描述

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.

输入

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.

输出

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.

样例输入

5
1 1
5 1
7 1
3 3
5 5

样例输出

1
2
1
1
0

提示

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

题目来源

Ural Collegiate 1999

做这题的时候一直超时。后来借了贞贞的代码看了,cjx更加混乱了,就两处地方改一改就不超时了。

实在搞不懂,求大神解答!

#include <stdio.h>
#include <string.h>
#define MAXN 32110 int n,m;
int L[];
int C[MAXN];
int lowbit(int t){
return t&(t^(t-));
}
int sum(int end){
int sum();
while(end>){
sum+=C[end];
end-=lowbit(end);
}
return sum;
} void add(int pos, int num){
while(pos<=MAXN){
C[pos]+=num;
pos+=lowbit(pos);
}
} int main(int argc, char *argv[])
{
int x,y;
scanf("%d",&n);
for(int i=; i<=n; i++){
scanf("%d%d",&x,&y);
x++;
add(x,);
L[sum(x)]++;
//----超时代码---
int level=sum(x);
L[level]++;
add(x,);
//---------------
}
for(int i=; i<=n; i++)
printf("%d\n",L[i]);
//----超时代码---
for(int i=; i<n; i++)
printf("%d\n",L[i]);
//---------------
return ;
}

TOJ 2711 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. TOJ 2776 CD Making

    TOJ 2776题目链接http://acm.tju.edu.cn/toj/showp2776.html 这题其实就是考虑的周全性...  贡献了好几次WA, 后来想了半天才知道哪里有遗漏.最大的问题 ...

  3. POJ 2352 Stars(树状数组)

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

  4. 【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:  ...

  5. 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 ...

  6. 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 ...

  7. POJ-2352 Stars 树状数组

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

  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 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 ...

随机推荐

  1. .net 基元类型,引用类型和值类型

    基元类型(primitive type): 编译器直接支持的数据类型称为基元类型(primitive type). string 与 String: 由于C#中的string (一个关键字)直接映射到 ...

  2. Android 自定义ViewGroup,实现侧方位滑动菜单

    侧方位滑动菜单 1.现在adnroid流行的应用当中很多都是用的侧方位滑动菜单如图:

  3. arcconf工具相关命令V1.0

    arcconf工具相关命令V1.0 清除当前所有raid配置 Arcconf  delete  1  array  all       #删除所有逻辑盘 Arcconf  uninit  1  all ...

  4. C# 文本输入限制类型,datagridview单元格输入验证

    1.只能输入double类型 private void textBoxX6_KeyPress(object sender, KeyPressEventArgs e) { { //数字0~9所对应的ke ...

  5. python-列表增删改查、排序、两个list合并、多维数组等

    一.list列表 数组 列表类型:list 下标从0开始,0,1,2... 二.列表增加元素 stus.append() 在列表末尾增加一个元素: stus.insert(,)  在指定位置添加一个元 ...

  6. ajax的get,post ,封装

    let ajax = new Object(); ajax.get = function(url,fn){ //创建ajax对象 let xhr = new XMLHttpRequest(); //与 ...

  7. Django 使用第三方服务发送电子邮件

    在 Django 网站中使用 mailgun 的邮件收发服务. 1.在 mailgun 官网上注册个账号(免费,免费账号每个月有10000条收发邮件的服务,对我来说已经完全够用了),注册完成后界面如图 ...

  8. vim文本编辑

    6.1 vi/vim 的使用 基本上 vi/vim 共分为三种模式,分别是命令模式(Command mode),插入模式(Insert mode)和底线命令模式(Last line mode). 6. ...

  9. VS2013安装及破解教程

    https://blog.csdn.net/qq_33742119/article/details/80075352 软件下载的百度云链接,也可以在官网直接下载 链接:https://pan.baid ...

  10. CentOS7.4关闭防火墙

    //临时关闭 systemctl stop firewalld //禁止开机启动 systemctl disable firewalld Removed symlink /etc/systemd/sy ...