stars)

    Stars Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 41521   Accepted: 18100 Descripti ...

随机推荐

  1. Java并发基础--ThreadLocal

    一.ThreadLocal定义 ThreadLocal是一个可以提供线程局部变量的类,ThreadLocal为解决多线程程序的并发问题提供了一种新的思路,通过为每个线程提供一个独立的变量副本解决了变量 ...

  2. ionic 获取input的值

    1.参数传递法 例子:获取input框内容 这里有个独特的地方,直接在input处使用 #定义参数的name值,注意在ts中参数的类型 在html页面中 <ion-input type=&quo ...

  3. html常用小知识

    请求重定向:加载页面之后,除了用js做重定向之外,我们还可以直接用<meta>标签做重定向. <meta http-equiv="refresh" content ...

  4. python常用函数—enumerate()

    enumerate() 对于一个可迭代的(iterable)/可遍历的对象(如列表.字符串),enumerate将其组成一个索引序列,利用它可以同时获得索引和值的元组. 使用拆包,可以单独获得索引和值 ...

  5. 四、oracle 用户管理二

    一.使用profile管理用户口令概述:profile是口令限制,资源限制的命令集合,当建立数据库时,oracle会自动建立名称为default的profile.当建立用户没有指定profile选项时 ...

  6. ZOJ 2760 How Many Shortest Path(最短路径+最大流)

    Description Given a weighted directed graph, we define the shortest path as the path who has the sma ...

  7. PCB各层介绍及AD软件画PCB时的规则

    好久没画过板了,最近因为工作关系,硬件软件全部得自己来,不得不重新打开闲置很久的AltiumDesigner.以前做过点乱七八糟的笔记,本来想回头翻看一下,结果哪儿也找不到,估计已经被不小心删掉了.  ...

  8. vim编辑器配置及常用命令

    最近工作不安分, 没有了刚入行时候的锐气, 不知道什么时候开始懈怠起来, 周末在电脑旁边看新闻, 搞笑图片, 追美剧, 一坐就是一天, 很是空虚. 我需要摆脱这种状态, 正好想学习一下安卓底层, An ...

  9. matlab中的静态变量

    persistent X Y Z 将X,Y,Z定义为在其声明处的函数的局部变量.然而,这些变量的值在函数调用期间在内存中保存(应该是堆区).Persistent 变量和global(全局)变量相似,因 ...

  10. Delphi中Sender对象的知识

    Sender是一个TObject类型的参数,它告诉Delphi哪个控件接收这个事件并调用相应的处理过程.你可以编写一个单一的事件处理句柄,通过Sender参数和IF…THEN…语句或者CASE语句配合 ...