BZOJ3940:[USACO]Censoring(AC自动机,栈)

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

Solution

一开始的确没啥思路……一看到有栈这个标签就明了了……
果然我还是水平太菜啊
不过一看到栈这个题就没什么思考难度了……就基本全靠细节了
首先把自动机建出来……话说我这几天才发现原来自己建的一直叫trie图
然后一位一位放到自动机上跑。
分两种情况：
1、栈空
   判断是否是根节点的一个儿子即可(即判断是否是单词的第一位)
2、栈不为空
   判断当前位能否和上一位匹配，如果不能的话就清空栈(因为有这一位挡着就已经前功尽弃了)
然后当当前位匹配到某一位的末尾后，就将这个单词从栈中清空(因为题目说了越靠前出现的越早清空)
清空栈的时候输出一下就好了(除了匹配到的单词不输出)

Code

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define N (100000+10)
using namespace std;
int Fail[N],Son[N][27],End[N];
int n,sz,maxn,stack[N],top;
char s[N],st[N],ch[N];
queue<int>q;

void Insert(char s[])
{
    int len=strlen(s),now=0;
    for (int i=0;i<len;++i)
    {
        int x=s[i]-'a';
        if (!Son[now][x]) Son[now][x]=++sz;
        now=Son[now][x];
        ch[now]=s[i];
    }
    End[now]=len;
}

void Build_Fail()
{
    for (int i=0;i<26;++i)
        if (Son[0][i])
            q.push(Son[0][i]);
    while (!q.empty())
    {
        int now=q.front(); q.pop();
        for (int i=0;i<26;++i)
        {
            if (!Son[now][i])
            {
                Son[now][i]=Son[Fail[now]][i];
                continue;
            }
            Fail[Son[now][i]]=Son[Fail[now]][i];
            q.push(Son[now][i]);
        }
    }
}

void Compare(char s[])
{
    int len=strlen(s);
    for (int i=0;i<len;++i)
    {
        int x=s[i]-'a';
        if (!top)
        {
            if (Son[0][x])
                stack[++top]=Son[0][x];
            else
                printf("%c",s[i]);
        }
        else
        {
            int son=Son[stack[top]][x];
            if (Son[stack[top]][x])
                stack[++top]=son;
            else
            {
                for (int j=1;j<=top;++j)
                    printf("%c",ch[stack[j]]);
                printf("%c",s[i]);
                top=0;
            }
        }
        if (End[stack[top]])
        {
            int t=End[stack[top]];
            for (int i=1;i<=t;++i)
                top--;
        }
    }
    for (int i=1;i<=top;++i)
        printf("%c",ch[stack[i]]);
}

int main()
{
    scanf("%s%d",s,&n);
    for (int i=1;i<=n;++i)
        scanf("%s",st),Insert(st);
    Build_Fail();
    Compare(s);
}