Codeforce 633C. Spy Syndrome 2
2 seconds
256 megabytes
standard input
standard output
After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such as Siddhant can't use a cipher as basic and ancient as Caesar cipher. After many weeks of observation of Siddhant’s sentences, Yash determined a new cipher technique.
For a given sentence, the cipher is processed as:
- Convert all letters of the sentence to lowercase.
- Reverse each of the words of the sentence individually.
- Remove all the spaces in the sentence.
For example, when this cipher is applied to the sentence
Kira is childish and he hates losing
the resulting string is
ariksihsidlihcdnaehsetahgnisol
Now Yash is given some ciphered string and a list of words. Help him to find out any original sentence composed using only words from the list. Note, that any of the given words could be used in the sentence multiple times.
The first line of the input contains a single integer n (1 ≤ n ≤ 10 000) — the length of the ciphered text. The second line consists of nlowercase English letters — the ciphered text t.
The third line contains a single integer m (1 ≤ m ≤ 100 000) — the number of words which will be considered while deciphering the text. Each of the next m lines contains a non-empty word wi (|wi| ≤ 1 000) consisting of uppercase and lowercase English letters only. It's guaranteed that the total length of all words doesn't exceed 1 000 000.
Print one line — the original sentence. It is guaranteed that at least one solution exists. If there are multiple solutions, you may output any of those.
30
ariksihsidlihcdnaehsetahgnisol
10
Kira
hates
is
he
losing
death
childish
L
and
Note
Kira is childish and he hates losing
12
iherehtolleh
5
HI
Ho
there
HeLLo
hello
HI there HeLLo
In sample case 2 there may be multiple accepted outputs, "HI there HeLLo" and "HI there hello" you may output any of them.
思路:暴力,字典树+DFS。感觉这题有点水过去的感觉。
说下我的思路,就是将下面给的单词到过来存进字典树中,然后用DFS正着搜。时间给了2s,140ms过的,也差点爆了内存
1 #include<stdio.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<string.h>
5 #include<math.h>
6 #include<queue>
7 #include<map>
8 using namespace std;
9 typedef long long LL;
10 char str[10005];
11 char trr[10005];
12 char strr[1005];
13 char tt[100005][1001];
14 int BB;
15 struct node
16 {
17 node*p[26];
18 int id;//记录单词的标号
19 node()
20 {
21 id=-2;
22 memset(p,0,sizeof(p));
23 }
24 };
25 int N=0;
26 int ii[100005];
27 int rr[100005];
28 node *head=new node();
29 void inserts(struct node*r,char *q,int v)//字典树的添加
30 {
31 int i,j,l;
32 node*NN=r;
33 l=strlen(q);
34 for(i=l-1; i>=0; i--)
35 {
36 int s=tolower(q[i])-'a';
37 if(NN->p[s]==NULL)
38 {
39 node*n=new node();
40 NN->p[s]=n;
41 NN=n;
42 }
43 else
44 NN=NN->p[s];
45 if(i==0)
46 NN->id=v;
47 }
48 }
49 struct node*ask(char p,struct node*r)//查询,返回当前的节点
50 {
51 int i,j;
52 struct node*dd=r;
53 int s=p-'a';
54 return dd->p[s];
55 }
56 void dfs(int id,int l,int k,node*d)
57 {
58 if(N)return ;
59 int i;
60 char tr[10005];
61 if(id==l&&N==0)
62 {for(i=0;;i++)
63 {
64 if(ii[i])
65 {
66 rr[i]=ii[i];
67 }
68 else break;
69 }
70 N=1;
71 return ;
72 }node *WW;
73 for(i=id; i<l; i++)
74 {
75 if(N)return ;
76 tr[i-id]=str[i];
77 tr[i-id+1]='\0';
78 if(i==id)WW=head;
79 node *nn=ask(tr[i-id],WW);
80 if(nn==NULL)//如果没有符合的直接跳出
81 {
82 return ;
83 }
84 if(nn->id!=-2)//当符合了进入下一层
85 {
86 ii[k]=nn->id;
87 dfs(i+1,l,k+1,nn);
88 ii[k]=0;
89 } WW=nn;//继续按当前的前缀串查找
90 }
91 }
92 int main(void)
93 {
94 int i,j,k,s;
95 N=0,BB=0;
96 scanf("%d",&k);
97 scanf("%s",str);
98 scanf("%d",&s);
99 int ans=0;
100 for(i=1; i<=s; i++)
101 {
102 scanf("%s",tt[i]);
103 int l=strlen(tt[i]);
104 inserts(head,tt[i],i);
105 }
106 dfs(0,k,0,head);
107 for(i=0;;i++)
108 {
109 if(rr[i]==0)
110 {
111 break;
112 }
113 }
114 int sl=i-1;
115 printf("%s",tt[rr[0]]);
116 for(i=1; i<=sl; i++)
117 {
118 printf(" %s",tt[rr[i]]);
119 }
120 return 0;
121 }
Codeforce 633C. Spy Syndrome 2的更多相关文章
- Codeforces 633C Spy Syndrome 2 | Trie树裸题
Codeforces 633C Spy Syndrome 2 | Trie树裸题 一个由许多空格隔开的单词组成的字符串,进行了以下操作:把所有字符变成小写,把每个单词颠倒过来,然后去掉单词间的空格.已 ...
- codeforces 633C. Spy Syndrome 2 hash
题目链接 C. Spy Syndrome 2 time limit per test 2 seconds memory limit per test 256 megabytes input stand ...
- CF#633C Spy Syndrome 2 DP+二分+hash
Spy Syndrome 2 题意 现在对某个英文句子,进行加密: 把所有的字母变成小写字母 把所有的单词反过来 去掉单词之间的空格 比如:Kira is childish and he hates ...
- Codeforces 633C Spy Syndrome 2(DP + Trie树)
题目大概说给一个加密的字符串,加密规则是把原文转化成小写字母,然后各个单词反转,最后去掉空格.现在给几个已知的单词,还原加密的字符串. 和UVa1401一个道理.. 用dp[i]表示加密字符前i个字符 ...
- Codeforces 633C Spy Syndrome 2 【Trie树】+【DFS】
<题目链接> 题目大意:给定一个只有小写字母组成的目标串和m个模式串(里面可能有大写字母),记目标串反过来后的串为S,让你从m个模式串中选出若干个组成S串(不区分大小写).输出任意一种方案 ...
- [codeforces] 633C Spy Syndrome 2
原题 Trie树+dp 首先,我们可以简单的想到一种dp方式,就是如果这一段可以匹配并且可以与前一段接上,那么更新dp[i]为当前字符串的编号,然后倒推就可以得到答案. 但是,显然我们不能O(m)比较 ...
- Codeforce 633.C Spy Syndrome 2
C. Spy Syndrome 2 time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...
- Manthan, Codefest 16 -C. Spy Syndrome 2
time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...
- Manthan, Codefest 16 C. Spy Syndrome 2 字典树 + dp
C. Spy Syndrome 2 题目连接: http://www.codeforces.com/contest/633/problem/C Description After observing ...
随机推荐
- 突破冯·诺依曼架构瓶颈!全球首款存算一体AI芯片诞生
过去70年,计算机一直遵循冯·诺依曼架构设计,运行时数据需要在处理器和内存之间来回传输. 随着时代发展,这一工作模式面临较大挑战:在人工智能等高并发计算场景中,数据来回传输会产生巨大的功耗:目前内存系 ...
- Spark(八)【广播变量和累加器】
目录 一. 广播变量 使用 二. 累加器 使用 使用场景 自定义累加器 在spark程序中,当一个传递给Spark操作(例如map和reduce)的函数在远程节点上面运行时,Spark操作实际上操作的 ...
- 位运算符在JS中的妙用
正文 位运算 JavaScript 中最臭名昭著的 Bug 就是 0.1 + 0.2 !== 0.3,因为精度的问题,导致所有的浮点运算都是不安全的,具体原因可详见<0.1 + 0.2不等于0. ...
- 3.5 Rust Generic Types, Traits, and Lifetimes
Every programming language has tools for effectively handling the duplication of concepts. In Rust, ...
- Output of C++ Program | Set 3
Predict the output of below C++ programs. Question 1 1 #include<iostream> 2 using namespace st ...
- 【Linux】【Database】【MySQL】使用percona搭建高可用的MySQL数据库
1. 简介 1.1. 官方文档: 数据库架构:https://docs.openstack.org/ha-guide/shared-database.html 1.2. 本次使用的的是Percona ...
- Linux单机安装Zookeeper
一.官网 https://zookeeper.apache.org/ 二.简介 Apache ZooKeeper致力于开发和维护开源服务器,实现高度可靠的分布式协调. ZooKeeper是一种集中式服 ...
- Linux 易错小结
修改文件夹(递归修改)权限 chmod -R 777 /html Linux查看进程的4种方法 第一种: ps aux ps命令用于报告当前系统的进程状态.可以搭配kill指令随时中断.删除不必要的程 ...
- 【Java 与数据库】How to Timeout JDBC Queries
How to Timeout JDBC Queries JDBC queries by default do not have any timeout, which means that a quer ...
- Windows下80端口被占用的解决方法(SQL Server)
查找80端口被谁占用的方法 进入命令提示行(WIN+R 输入 CMD),输入命令 netstat -ano|findstr 80 (显示包含:80的网络连接) ,就可以看到本机所有端口的使用情况,一般 ...