3940: [Usaco2015 Feb]Censoring

Time Limit: 10 Sec  Memory Limit: 128 MB

Description

Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so they have plenty 
of material to read while waiting around in the barn during milking sessions. Unfortunately, the latest 
issue contains a rather inappropriate article on how to cook the perfect steak, which FJ would rather his 
cows not see (clearly, the magazine is in need of better editorial oversight).
FJ has taken all of the text from the magazine to create the string S of length at most 10^5 characters. 
He has a list of censored words t_1 ... t_N that he wishes to delete from S. To do so Farmer John finds 
the earliest occurrence of a censored word in S (having the earliest start index) and removes that instance 
of the word from S. He then repeats the process again, deleting the earliest occurrence of a censored word 
from S, repeating until there are no more occurrences of censored words in S. Note that the deletion of one 
censored word might create a new occurrence of a censored word that didn't exist before.
Farmer John notes that the censored words have the property that no censored word appears as a substring of 
another censored word. In particular this means the censored word with earliest index in S is uniquely 
defined.Please help FJ determine the final contents of S after censoring is complete.
FJ把杂志上所有的文章摘抄了下来并把它变成了一个长度不超过10^5的字符串S。他有一个包含n个单词的列表,列表里的n个单词
记为t_1...t_N。他希望从S中删除这些单词。 
FJ每次在S中找到最早出现的列表中的单词(最早出现指该单词的开始位置最小),然后从S中删除这个单词。他重复这个操作直到S中
没有列表里的单词为止。注意删除一个单词后可能会导致S中出现另一个列表中的单词 
FJ注意到列表中的单词不会出现一个单词是另一个单词子串的情况,这意味着每个列表中的单词在S中出现的开始位置是互不相同的 
请帮助FJ完成这些操作并输出最后的S

Input

The first line will contain S. The second line will contain N, the number of censored words. The next N lines contain the strings t_1 ... t_N. Each string will contain lower-case alphabet characters (in the range a..z), and the combined lengths of all these strings will be at most 10^5.
第一行包含一个字符串S 
第二行包含一个整数N 
接下来的N行,每行包含一个字符串,第i行的字符串是t_i

Output

The string S after all deletions are complete. It is guaranteed that S will not become empty during the deletion process.
一行,输出操作后的S
 
 

Sample Input

begintheescapexecutionatthebreakofdawn
2
escape
execution

Sample Output

beginthatthebreakofdawn

HINT

 

Source

Gold

#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
char s[N],s1[N],ans[N];
int fail[N],ch[N][],dep[N];
int n,rt=,tot=,q[N],w[N];
void build()
{
int h=,t=,x;
fail[rt]=rt;
for(int i=;i<;i++)
if(ch[rt][i]==-) ch[rt][i]=rt;
else
{
fail[ch[rt][i]]=rt;
q[++t]=ch[rt][i];
}
while(h<t)
{
x=q[++h];
for(int i=;i<;i++)
if(ch[x][i]==-) ch[x][i]=ch[fail[x]][i];
else
{
fail[ch[x][i]]=ch[fail[x]][i];
q[++t]=ch[x][i];
}
}
}
int main()
{
scanf("%s",s+);
memset(ch,-,sizeof(ch));
scanf("%d",&n);
int len,x,t;
for(int i=;i<=n;i++)
{
scanf("%s",s1+);
len=strlen(s1+);x=rt;
for(int i=;i<=len;i++)
{
t=s1[i]-'a';
if(ch[x][t]==-) ch[x][t]=++tot;
x=ch[x][t];
}
dep[x]=len;
}
build();
x=rt;len=strlen(s+);tot=;w[]=rt;
for(int i=;i<=len;i++)
{
ans[++tot]=s[i];
x=ch[x][s[i]-'a'];w[tot]=x;
if(dep[x]){tot-=dep[x];x=w[tot];}
}
for(int i=;i<=tot;i++) printf("%c",ans[i]);puts("");
return ;
}

bzoj 3940: [Usaco2015 Feb]Censoring -- AC自动机的更多相关文章

  1. BZOJ 3940: [Usaco2015 Feb]Censoring AC自动机_栈

    Description Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so ...

  2. BZOJ 3940: [Usaco2015 Feb]Censoring

    3940: [Usaco2015 Feb]Censoring Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 367  Solved: 173[Subm ...

  3. 【BZOJ3940】【BZOJ3942】[Usaco2015 Feb]Censoring AC自动机/KMP/hash+栈

    [BZOJ3942][Usaco2015 Feb]Censoring Description Farmer John has purchased a subscription to Good Hoov ...

  4. [Usaco2015 Feb]Censoring --- AC自动机 + 栈

    bzoj 3940 Censoring 题目描述 FJ把杂志上所有的文章摘抄了下来并把它变成了一个长度不超过10^5的字符串S. 他有一个包含n个单词的列表,列表里的n个单词记为T1......Tn. ...

  5. 【bzoj3940】[Usaco2015 Feb]Censoring AC自动机

    题目描述 Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so they h ...

  6. BZOJ3940: [Usaco2015 Feb]Censoring (AC自动机)

    题意:在文本串上删除一些字符串 每次优先删除从左边开始第一个满足的 删除后剩下的串连在一起重复删除步骤 直到不能删 题解:建fail 用栈存当前放进了那些字符 如果可以删 fail指针跳到前面去 好菜 ...

  7. Bzoj 3942: [Usaco2015 Feb]Censoring(kmp)

    3942: [Usaco2015 Feb]Censoring Description Farmer John has purchased a subscription to Good Hooveske ...

  8. [BZOJ 3942] [Usaco2015 Feb] Censoring 【KMP】

    题目链接:BZOJ - 3942 题目分析 我们发现,删掉一段 T 之后,被删除的部分前面的一段可能和后面的一段连接起来出现新的 T . 所以我们删掉一段 T 之后应该接着被删除的位置之前的继续向后匹 ...

  9. BZOJ 3942: [Usaco2015 Feb]Censoring

    Description 有两个字符串,每次用一个中取出下一位,放在一个字符串中,如果当前字符串的后缀是另一个字符串就删除. Sol KMP+栈. 用一个栈来维护新加的字符串就可以了.. 一开始我非常的 ...

随机推荐

  1. 【IDEA】IDEA中maven项目pom.xml依赖不生效解决

    问题: 今天在web项目中需要引入poi相关jar包.查看之下才发现pom.xml中的依赖虽然已经下载到了本地仓库 repository,但是却没有加入到项目路径的 Extenal Libraries ...

  2. Zabbix3.0 API调用

    Zabbix API 是什么? API简单来说是服务对外开放的一个接口,用户通过该接口传递请求,完成操作.API的背后是一组方法的集合,这些方法实现了服务对应的不同功能,调用API实际上就是换了一种方 ...

  3. Java集合里的一些“坑”

    这里主要谈下Java集合在使用中容易被忽略.又容易出现的两个“坑”,一个是集合与数组互相转换,另一个是集合遍历删除.主要通过代码演示. 一.集合与数组互相转换中的“坑” //Test1.java pa ...

  4. java版云笔记(五)

    下来是创建笔记本,创建笔记,这个没什么难点和前面是一样的. 创建笔记本 首先点击"+"弹出添加笔记的对话框,然后点击确定按钮创建笔记本. //点击"+"弹出添加 ...

  5. hdu 5895(矩阵快速幂+欧拉函数)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5895 f(n)=f(n-2)+2*f(n-1) f(n)*f(n-1)=f(n-2)*f(n-1)+2 ...

  6. Linux 基础——关机重启命令shutdown、reboot等

    一.关机重启命令的作用 相信对于接触过电脑的人来说,特别是对于windows系统来说,如果长时间使用不经重启的话会出现一点点卡顿的感觉.但是当重启整个系统后,这点点卡顿的感觉好像又没了,重启后wind ...

  7. ***PHP5.6.x SSL3_GET_SERVER_CERTIFICATE:certificate verify failed 解决方案

    centos: 在php.ini中增加一行 1 openssl.cafile=/etc/pki/tls/certs/ca-bundle.crt 重启服务器使修改生效

  8. 如何去除decimal后面的零?

    如何去除decimal后面的零? 1.260000m.ToString("G29") 不显示科学记数法? decimal.Parse("0.0000001",S ...

  9. IEEEXtreme 10.0 - Game of Stones

    这是 meelo 原创的 IEEEXtreme极限编程大赛题解 Xtreme 10.0 - Game of Stones 题目来源 第10届IEEE极限编程大赛 https://www.hackerr ...

  10. day1作业一:编写登陆接口

    作业一:编写登陆接口 1.输入用户名和密码 2.认证成功后显示欢迎信息 3.输错三次后锁定 Readme: (1)提示用户输入用户名: (2)用户名验证,验证是否已经锁定: (3)是否锁定:已锁定告诉 ...