1016 - Brush (II)
Time Limit: 2 second(s) | Memory Limit: 32 MB |
After the long contest, Samee returned home and got angry after seeing his room dusty. Who likes to see a dusty room after a brain storming programming contest? After checking a bit he found a brush in his room which has width w. Dusts are defined as 2D points. And since they are scattered everywhere, Samee is a bit confused what to do. So, he attached a rope with the brush such that it can be moved horizontally (in X axis) with the help of the rope but in straight line. He places it anywhere and moves it. For example, the y co-ordinate of the bottom part of the brush is 2 and its width is 3, so the y coordinate of the upper side of the brush will be 5. And if the brush is moved, all dusts whose y co-ordinates are between 2 and 5 (inclusive) will be cleaned. After cleaning all the dusts in that part, Samee places the brush in another place and uses the same procedure. He defined a move as placing the brush in a place and cleaning all the dusts in the horizontal zone of the brush.
You can assume that the rope is sufficiently large. Now Samee wants to clean the room with minimum number of moves. Since he already had a contest, his head is messy. So, help him.
Input
Input starts with an integer T (≤ 15), denoting the number of test cases.
Each case starts with a blank line. The next line contains two integers N (1 ≤ N ≤ 50000) and w (1 ≤ w ≤ 10000), means that there are N dust points. Each of the next N lines will contain two integers: xi yi,denoting coordinates of the dusts. You can assume that (-109 ≤ xi, yi ≤ 109) and all points are distinct.
Output
For each case print the case number and the minimum number of moves.
Sample Input |
Output for Sample Input |
2 3 2 0 0 20 2 30 2 3 1 0 0 20 2 30 2 |
Case 1: 1 Case 2: 2 |
Note
Data set is huge, so use faster I/O methods.
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<iostream>
4 #include<algorithm>
5 #include<string.h>
6 #include<queue>
7 #include<stack>
8 #include<math.h>
9 #include<vector>
10 using namespace std;
11 typedef struct pp
12 {
13 int x;
14 int y;
15 }ss;
16 ss dd[500006];
17 int id[500006];
18 int dp[500006];
19 bool cmp(pp p,pp q)
20 {
21 return p.y<q.y;
22 }
23 int main(void)
24 {
25 int i,j,k;
26 scanf("%d",&k);
27 int s;
28 for(s=1;s<=k;s++)
29 { int n,m;
30 scanf("%d %d",&n,&m);
31 for(i=1;i<=n;i++)
32 {
33 scanf("%d %d",&dd[i].x,&dd[i].y);
34 }
35 sort(dd+1,dd+n+1,cmp);
36 id[0]=0;id[1]=1;
37 for(i=2;i<=n;i++)
38 {
39 int l=0;int r=i;
40 int ik=-1;
41 while(l<=r)
42 {
43 int mid=(l+r)/2;
44 if(dd[mid].y+m>=dd[i].y)
45 {
46 ik=mid;
47 r=mid-1;
48 }
49 else l=mid+1;
50 }
51 id[i]=ik;
52 }
53 fill(dp,dp+500006,1e9);
54 dp[0]=0;
55 for(i=1;i<=n;i++)
56 {
57 dp[i]=dp[id[i]-1]+1;
58 }
59 printf("Case %d: %d\n",s,dp[n]);
60 }
61 return 0;
62 }
1016 - Brush (II)的更多相关文章
- Lightoj 1016 - Brush (II)
After the long contest, Samee returned home and got angry after seeing his room dusty. Who likes to ...
- lightoj刷题日记
提高自己的实力, 也为了证明, 开始板刷lightoj,每天题量>=1: 题目的类型会在这边说明,具体见分页博客: SUM=54; 1000 Greetings from LightOJ [简单 ...
- one recursive approach for 3, hdu 1016 (with an improved version) , permutations, N-Queens puzzle 分类: hdoj 2015-07-19 16:49 86人阅读 评论(0) 收藏
one recursive approach to solve hdu 1016, list all permutations, solve N-Queens puzzle. reference: t ...
- 洛谷 1016 / codevs 1046 旅行家的预算
https://www.luogu.org/problem/show?pid=1016 http://codevs.cn/problem/1046/ 题目描述 Description 一个旅行家想驾驶 ...
- Leetcode 笔记 113 - Path Sum II
题目链接:Path Sum II | LeetCode OJ Given a binary tree and a sum, find all root-to-leaf paths where each ...
- Leetcode 笔记 117 - Populating Next Right Pointers in Each Node II
题目链接:Populating Next Right Pointers in Each Node II | LeetCode OJ Follow up for problem "Popula ...
- 【转】c#、wpf 字符串,color,brush之间的转换
转自:http://www.cnblogs.com/wj-love/archive/2012/09/14/2685281.html 1,将#3C3C3C 赋给background this.selec ...
- 函数式Android编程(II):Kotlin语言的集合操作
原文标题:Functional Android (II): Collection operations in Kotlin 原文链接:http://antonioleiva.com/collectio ...
- 统计分析中Type I Error与Type II Error的区别
统计分析中Type I Error与Type II Error的区别 在统计分析中,经常提到Type I Error和Type II Error.他们的基本概念是什么?有什么区别? 下面的表格显示 b ...
随机推荐
- .net与java建立WebService再互相调用
A: .net建立WebService,在java中调用. 1.在vs中新建web 简单修改一下Service.cs的[WebMethod]代码: using System; using System ...
- C/C++ Qt 数据库与ComBox多级联动
Qt中的SQL数据库组件可以与ComBox组件形成多级联动效果,在日常开发中多级联动效果应用非常广泛,例如当我们选择指定用户时,我们让其在另一个ComBox组件中列举出该用户所维护的主机列表,又或者当 ...
- 备忘录:关于.net程序连接Oracle数据库
目录 关于使用MSSM访问Oracle数据库 关于. net 程序中连接Oracle数据库 志铭-2021年12月7日 21:22:15 关于使用MSSM访问Oracle数据库 安装访问接口组件:Or ...
- 【Android】安装插件 + 改变文字大小、颜色 + 隐藏代码区块的直线
安装插件 可以在搜寻框里面填入关键字搜寻,具体的插件,网上有很多介绍了 改变文字大小.颜色 隐藏代码区块的直线
- Oracle参数文件—pfile与spfile
oracle的参数文件:pfile和spfile 1.pfile和spfile Oracle中的参数文件是一个包含一系列参数以及参数对应值的操作系统文件.它们是在数据库实例启动时候加载的, ...
- oracle 拆分字符串
WITH t AS (SELECT '1-2-3-4' a FROM dual)SELECT Regexp_Substr(a, '[^-]+', 1, LEVEL) i FROM tCONNECT B ...
- ORACLE lag,lead
oracle中想取对应列前几行或者后几行的数据时可以使用lag和lead分析函数 lag:是滞后的意思,表示本行数据是要查询的数据后面,即查询之前行的记录. lead:是领队的意思,表示本行数据是要查 ...
- 实现new Date(), 获取当前时间戳
JS 获取时间戳: 我相信大家找了很久了吧! 希望我写的这个对您有些帮助哦~ 大家是不是以为时间戳是关于时间的,都去 new Date() 里面找方法了啊,我来告诉你们正确的吧 其实大家用 JS 里的 ...
- 【Linux卷管理】LVM创建与管理
安装LVM 首先确定系统中是否安装了lvm工具: [root@jetsen ~]# rpm -qa|grep lvm system-config-lvm-1.1.5-1.0.el5 lvm2-2.02 ...
- Docker的常用命令总结
一.普通指令 启动 Docker sudo systemctl start docker 停止 Docker sudo systemctl stop docker 普通重启 Docker sudo s ...