codeforces 985E Pencils and Boxes(dp+思维)
2 seconds
256 megabytes
standard input
standard output
Mishka received a gift of multicolored pencils for his birthday! Unfortunately he lives in a monochrome world, where everything is of the same color and only saturation differs. This pack can be represented as a sequence a1, a2, ..., an of n integer numbers — saturation of the color of each pencil. Now Mishka wants to put all the mess in the pack in order. He has an infinite number of empty boxes to do this. He would like to fill some boxes in such a way that:
- Each pencil belongs to exactly one box;
- Each non-empty box has at least k pencils in it;
- If pencils i and j belong to the same box, then |ai - aj| ≤ d, where |x| means absolute value of x. Note that the opposite is optional, there can be pencils i and j such that |ai - aj| ≤ d and they belong to different boxes.
Help Mishka to determine if it's possible to distribute all the pencils into boxes. Print "YES" if there exists such a distribution. Otherwise print "NO".
The first line contains three integer numbers n, k and d (1 ≤ k ≤ n ≤ 5·105, 0 ≤ d ≤ 109) — the number of pencils, minimal size of any non-empty box and maximal difference in saturation between any pair of pencils in the same box, respectively.
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109) — saturation of color of each pencil.
Print "YES" if it's possible to distribute all the pencils into boxes and satisfy all the conditions. Otherwise print "NO".
6 3 10
7 2 7 7 4 2
YES
6 2 3
4 5 3 13 4 10
YES
3 2 5
10 16 22
NO
In the first example it is possible to distribute pencils into 2 boxes with 3 pencils in each with any distribution. And you also can put all the pencils into the same box, difference of any pair in it won't exceed 10.
In the second example you can split pencils of saturations [4, 5, 3, 4] into 2 boxes of size 2 and put the remaining ones into another box.
题意:有n个铅笔,每只铅笔有个value,现在把他们分发到几个笔筒中,每个笔筒最少放k个笔,且放入的每支笔的value差不能大于d。求解能不能找到一种方案满足。
思路:
dp。先排序,判断最后一个数能否作为一个序列的结尾。
#include <iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<deque>
#include<vector>
#define ll long long
#define inf 0x3f3f3f3f
#define mod 1000000007;
using namespace std;
ll a[];
bool dp[];
int main()
{
int n,k;
ll d;
scanf("%d%d%I64d",&n,&k,&d);
for(int i=;i<=n;i++)
{
scanf("%I64d",&a[i]);
}
sort(a+,a++n);
memset(dp,,sizeof(dp));
dp[]=;
int p=;
for(int i=;i<=n;i++)
{
if(dp[i])//它是结尾的基础上,从下一个开始判断哪些可以作为结尾
{
p=max(p,i+k);//算过的就不用再重复算了,以防超时
//经典样例 50000 10 0
//1 1 1 1 1 1 1 1 1 1 1 ……
while(p<=n&&a[p]-a[i+]<=d)
{//如果它可以作为结尾
dp[p]=;
p++;
}
}
}
printf(dp[n]?"YES\n":"NO\n");
return ;
}
codeforces 985E Pencils and Boxes(dp+思维)的更多相关文章
- codeforces 985E Pencils and Boxes
题意: 把一个数组分成若干组,保证每组的size >= k并且一组中任意两个数字的差的绝对值 <= d,问存不存在这样的分法. 思路: 线性dp. 用dp[i]表示前i个数是否有分法. 设 ...
- codeforces 985 E. Pencils and Boxes (dp 树状数组)
E. Pencils and Boxes time limit per test 2 seconds memory limit per test 256 megabytes input standar ...
- codeforces 572 D. Minimization(dp+ 思维)
题目链接:http://codeforces.com/contest/572/problem/D 题意:给出一个序列,可以任意调整序列的顺序,使得给出的式子的值最小 题解:显然要先排一下序,然后取相邻 ...
- Codeforces Educational Codeforces Round 44 (Rated for Div. 2) E. Pencils and Boxes
Codeforces Educational Codeforces Round 44 (Rated for Div. 2) E. Pencils and Boxes 题目连接: http://code ...
- Codeforces 985 E - Pencils and Boxes
E - Pencils and Boxes 思路: dp 先排个序,放进一个袋子里的显然是一段区间 定义状态:pos[i]表示小于等于i的可以作为(放进一个袋子里的)一段区间起点的离i最近的位置 显然 ...
- [BZOJ 3625] [Codeforces 438E] 小朋友的二叉树 (DP+生成函数+多项式开根+多项式求逆)
[BZOJ 3625] [Codeforces 438E] 小朋友的二叉树 (DP+生成函数+多项式开根+多项式求逆) 题面 一棵二叉树的所有点的点权都是给定的集合中的一个数. 让你求出1到m中所有权 ...
- 【CF1256】Codeforces Round #598 (Div. 3) 【思维+贪心+DP】
https://codeforces.com/contest/1256 A:Payment Without Change[思维] 题意:给你a个价值n的物品和b个价值1的物品,问是否存在取物方案使得价 ...
- Codeforces Round #533 (Div. 2) C.思维dp D. 多源BFS
题目链接:https://codeforces.com/contest/1105 C. Ayoub and Lost Array 题目大意:一个长度为n的数组,数组的元素都在[L,R]之间,并且数组全 ...
- Codeforces 407B Long Path(好题 DP+思维)
题目链接:http://codeforces.com/problemset/problem/407/B 题目大意:一共n+1个房间,一个人从1走到n+1,每次经过房间都会留下一个标记,每个房间有两扇门 ...
随机推荐
- 使用springmvc时报错org.springframework.beans.NullValueInNestedPathException: Invalid property 'department' of bean class [com.atguigu.springmvc.crud.entities.Employee]:
使用springmvc时报错 org.springframework.beans.NullValueInNestedPathException: Invalid property 'departmen ...
- python学习笔记(接口自动化框架 V1.0)
之前是利用python自带的unittest测试框架 这次自己设计一个 之后再一点点往里面加功能 (ps:当然这个框架真的是很简单..很简单...很简单...) excel文件格式: #!/usr/b ...
- GO学习笔记:函数作为值、类型
在Go中函数也是一种变量,我们可以通过type来定义它,它的类型就是所有拥有相同的参数,相同的返回值的一种类型: type typeName func(input1 inputType1 , inpu ...
- 委托,lambda,匿名方法
lambda表达式其实就是匿名方法的变体或者说简写. 原来我们用 delegate void Del(int x); Del d = delegate(int x) { return x + 1; } ...
- SimpleDateFormat函数语法
SimpleDateFormat函数语法: G 年代标志符 y 年 M 月 d 日 h 时 在上午或下午 (1~12) ...
- 一些开源搜索引擎实现——倒排使用原始文件,列存储Hbase,KV store如levelDB、mongoDB、redis,以及SQL的,如sqlite或者xxSQL
本文说明:除开ES,Solr,sphinx系列的其他开源搜索引擎汇总于此. A search engine based on Node.js and LevelDB A persistent, n ...
- Cassandra 数据模型设计,根据你的查询来制定设计——反范式设计本质:空间换时间
转自:http://www.infoq.com/cn/articles/best-practice-of-cassandra-data-model-design 不要把Cassandra model想 ...
- .NET中 数据库连接
(转自:http://www.iwms.net/n459c12.aspx) SQL Server ODBC Standard Security:"Driver={SQL Server};S ...
- c#中绝对路径和相对路径
文件操作涉及一个非常重要的概念——文件路径.文件路径是指用来标识系统中文件存放位置的字符串.如:D:\\test.txt,表示在D盘根目录下存入test.txt文件. 文件路径分为绝对路径和相对路径. ...
- PostgreSQL.conf文件配置详解[转]
一.连接配置与安全认证 1.连接Connection Settings listen_addresses (string) 这个参数只有在启动数据库时,才能被设置.它指定数据库用来监听客户端连接的 ...