Family View

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 175    Accepted Submission(s): 20

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
2016 ACM/ICPC Asia Regional Qingdao Online
/*
hdu 5880 AC自动机 problem:
给你一些子串,然后在文章中将这些子串屏蔽掉. solve:
用AC自动机扫一扫,然后给需要屏蔽的地方打下标记 hhh-2016-09-17 20:51:55
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue> using namespace std;
const int maxn = 1001000; char s[maxn];
int dis[maxn];
int ans[maxn]; struct Tire
{
int nex[1001000][27],fail[1001000],ed[1001000];
int root,L;
int newnode()
{
for(int i = 0; i < 26; i++)
nex[L][i] = -1;
ed[L++] = 0;
return L-1;
} void ini()
{
L = 0,root = newnode();
} int cha(char x)
{
return x-'a';
} void inser(char buf[])
{
int len = strlen(buf);
int now = root;
for(int i = 0; i < len; i++)
{
int ta = cha(buf[i]);
if(nex[now][ta] == -1)
nex[now][ta] = newnode();
now = nex[now][ta];
}
ed[now] = 1;
dis[now] = len;
} void build()
{
queue<int >q;
fail[root] = root;
for(int i = 0; i < 26; i++)
if(nex[root][i] == -1)
nex[root][i] = root;
else
{
fail[nex[root][i]] = root;
q.push(nex[root][i]);
}
while(!q.empty())
{
int now = q.front();
// if(ed[fail[now]])
// ed[now] = 1;
q.pop();
for(int i = 0; i < 26; i++)
{
if(nex[now][i] == -1)
nex[now][i] = nex[fail[now]][i];
else
{
fail[nex[now][i]] = nex[fail[now]][i];
q.push(nex[now][i]);
}
}
}
} void solve(char* str)
{
int cur = root;
int len = strlen(str);
int index;
for(int i = 0; i < len; i++)
{
if(str[i]>='A'&&str[i]<='Z')
{
index = str[i] - 'A';
}
else if(str[i]>='a'&&str[i]<='z')
{
index = str[i] - 'a';
}
else{
cur = root; //a,,,b,,,c,,,d
continue;
}
cur = nex[cur][index];
int tp = cur; while(tp != root)
{
if(ed[tp]){
ans[i+1] -= 1;
ans[i - dis[tp]+1] += 1;
break;
}
tp = fail[tp];
}
}
}
}; Tire ac; int main()
{
int t;
int n;
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
ac.ini();
for(int i=0; i<n; i++)
{
scanf("%s", s);
ac.inser(s);
}
ac.build();
getchar();
gets(s);
// puts(s);
memset(ans, 0, sizeof(ans));
ac.solve(s);
int len = strlen(s);
long long int tans = 0; for(int i=0; i<len; i++)
{
tans += ans[i];
if(tans <= 0) printf("%c", s[i]);
else printf("*"); }
printf("\n");
}
return 0;
}

  

hdu 5880 AC自动机的更多相关文章

  1. hdu 2896 AC自动机

    // hdu 2896 AC自动机 // // 题目大意: // // 给你n个短串,然后给你q串长字符串,要求每个长字符串中 // 是否出现短串,出现的短串各是什么 // // 解题思路: // / ...

  2. hdu 3065 AC自动机

    // hdu 3065 AC自动机 // // 题目大意: // // 给你n个短串,然后给你一个长串,问:各个短串在长串中,出现了多少次 // // 解题思路: // // AC自动机,插入,构建, ...

  3. hdu 2296 aC自动机+dp(得到价值最大的字符串)

    Ring Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submis ...

  4. hdu 2825 aC自动机+状压dp

    Wireless Password Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others ...

  5. hdu 3065 AC自动机(各子串出现的次数)

    病毒侵袭持续中 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Sub ...

  6. HDU 5384 AC自动机

    链接:http://acm.hdu.edu.cn/showproblem.php?pid=5384 题意:给n个母串,给m个匹配串,求每个母串依次和匹配串匹配,能得到的数目和. 分析:之前并不知道AC ...

  7. HDU 2222 AC自动机模板题

    题目: http://acm.hdu.edu.cn/showproblem.php?pid=2222 AC自动机模板题 我现在对AC自动机的理解还一般,就贴一下我参考学习的两篇博客的链接: http: ...

  8. HDU 2846 (AC自动机+多文本匹配)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2846 题目大意:有多个文本,多个模式串.问每个模式串中,有多少个文本?(匹配可重复) 解题思路: 传统 ...

  9. HDU 3065 (AC自动机模板题)

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3065 题目大意:多个模式串,范围是大写字母.匹配串的字符范围是(0~127).问匹配串中含有哪几种模 ...

随机推荐

  1. Software Engineering-HW1

    title: Software Engineering-HW1 date: 2017-09-13 15:41:13 tags: HW --- 阅读随笔 在<徐宥:掉进读书的兔子洞>里面, ...

  2. 201621123050 《Java程序设计》第9周学习总结

    1. 本周学习总结 1.1 以你喜欢的方式(思维导图或其他)归纳总结集合与泛型相关内容. 1.2 选做:收集你认为有用的代码片段 ①foreach循环 for (String e : map.keyS ...

  3. 2017-2018-1 1623 bug终结者 冲刺006

    bug终结者 冲刺006 by 20162328 蔡文琛 今日任务:音频素材添加 又是新的一天,小组项目有了很大的起色,已经可以在手机上试玩了. 添加背景音乐能使我们的游戏锦上添花. 音频资源需求 需 ...

  4. APP案例分析

    产品 蓝叠安卓模拟器 选择理由     看了一眼桌面,就这个比较有意思.现在很多人喜欢玩手游,经常喜欢开个小号搞事情.这时候身边又没有多余的手机,怎么办?安卓模拟器下一个.手机屏幕太小玩起来没意思怎么 ...

  5. nodeJS基于smtp发邮件

    邮件的协议smtp是tcp/ip族中的一个协议,所以我们这次考虑使用net模块来发送邮件. const net = require('net') const assert = require('ass ...

  6. Web移动端适配总结

    移动端适配的相关概念以及几种方案总结 适配相关概念 布局视口(layout viewport):html元素的上一级容器即顶级容器,用于解决页面在手机上显示的问题.大部分移动设备都将这个视口分辨率设置 ...

  7. kubernetes入门(02)kubernetes的架构

    一.kubernetes的主从架构 kubectl,全称 Kubernetes Control Plane,它表示Kubernetes为了实现最终的目标而构建的一组集群范围内的进程,这组进程相互协调, ...

  8. Jmeter入门(01)Jmeter的下载和安装

    一.什么是Jmeter 1.一款优秀的.开源的.免费的.功能测试和性能测试 工具 Jmeter ,使用Java开发的一款优秀的开源免费测试工具,主要用来做功能测试和性能测试(压力测试/负载测试),用J ...

  9. 新概念英语(1-1)Excuse me!

    Excuse me!Whose handbag is it? A:Excuse me! B:Yes? A:Is this your handbag? B:Pardon? A:Is this your ...

  10. 新概念英语(1-47)A cup of coffee

    新概念英语(1-47)A cup of coffee How does Ann like her coffee? A:Do you like coffee, Ann? B:Yes, I do. A:D ...