emm,ssy说可以直接CDQ分治...%%%但是注意到y是有序的,所以可以直接求一下前缀和就行了.

题干:

    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  is equal to  (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 (<=N<=). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, <=X,Y<=). 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 , the second does amount of stars of the level and so on, the last line contains amount of stars of the level N-.

代码:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<queue>
#include<algorithm>
#include<map>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(register int i = a;i <= n;i++)
#define lv(i,a,n) for(register int i = a;i >= n;i--)
#define clean(a) memset(a,0,sizeof(a))
const int INF = << ;
typedef long long ll;
typedef double db;
template <class T>
void read(T &x)
{
char c;
bool op = ;
while(c = getchar(), c < '' || c > '')
if(c == '-') op = ;
x = c - '';
while(c = getchar(), c >= '' && c <= '')
x = x * + c - '';
if(op) x = -x;
}
template <class T>
void write(T x)
{
if(x < ) putchar('-'), x = -x;
if(x >= ) write(x / );
putchar('' + x % );
}
const int maxn = ;
int tree[];
int ans[];
int n;
int lowbit(int x)
{
return x & -x;
}
void change(int x)
{
while(x <= maxn)
{
tree[x]++;
x += lowbit(x);
}
}
int query(int x)
{
int tot = ;
while(x > )
{
tot += tree[x];
x -= lowbit(x);
}
return tot;
}
int main()
{
read(n);
duke(i,,n)
{
int x,y;
read(x);read(y);
ans[query(x + )]++;
/*duke(j,1,n)
{
printf("%d ",query(j));
}
puts("");*/
change(x + );
}
duke(i,,n - )
{
printf("%d\n",ans[i]);
}
return ;
}

POJ2352 Stars 树状数组的更多相关文章

  1. POJ-2352 Stars 树状数组

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

  2. POJ2352 Stars [树状数组模板]

    题意:输入一n颗星星的x,y坐标,给定判断level的标准,即某颗星星左下边(不高于它,不超过他,相当于以他为基准的第三象限)星星的数目为level, 输出level从0到n的星星个数. //poj2 ...

  3. Stars(树状数组或线段树)

    Stars Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 37323 Accepted: 16278 Description A ...

  4. Stars(树状数组)

    http://acm.hdu.edu.cn/showproblem.php?pid=1541 Stars Time Limit: 2000/1000 MS (Java/Others)    Memor ...

  5. Stars(树状数组单点更新)

    Astronomers often examine star maps where stars are represented by points on a plane and each star h ...

  6. HDU-1541 Stars 树状数组

    题目链接:https://cn.vjudge.net/problem/HDU-1541 题意 天上有许多星星 现给天空一个平面坐标轴,统计每个星星的level, level是指某一颗星星的左下角(x& ...

  7. poj2352(树状数组)

    题目链接:https://vjudge.net/problem/POJ-2352 题意:在直角坐标系中给出n个点的 (x,y),(0<=x,y<=32000),定义每个点的level为(x ...

  8. 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. ~~~~~~~~~~~~~~~~~~~ ...

  9. hdu1541 Stars 树状数组

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1541 题目大意就是统计其左上位置的星星的个数 由于y已经按升序排列,因此只用按照x坐标生成一维树状数组 ...

随机推荐

  1. JavaScipt30(第四个案例)(主要知识点:数组原型链上的一些方法)

    承接上文,下面是第四个案例 附上项目链接: https://github.com/wesbos/JavaScript30 const inventors = [ { first: 'Albert', ...

  2. ubuntu 16.04 添加网卡

    root@ubuntu:~# ls /sys/class/net/ enp0s3 enp0s8 lo root@ubuntu:~# vim /etc/network/interfaces # This ...

  3. 还没更换RubyGems镜像?

    相信用过Ruby的人都知道 gem install 命令,但是在国内该命令安装的速度甚是不稳定(你懂的),导致尝试数次便是等待数时,记得之前在安装redmine的时候便是如此,之前不懂什么意思,还以为 ...

  4. 第3章 从Flux到Redux

    第3章 从Flux到Redux 3.1 Flux 单向数据流,React是用来替换Jquery的,Flux是以替换Backbone.js.Ember.js等MVC框架为主的. actionTypes. ...

  5. 安装ubuntu系统空间分配问题

    以下是我安装linux系统(ubuntu)时的系统空间配置,以50G为例: 挂载点 大小 格式 分区类型 / 15G Ext4 主分区 /home 30G Ext4 逻辑分区 /boot 1G Ext ...

  6. 随机数生成工具类(中文姓名,性别,Email,手机号,住址)

    public class RandomValueUtil { public static String base = "abcdefghijklmnopqrstuvwxyz012345678 ...

  7. 输入框点击下滑Ztree菜单

    记录一个功能实现代码,我这边前端用的是layui,需要实现的效果如下: 需求:当点击选择地区的时候会出现如上图的下拉菜单. 分析:首先肯定给这个输入框加监听,click方法,然后将ztree的div显 ...

  8. eclipse 中常用快捷键

    * 字母大小写转换 ctrl+shift+x   转为大写 ctrl+shift+y   转为小写 * eclipse 自动生成对象来接收方法的返回值的快捷键 说明:光标一定要定位到要自动生成返回值对 ...

  9. idea14远程调试linux下的tomcat

    进入到idea的tomcat的run/debug配置,新建个remote tomcat,然后填写相关信息,如上图(注意远程调试端口). 再选择Startup/Connection,如下图所示: 到此, ...

  10. 全面了解cookie和session

    http协议: http即超文本传输协议(万维网定义的),一种基于浏览器请求与服务器响应的链接,它是一个很纯粹的传输协议.http协议主要的特征就是它是一种无状态的协议(只针对cookie与sessi ...