地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=5880

题目:

Family View

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2109    Accepted Submission(s): 445

Problem Description
Steam is a digital distribution platform developed by Valve Corporation offering digital rights management (DRM), multiplayer gaming and social networking services. A family view can help you to prevent your children access to some content which are not suitable for them.

Take an MMORPG game as an example, given a sentence T, and a list of forbidden words {P}, your job is to use '*' to subsititute all the characters, which is a part of the substring matched with at least one forbidden word in the list (case-insensitive).

For example, T is: "I love Beijing's Tiananmen, the sun rises over Tiananmen. Our great leader Chairman Mao, he leades us marching on."

And {P} is: {"tiananmen", "eat"}

The result should be: "I love Beijing's *********, the sun rises over *********. Our gr*** leader Chairman Mao, he leades us marching on."

 
Input
The first line contains the number of test cases. For each test case:
The first line contains an integer n, represneting the size of the forbidden words list P. Each line of the next n lines contains a forbidden words Pi (1≤|Pi|≤1000000,∑|Pi|≤1000000) where Pi only contains lowercase letters.

The last line contains a string T (|T|≤1000000).

 
Output
For each case output the sentence in a line.
 
Sample Input
1
3
trump
ri
o
Donald John Trump (born June 14, 1946) is an American businessman, television personality, author, politician, and the Republican Party nominee for President of the United States in the 2016 election. He is chairman of The Trump Organization, which is the principal holding company for his real estate ventures and other business interests.
 
Sample Output
D*nald J*hn ***** (b*rn June 14, 1946) is an Ame**can businessman, televisi*n pers*nality, auth*r, p*litician, and the Republican Party n*minee f*r President *f the United States in the 2016 electi*n. He is chairman *f The ***** *rganizati*n, which is the p**ncipal h*lding c*mpany f*r his real estate ventures and *ther business interests.
 
Source

思路:

  裸ac自动机题,走到一个节点看下有匹配的没,有的话用最长串的长度把位置标记一下,然后输出即可。

  ac自动机中实际节点数并不需要开到27*1e6!!!数据并没那么大
  

  只需开到27*1e5

 #include <queue>
#include <cstring>
#include <cstdio>
using namespace std; struct AC_auto
{
const static int LetterSize = ;
const static int TrieSize = * ( 1e5 + ); int tot,root,fail[TrieSize],end[TrieSize],next[TrieSize][LetterSize]; int newnode(void)
{
memset(next[tot],-,sizeof(next[tot]));
end[tot] = ;
return tot++;
} void init(void)
{
tot = ;
root = newnode();
} int getidx(char x)
{
if(x<='z'&&x>='a') return x - 'a';
if(x<='Z'&&x>='A') return x - 'A';
return ;
} void insert(char *ss)
{
int len = strlen(ss);
int now = root;
for(int i = ; i < len; i++)
{
int idx = getidx(ss[i]);
if(next[now][idx] == -)
next[now][idx] = newnode();
now = next[now][idx];
}
end[now] = len;
} void build(void)
{
queue<int>Q;
fail[root] = root;
for(int i = ; i < LetterSize; i++)
if(next[root][i] == -)
next[root][i] = root;
else
fail[next[root][i]] = root,Q.push(next[root][i]);
while(Q.size())
{
int now = Q.front();Q.pop();
for(int i = ; i < LetterSize; i++)
if(next[now][i] == -) next[now][i] = next[fail[now]][i];
else
fail[next[now][i]] = next[fail[now]][i],Q.push(next[now][i]);
}
} void match(char *ss,int *cnt)
{
int len,now;
len = strlen(ss),now = root;
for(int i = ; i < len; i++)
{
int idx = getidx(ss[i]);
int tmp = now = next[now][idx], ret = ;
while(tmp)
{
ret = max( ret, end[tmp]);
tmp = fail[tmp];
}
if(ret)
cnt[i-ret+]++,cnt[i+]--;
}
}
void debug()
{
for(int i = ;i < tot;i++)
{
printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
for(int j = ;j < LetterSize;j++)
printf("%3d",next[i][j]);
printf("]\n");
}
}
}ac; char ss[];
int cnt[]; int main(void)
{
int t,n;
scanf("%d",&t);
while(t--)
{
ac.init();
scanf("%d",&n);
for(int i=;i<=n;i++) scanf("%s",ss),ac.insert(ss);
getchar(),gets(ss);
ac.build();
ac.match(ss,cnt);
for(int i=,ret=,len=strlen(ss);i<len;i++)
{
ret+=cnt[i],cnt[i]=;
if(ret>) ss[i]='*';
}
printf("%s\n",ss);
}
return ;
}

hdu5880 Family View的更多相关文章

  1. HDU5880 Family View ac自动机第二题

    Steam is a digital distribution platform developed by Valve Corporation offering digital rights mana ...

  2. HDU5880 Family View(2016青岛网络赛 AC自动机)

    题意:将匹配的串用'*'代替 tips: 1 注意内存的使用,据说g++中指针占8字节,c++4字节,所以用g++交会MLE 2 注意这种例子, 12abcdbcabc 故失败指针要一直往下走,否则会 ...

  3. 虾扯蛋:Android View动画 Animation不完全解析

    本文结合一些周知的概念和源码片段,对View动画的工作原理进行挖掘和分析.以下不是对源码一丝不苟的分析过程,只是以搞清楚Animation的执行过程.如何被周期性调用为目标粗略分析下相关方法的执行细节 ...

  4. MVVM模式解析和在WPF中的实现(五)View和ViewModel的通信

    MVVM模式解析和在WPF中的实现(五) View和ViewModel的通信 系列目录: MVVM模式解析和在WPF中的实现(一)MVVM模式简介 MVVM模式解析和在WPF中的实现(二)数据绑定 M ...

  5. Android 判断一个 View 是否可见 getLocalVisibleRect(rect) 与 getGlobalVisibleRect(rect)

    Android 判断一个 View 是否可见 getLocalVisibleRect(rect) 与 getGlobalVisibleRect(rect) [TOC] 这两个方法的区别 View.ge ...

  6. android 使用Tabhost 发生could not create tab content because could not find view with id 错误

    使用Tabhost的时候经常报:could not create tab content because could not find view with id 错误. 总结一下发生错误的原因,一般的 ...

  7. SAP CRM 树视图(TREE VIEW)

    树视图可以用于表示数据的层次. 例如:SAP CRM中的组织结构数据可以表示为树视图. 在SAP CRM Web UI的术语当中,没有像表视图(table view)或者表单视图(form view) ...

  8. 深入理解 Android 之 View 的绘制流程

    概述 本篇文章会从源码(基于Android 6.0)角度分析Android中View的绘制流程,侧重于对整体流程的分析,对一些难以理解的点加以重点阐述,目的是把View绘制的整个流程把握好,而对于特定 ...

  9. Android listview和gridview以及view的区别

    GridView 可以指定显示的条目的列数. listview一般显示的条目的列数都是一列 如果是列表(单列多行形式)的使用ListView,如果是多行多列网状形式的优先使用GridView andr ...

随机推荐

  1. MySQL 密码设置

    如何修改 MySQL 密码: [root@localhost ~]$ mysqladmin -uroot password 'newPass' # 在无密码的情况下设置密码 [root@localho ...

  2. oracle查锁表

    查锁表 select LOCK_INFO.OWNER || '.' || LOCK_INFO.OBJ_NAME as OBJ_NAME, -- 对象名称(已经被锁住) LOCK_INFO.SUBOBJ ...

  3. oracle 与mysql 的当前时间比较

    select p.id,p.order_Num,p.image_url,p.url,p.image_topic, p.is_download, p.big_image_url, p.begin_tim ...

  4. combobox组合框

    最近在改BUG的时候发现,combobox组合框如果选择的是Dropdown模式在初始化combobox对象时候有如下操作 1.SetDlgItemInt(IDC_WB_FONTSIZECOMBOX, ...

  5. poj_1390 动态规划

    题目大意 将一些连续的序列根据颜色分为N段,每段有颜色 为 Ci, 长度为 Li.每次点击其中的一段 i ,则可以将该段i消除,该段相邻的两段自动连接到一起,如果连接到一起的两段之前的颜色相同,则更新 ...

  6. xcode 4.6 破解及真机调试

    从安卓到IOS,从  eclipse 到xcode跨度还是比较大的.在研究的过程中发现,许多时候不仅仅是C,C++,JAVA和OBJECT-C的区别,相对于编程语言来说,操作习惯和开发工具带来的困惑要 ...

  7. 点击button,button背景图片变化

    1.设置背景渐变效果,在drawable目录下建buttonshape.xml文件, 内容为: <?xml version="1.0" encoding="utf- ...

  8. java基础---->多线程之Runnable(一)

    java线程的创建有两种方式,这里我们通过简单的实例来学习一下.一切都明明白白,但我们仍匆匆错过,因为你相信命运,因为我怀疑生活. java中多线程的创建 一.通过继承Thread类来创建多线程 pu ...

  9. MyBatis——Mapper XML 文件

    Mapper XML 文件 MyBatis 的真正强大在于它的映射语句,也是它的魔力所在.由于它的异常强大,映射器的 XML 文件就显得相对简单.如果拿它跟具有相同功能的 JDBC 代码进行对比,你会 ...

  10. 域名与IP对应,解决只能IP访问不能域名访问的问题

    sudo vim /etc/hosts 127.0.0.1 localhost 127.0.1.1 ubuntu 192.168.1.60 api.sscmp.com