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. ORACLE高级部分内容

    1.pl/sql基本语句 DECLARE BEGIN END; / 循环语句 DECLARE I  NUMBER(2):=1; BEGIN WHILE I<100 LOOP I:=I+1; EN ...

  2. [SHELL]退出脚本

    一,退出状态码 1,范围:0~255 2,查看退出状态码:必须在命令执行之后立即执行 ,显示的是脚本最后一条命令的退出状态码 echo $? 若f返回值为0,则表示正常 有异常为正值 二,exit 脚 ...

  3. 90 [LeetCode] Subsets2

    Given a collection of integers that might contain duplicates, nums, return all possible subsets (the ...

  4. Python3 Tkinter-Frame

    1.创建 from tkinter import * root=Tk() for fm in ['red','blue','yellow','green','white','black']: Fram ...

  5. Python练习—循环

    1.输入n的值,求出n的阶乘. s=1 n = int(input("请输入一个数")) for i in range(1,n+1): s=s*i print(s) 2.折纸上月球 ...

  6. Thunder团队第五周 - Scrum会议1

    Scrum会议1 小组名称:Thunder 项目名称:i阅app Scrum Master:杨梓瑞 工作照片: 邹双黛在照相,所以图片中没有该同学. 参会成员: 王航:http://www.cnblo ...

  7. Python学习之路5 - 函数

    函数 定义方式: def func(): "这里面写函数的描述" 这里写代码 return x #如果没有返回值就叫"过程",函数和过程的区别就是有无返回值 实 ...

  8. Java微笔记(3)

    Java 中的 static 使用之静态变量 Java 中被 static 修饰的成员称为静态成员或类成员. 它属于整个类所有,而不是某个对象所有,即被类的所有对象所共享. 静态成员可以使用类名直接访 ...

  9. LintCode-112.删除排序链表中的重复元素

    删除排序链表中的重复元素 给定一个排序链表,删除所有重复的元素每个元素只留下一个. 样例 给出 1->1->2->null,返回 1->2->null 给出 1-> ...

  10. Scala快速入门-基础

    HelloWorld 从HelloWorld开始,使用scala IDE编辑器. 新建scala project 新建scala object 编写HelloWorld run as scala ap ...