POJ-4052-ac自动机

先把匹配上的串都标记。然后把每个匹配上的串的子串变成非结束。

直接会超时。优化找失败指针的地方。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int maxn = 250009;

char str1[5100009], str2[5100009]; 
char s[2509][1109];
char ss[1109];
int vis[maxn];
int cnt, n;

struct trie
{
    trie *ch[26];
    trie *fail;
    int end, key;
    void init()
    {
        memset(ch, NULL, sizeof(ch));
        fail = NULL;
        key = 0;
    }
}*root, t[maxn];

void insert(char *s, int id) //正常的插入
{
    trie *p = root;
    for(int i=0; s[i]; i++)
    {
        int k = s[i]-'A';
        if(p->ch[k]==NULL)
        {
            t[cnt].init();
            p->ch[k] = &t[cnt++];
        }
        p = p->ch[k];
    }
    p->end = id;
    p->key = 1;
}

void reinsert(char *a) //将其子串删掉
{
    trie *p = root;
    for(int i=0; a[i]; i++)
    {
        int k = a[i]-'A';
        while(p->ch[k]==NULL && p!=root) p=p->fail;
        p = p->ch[k];
        p = (p==NULL)? root : p;
        trie *tmp = p;

        while(tmp!=root && tmp->key!=-3)
        {
            if(abs(tmp->key)==1 && vis[tmp->end])
            {
                vis[tmp->end] = 0;
            }
            tmp->key = -3;
            tmp = tmp->fail;
        }
    }
}

queue<trie*>q;
void build_fail()
{
    q.push(root);
    root->fail = root;
    while(!q.empty())
    {
        trie *p = q.front();
        q.pop();
        for(int k=0; k<26; k++)
        if(p->ch[k]!=NULL)
        {
            q.push(p->ch[k]);
            if(p==root)
            {
                p->ch[k]->fail = root;
                continue;
            }
            trie *tmp = p->fail;
            while( tmp!=root && tmp->ch[k]==NULL)
                tmp = tmp->fail;
            if(tmp->ch[k]) p->ch[k]->fail = tmp->ch[k];
            else p->ch[k]->fail = root;
        }
    }
}

void query(char *a) //标记出现过的所有串
{
    trie *p = root;
    for(int i=0; a[i]; i++)
    {
        int k = a[i]-'A';
        while(p->ch[k]==NULL && p!=root) p=p->fail;
        p = p->ch[k];
        p = (p==NULL)? root : p;
        trie *tmp = p;

        while(tmp!=root && tmp->key>=0)
        {
            if(tmp->key==1)
            {
                vis[tmp->end] = 1;
                tmp->key = -1;
            }
            else tmp->key = -2;
            tmp = tmp->fail;
        }
    }
}

int requery(char *a) 
{
    int ans = 0;
    trie *p = root;
    for(int i=0; a[i]; i++)
    {
        int k = a[i]-'A';
        while(p->ch[k]==NULL && p!=root) p=p->fail;
        p = p->ch[k];
        p = (p==NULL)? root : p;
        trie *tmp = p;
        while(tmp!=root && tmp->key!=-4)
        {
            if(abs(tmp->key)==1) ans++;
            tmp->key = -4;
            tmp = tmp->fail;
        }
    }
    return ans;
}

int main()
{
    int tt, i, j, k;
    scanf("%d", &tt);
    while(tt--)
    {
        cnt = 1;
        t[0].init();
        root = &t[0];
        scanf("%d", &n);
        for(i = 1; i<=n; i++)
        {
            scanf("%s", ss);
            for(k=0, j=0; ss[k]; k++)
            {
                if(ss[k]>='A' && ss[k]<='Z')
                {
                    s[i][j] = ss[k];
                    j++;
                }
                else
                {
                    k++;
                    int kk = ss[k]-'0';
                    k++;
                    while(ss[k]>='0'&&ss[k]<='9') kk = kk*10+ ss[k++]-'0';
                    while(kk--) s[i][j++] = ss[k];
                    k++;
                }
            }
            s[i][j] = '\0';
            insert(s[i], i);
        }
        build_fail();

        scanf("%s", str1);
        for(i=0, j=0; str1[i]; i++)
        {
            if(str1[i]>='A' && str1[i]<='Z')
            {
                str2[j] = str1[i];
                j++;
            }
            else
            {
                i++;
                int k = str1[i]-'0';
                i++;
                while(str1[i]>='0'&&str1[i]<='9') k = k*10+ str1[i++]-'0';
                while(k--) str2[j++] = str1[i];
                i++;
            }
        }
        str2[j] = '\0';
        memset(vis, 0, sizeof(vis));
        query(str2);

        for(int i=1; i<=n; i++)
        if(vis[i])
        {
            reinsert( s[i] );
            insert( s[i], i );
        }

        int ans = requery( str2 );
        printf("%d\n", ans);
    }
	return 0;
}