A. Censoring

题目描述

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 10510^510​5​​ characters. He has a list of censored words t1t_1t​1​​ ... tNt_Nt​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10510^510​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t1t_1t​1​​...tNt_Nt​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

输入格式

The first line will contain S.
The second line will contain N, the number of censored words.
The next N lines contain the strings t1t_1t​1​​ ... tNt_Nt​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 10510^510​5​​.
第一行包含一个字符串S
第二行包含一个整数N N<2000 接下来的N行，每行包含一个字符串，第i行的字符串是tit_it​i​​

输出格式

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

样例

样例输入

begintheescapexecutionatthebreakofdawn
2
escape
execution

样例输出

beginthatthebreakofdawn

AC自动机+栈，和之前kmp一样，不过这个是多模式串用到AC自动机。然而标准的AC自动机会TLE，所以我又写了一个trie图。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<time.h>
using namespace std;
struct trie
{
    int count;
    trie *nxt[26],*fail;
    trie()
    {
        count=0;
        fail=NULL;
        for(int i=0;i<26;i++)nxt[i]=NULL;
    }
}*q[2000000],*root=new trie();
int head,tail;

char str[100010];
char key[100010];
int n;
char stack[100010];
trie *ff[100010];
int top;

void insert(char s[],trie *root)
{
    int len=strlen(s);
    trie *p=root;
    int i=0,index;
    while(s[i])
    {
        index=s[i]-'a';
        if(p->nxt[index]==NULL)p->nxt[index]=new trie();
        p=p->nxt[index];
        i++;
    }
    p->count=len;
}
void build_ac(trie *root)
{
    q[++tail]=root;
    while(head!=tail)
    {
        trie *now=q[++head];
        for(int i=0;i<26;++i)
        {
            if(now->nxt[i])
            {
                if(now==root) now->nxt[i]->fail=now;
                else now->nxt[i]->fail=now->fail->nxt[i];
                q[++tail]=now->nxt[i];
            }
            else
            {
                if(now==root) now->nxt[i]=now;
                else now->nxt[i]=now->fail->nxt[i]; 
            }
        }
    }
} 
inline int read()
{
    int s=0;char a=getchar();
    while(a<'0'||a>'9')a=getchar();
    while(a>='0'&&a<='9'){s=s*10+a-'0';a=getchar();}
    return s;
}
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("data14.in","r",stdin);
//    freopen("cen.out","w",stdout);

    scanf("%s",str);
    int len=strlen(str);
    n=read();
    for(int i=1;i<=n;i++)
    {
        scanf("%s",key);
        insert(key,root);
    }
    build_ac(root);
    
    int index;
    trie *p=root;
    trie *temp=NULL;
    for(int i=0;i<len;i++)
    {
        stack[++top]=str[i];
        index=str[i]-'a';
        ff[top]=p;
        p=p->nxt[index];
        if(p==NULL)p=root;
        if(p->count)
        {
            top-=p->count;
            p=ff[top+1];
        }
        
    }
    for(int i=1;i<=top;i++)
        putchar(stack[i]);
    return 0;    
}

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