CF724D. Dense Subsequence[贪心 字典序!]
2 seconds
256 megabytes
standard input
standard output
You are given a string s, consisting of lowercase English letters, and the integer m.
One should choose some symbols from the given string so that any contiguous subsegment of length m has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.
Then one uses the chosen symbols to form a new string. All symbols from the chosen position should be used, but we are allowed to rearrange them in any order.
Formally, we choose a subsequence of indices 1 ≤ i1 < i2 < ... < it ≤ |s|. The selected sequence must meet the following condition: for every j such that 1 ≤ j ≤ |s| - m + 1, there must be at least one selected index that belongs to the segment [j, j + m - 1], i.e. there should exist a k from 1 to t, such that j ≤ ik ≤ j + m - 1.
Then we take any permutation p of the selected indices and form a new string sip1sip2... sipt.
Find the lexicographically smallest string, that can be obtained using this procedure.
The first line of the input contains a single integer m (1 ≤ m ≤ 100 000).
The second line contains the string s consisting of lowercase English letters. It is guaranteed that this string is non-empty and its length doesn't exceed 100 000. It is also guaranteed that the number m doesn't exceed the length of the string s.
Print the single line containing the lexicographically smallest string, that can be obtained using the procedure described above.
3
cbabc
a
2
abcab
aab
3
bcabcbaccba
aaabb
In the first sample, one can choose the subsequence {3} and form a string "a".
In the second sample, one can choose the subsequence {1, 2, 4} (symbols on this positions are 'a', 'b' and 'a') and rearrange the chosen symbols to form a string "aab".
题意:至少每m个选一个字符使得选择的字符字典序最小的排列的字典序最小
和那个经典问题很想,就是多了选择的字符可以任意排列
要深入理解字典序
考虑只选a,可行就只选a,否则就要选b,此时a一定全选(字典序最小),b尽量少的选
枚举选到哪个字符就行了
PS:貌似可以二分,然而才26个字符并不需要
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=1e5+,INF=1e9+; int n,m,a[];
char s[N];
bool sol(char c){
int last=,cnt=,lc=;
for(int i=;i<=n;i++){
if(s[i]==c) lc=i;
if(s[i]<c) last=i;
if(i-last>=m){
if(i-lc<m) last=lc,cnt++;
else return false;
}
} for(int i=;i<=n;i++) a[s[i]]++;
for(char now='a';now<c;now++){
while(a[now]--) putchar(now);
}
while(cnt--) putchar(c);
return true;
}
int main(){
scanf("%d%s",&m,s+);
n=strlen(s+);
for(char c='a';c<='z';c++)
if(sol(c)) break;
}
CF724D. Dense Subsequence[贪心 字典序!]的更多相关文章
- Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) D. Dense Subsequence 暴力
D. Dense Subsequence 题目连接: http://codeforces.com/contest/724/problem/D Description You are given a s ...
- Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined)D Dense Subsequence
传送门:D Dense Subsequence 题意:输入一个m,然后输入一个字符串,从字符串中取出一些字符组成一个串,要求满足:在任意长度为m的区间内都至少有一个字符被取到,找出所有可能性中字典序最 ...
- Dense Subsequence
Dense Subsequence time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...
- [codeforces724D]Dense Subsequence
[codeforces724D]Dense Subsequence 试题描述 You are given a string s, consisting of lowercase English let ...
- Intel Code Challenge Final Round D. Dense Subsequence 二分思想
D. Dense Subsequence time limit per test 2 seconds memory limit per test 256 megabytes input standar ...
- AtCoder Grand Contest 026 (AGC026) E - Synchronized Subsequence 贪心 动态规划
原文链接https://www.cnblogs.com/zhouzhendong/p/AGC026E.html 题目传送门 - AGC026E 题意 给定一个长度为 $2n$ 的字符串,包含 $n$ ...
- 【29.41%】【codeforces 724D】Dense Subsequence
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 2016.10.08--Intel Code Challenge Final Round--D. Dense Subsequence
time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...
- CF Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined)
1. Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) B. Batch Sort 暴力枚举,水 1.题意:n*m的数组, ...
随机推荐
- Oracle11g 配置 ST_GEOMETRY
安装环境:ArcGIS Desktop10.2.1 .ArcSDE10.2.134940. Oracle11.2.0.1 操作系统:Windows Server 2012R2 DataCenter 安 ...
- window下 Sublime Text 3 安装
1.下载 官网 http://www.sublimetext.com/ 安装时,注意勾选Add to explorer context menu,这样Sublime Text可以被添加到右键中,在右键 ...
- C++序列化、反序列化
几个常见的库 http://stackoverflow.com/questions/3637581/fastest-c-serialization Boost: Fast, assorted C++ ...
- iOS APP上架过程常见问题
1.生产证书失效 2.上图中的蓝色选中部分的Provising Profile文件页需要导入,(调试证书(Developer).发布证书(distribution).还有Provising Profi ...
- iOS 用 SDWebImage 清理图片缓存
效果图如下: 1.找到 SDWebImage找到SDImageCache类 2.添加如下方法 - (float)checkTmpSize { ; NSDirectoryEnumerator *file ...
- 解决PKIX:unable to find valid certification path to requested target 的问题
这两天在twitter服务器上忽然遇到这样的异常: e: sun.security.validator.ValidatorException: PKIX path building failed: s ...
- 启动CLR
前面提到在SSCLI环境里运行.NET程序的时候,执行的命令类似java程序的执行过程,即通过clix程序解释执行.net程序.这个过程看起来跟在windows环境下执行.net程序表面上看起来不一样 ...
- CentOS 6编译安装lamp,并分别安装event模块方式和FPM方式的PHP
任务目标: 编译安装LAMP 要求(1) 安装一个模块化的PHP 要求(2) 安装一个FPM的PHP 注意PHP需要最后一个安装,因为需要前两者的支持. 所以这里的安装次序为 1.httpd 2.Ma ...
- MongoDB学习笔记~使用原生语句实现三层集合关系的更新
回到目录 MongoDB的文档型数据结构使得它在存储数据上很像JSON,在可读性方面很强,然而这种复杂的结构在update时相对麻烦一些,而对于官方给出的文档说的也不够细致,有些东西也是模棱两可的态度 ...
- 2-3 Linux文件管理命令详解
1. 复制文件 cp: copy 格式 SRCFILE DEST 一个文件到一个文件 多个文件到一个目录 注,cd 后面什么都不加,是进入用户的家目录 如果目标文件不存在,则先创建文 ...