AC自动机。为什么洛谷水题赛会出现这种题然而并不会那么题意就不说啦 。终于会写AC自动机判断是否是子串啦。。。用到kmp的就可以用AC自动机水过去啦

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
const int nmax=100005;
const int maxn=27;
const int inf=0x7f7f7f7f;
int ch[nmax][maxn],fail[nmax],F[nmax],T[nmax],pt=0;
char s[nmax],t[nmax],S[nmax];
void insert(int x){
int len=strlen(s),t=0;
rep(i,0,len-1) {
if(!ch[t][s[i]-'a']) ch[t][s[i]-'a']=++pt;
t=ch[t][s[i]-'a'];
}
F[t]=len;
}
void getfail(){
queue<int>q;fail[0]=0;q.push(0);
while(!q.empty()){
int x=q.front();q.pop();
rep(i,0,25) {
if(ch[x][i]) q.push(ch[x][i]),fail[ch[x][i]]=x==0?0:ch[fail[x]][i];
else ch[x][i]=x==0?0:ch[fail[x]][i];
}
}
}
int main(){
scanf("%s",t+1);
int n;scanf("%d",&n);
rep(i,1,n) scanf("%s",s),insert(i);
getfail();
//rep(i,0,pt) printf("%d ",F[i]);printf("\n");
//rep(i,0,pt) printf("%d ",fail[i]);printf("\n");
int len=strlen(t+1),x=0,top=0;
rep(i,1,len){
S[++top]=t[i];
x=ch[x][t[i]-'a'];T[top]=x;
if(F[x]) top-=F[x],x=T[top];
//printf("%d",x);
}
rep(i,1,top) printf("%c",S[i]);printf("\n");
return 0;
}

  

3940: [Usaco2015 Feb]Censoring

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 253  Solved: 126
[Submit][Status][Discuss]

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

[Submit][Status][Discuss]

bzoj3940: [Usaco2015 Feb]Censoring的更多相关文章

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

    题目传送门 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105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.他希望从S中删除这些单词.FJ每次在S中 ...

  2. 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 ...

  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. 【BZOJ3940】[USACO2015 Feb] Censoring (AC自动机的小应用)

    点此看题面 大致题意: 给你一个文本串和\(N\)个模式串,要你将每一个模式串从文本串中删去.(此题是[BZOJ3942][Usaco2015 Feb]Censoring的升级版) \(AC\)自动机 ...

  5. 3942: [Usaco2015 Feb]Censoring [KMP]

    3942: [Usaco2015 Feb]Censoring Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 375  Solved: 206[Subm ...

  6. 3942: [Usaco2015 Feb]Censoring

    3942: [Usaco2015 Feb]Censoring Time Limit: 10 Sec Memory Limit: 128 MB Submit: 964 Solved: 480 [Subm ...

  7. bzoj 3940: [Usaco2015 Feb]Censoring -- AC自动机

    3940: [Usaco2015 Feb]Censoring Time Limit: 10 Sec  Memory Limit: 128 MB Description Farmer John has ...

  8. BZOJ 3940: [Usaco2015 Feb]Censoring

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

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

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

随机推荐

  1. JavaScript的常见事件和Ajax小结

    一.常见事件类型 1.鼠标事件 事件名称 说明 onclick 鼠标单击时触发 ondbclick 鼠标双击时触发 onmousedown 鼠标左键按下时触发 onmouseup 鼠标释放时触发 on ...

  2. CR0,CR3寄存器

    驱动在hook系统函数的时候通常要将只读属性暂时的屏蔽掉,主要有三种方法 1.修改CR0寄存器的WP位,使只读属性失效(这是网上用的最多的方法),切忌使用完之后立马修改回来 2.只读的虚拟地址,通过C ...

  3. Beaglebone Back学习一(开发板介绍)

    随着开源软件的盛行.成熟,开源硬件也迎来了春天,先有Arduino,后有Raspherry Pi,到当前的Beaglebone .相信在不久的将来,开源项目将越来越多,越来越走向成熟.         ...

  4. WPF中利用后台代码实现窗口分栏动态改变

    在WPF中实现窗口分栏并能够通过鼠标改变大小已经非常容易,例如将一个GRID分成竖排三栏显示,就可以将GRID先分成5列,其中两个固定列放GridSplitter. <Grid Backgrou ...

  5. fedora gnome extension

    如果想在gnome-shell桌面放点个性化的应用,可以在https://extensions.gnome.org网站上安装扩展(记得使用firefox). 下面我记录几个我们觉得还不错的扩展: 1. ...

  6. SYBASE时间计算

    以摘录了计算月初,月末,上月同日,下月同日,上年同日,下年同日(年有闰月问题),各种函数输出格式. 可以写到存储过程中也可单独使用. Sybase日期函数 文章分类:数据库 关键字: sybase日期 ...

  7. Keil V5.1x命令“Build Target”重新编译所有文件

    网上的解决办法有多种,但不知道哪一种能对症,以下是我的解决方法:

  8. having与where区别

    having后可以跟组函数如avg(sal)而where后不可以有, 如果条件不是必须使用组函数最好还是使用where

  9. Notepad++ 书签

      Notepad++,有一个书签功能,指定书签是Ctrl+F2,在书签之间移动是按F2来切换,这个可以在几个想查看的数据之间进行快速切换,所以看起来就很方便.

  10. 【贪心】 BZOJ 3252:攻略

    3252: 攻略 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 261  Solved: 90[Submit][Status][Discuss] De ...