CoderForces 163E e-Government(AC自动机+树状数组维护fail树的dfs序)

E. e-Government

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The best programmers of Embezzland compete to develop a part of the project called "e-Government" — the system of automated statistic collecting and press analysis.

We know that any of the k citizens can become a member of the Embezzland government. The citizens' surnames are a1, a2, ..., ak. All surnames are different. Initially all k citizens from this list are members of the government. The system should support the following options:

• Include citizen ai to the government.
• Exclude citizen ai from the government.
• Given a newspaper article text, calculate how politicized it is. To do this, for every active government member the system counts the number of times his surname occurs in the text as a substring. All occurrences are taken into consideration, including the intersecting ones. The degree of politicization of a text is defined as the sum of these values for all active government members.

Implement this system.

Input

The first line contains space-separated integers n and k (1 ≤ n, k ≤ 105) — the number of queries to the system and the number of potential government members.

Next k lines contain the surnames a1, a2, ..., ak, one per line. All surnames are pairwise different.

Next n lines contain queries to the system, one per line. Each query consists of a character that determines an operation and the operation argument, written consecutively without a space.

Operation "include in the government" corresponds to the character "+", operation "exclude" corresponds to "-". An argument of those operations is an integer between 1 and k — the index of the citizen involved in the operation. Any citizen can be included and excluded from the government an arbitrary number of times in any order. Including in the government a citizen who is already there or excluding the citizen who isn't there changes nothing.

The operation "calculate politicization" corresponds to character "?". Its argument is a text.

All strings — surnames and texts — are non-empty sequences of lowercase Latin letters. The total length of all surnames doesn't exceed 106, the total length of all texts doesn't exceed 106.

Output

For any "calculate politicization" operation print on a separate line the degree of the politicization of the given text. Print nothing for other operations.

Examples

input

Copy

7 3
a
aa
ab
?aaab
-2
?aaab
-3
?aaab
+2
?aabbaa

output

Copy

6
4
3
6

---

题意：

给出n个字符串，表示n个人名，有两种操作：

?string ，统计字符串string中出现的属于城市居民的次数。

+id，把编号为id的人变为城市居民，如果已经是忽略。

-id，把编号为id的人变为不是城市居民，如果已经不是的话忽略。

现有m个操作，对于?输出结果。

思路：

我们在统计的时候其实就是沿着fail指针走，把所有的标记叠加起来，而fail指针构成了一棵fail树，所以我们在求当前节点的fail指针方向有多少个标记的时候不必一层层的fail上去了，对于每个点维护其到根的有效节点的个数即可，当更新某个点的时候，就相当于这个点的子树到根的有效节点的个数都发生了变化，将树形结构变成线性结构，在树状数组中更新即可。

 

参考代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define lowbit(x) (x&-x)
#define maxn 1000010
int n,m,tot;
char s[maxn];
vector<int> vec[maxn];
int ch[maxn][],fail[maxn],val[maxn],last[maxn];
int c[maxn],in[maxn],out[maxn],tim,id[maxn],use[maxn];

inline void Modify(int x,int num)
{
    if(x==) return ;
    while(x<maxn)
    {
        c[x]+=num;
        x+=lowbit(x);
    }
}
inline void Add(int x,int y,int num)
{
    Modify(x,num);Modify(y,-num);
}
inline int Query(int x)
{
    int res=;
    while(x>)
    {
        res+=c[x];
        x-=lowbit(x);
    }
    return res;
}

inline void Init()
{
    tot=;tim=;
    memset(ch[],,sizeof(ch[]));
    memset(val,,sizeof(val));
    memset(use,,sizeof(use));
}
inline int idx(char c){ return c-'a';}
inline void Insert(char*s,int x)
{
    int u=,len=strlen(s);
    for(int i=;i<len;++i)
    {
        int c=idx(s[i]);
        if(!ch[u][c])
        {
            memset(ch[tot],,sizeof(ch[tot]));
            val[tot]=;
            ch[u][c]=tot++;
        }
        u=ch[u][c];
    }
    val[u]=x;
    id[x]=u;
}
inline void GetFail()
{
    queue<int> q;
    fail[]=;
    for(int c=;c<;++c)
    {
        int u=ch[][c];
        if(u){ fail[u]=;q.push(u);last[u]=; }
    }
    //cout<<"cnt "<<cnt<<endl;
    while(!q.empty())
    {
        int r=q.front(); q.pop();
        vec[fail[r]].push_back(r);
        for(int c=;c<;++c)
        {
            int u=ch[r][c];
            if(!u){ch[r][c]=ch[fail[r]][c];continue;}
            q.push(u);
            int v=fail[r];
            fail[u]=ch[v][c];
            last[u] = val[fail[u]]?fail[u]:last[fail[u]];
        }
    }
}
inline void dfs(int u)
{
    in[u]=++tim;
    for(int i=,len=vec[u].size();i<len;++i)
        dfs(vec[u][i]);
    out[u]=tim;
}

inline void clac(int x,int num)
{
    Add(in[id[x]],out[id[x]]+,num);
}

inline void work()
{
    ll ans=;
    int u=,len=strlen(s);
    for(int i=;i<len;++i)
    {
        int r=idx(s[i]);
        u=ch[u][r];
        ans+=Query(in[u]);
    }
    printf("%lld\n",ans);
}

int main()
{
    scanf("%d%d",&m,&n);
    Init();
    for(int i=;i<=n;++i)
        scanf("%s",s),Insert(s,i),use[i]=;
    GetFail();
    dfs();

    for(int i=;i<=n;++i) clac(i,);
     while(m--)
    {
        scanf("%s",s);
        if(s[]=='?') work();
        else
        {
            int x;
            sscanf(s+,"%d",&x);
            if(use[x]&&s[]=='-')
            {
                use[x] = ;
                clac(x,-);
            }
            else if(!use[x]&&s[]=='+')
            {
                use[x] = ;
                clac(x,);
            }
        }
    }

    return ;
}
/*
7 3
a
aa
ab
?aaab
-2
?aaab
-3
?aaab
+2
?aabbaa
*/