HDU 3695 Computer Virus on Planet Pandora （AC自动机）

题意：有n种病毒序列（字符串），一个模式串，问这个字符串包含几种病毒。

包含相反的病毒也算，字符串中[qx]表示有q个x字符。详细见案列。

0 < q <= 5,000,000尽然不会超，无解

32ABDCBDACB3ABCCDEGHIABCCDEFIHG4ABBACDEEBBBFEEEA[2B]CD[4E]F

 

Sample Output

032
Hint
In the second case in the sample input, the reverse of the program is ‘GHIFEDCCBA’, and ‘GHI’ is a substring of the reverse, so the program is infected by virus ‘GHI’.

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<string>#include<queue>#include<math.h>#include<algorithm>#include<iostream>using namespace std;const int kind = 26;const int maxn = 250*1000; //注意RE,单词长度*单词个数const int M = 5100000;struct node{    node *fail;    node *next[kind];    int count;    node()    {        fail = NULL;        count = 0;        memset(next,0,sizeof(next));    }}*q[maxn];char keyword[1010],str[M],str1[M];int head,tail;void insert(char *str,node *root){    node *p=root;    int i=0,index;    while(str[i])    {        index = str[i] - 'A';        if(p->next[index]==NULL)            p->next[index] = new node();        p = p->next[index];        i++;    }    p->count++;}void build_ac(node *root){    int i;    root->fail=NULL;    q[head++]=root;    while(head!=tail)    {        node *temp = q[tail++];        node *p=NULL;        for(i=0;i<26;i++)        {            if(temp->next[i]!=NULL)            {                if(temp==root)                        temp->next[i]->fail=root;//失败指针指向root                else                 {                    p = temp->fail;                    while(p!=NULL)                    {                        if(p->next[i]!=NULL)                        {                            temp->next[i]->fail=p->next[i];                            break;                        }                        p=p->fail;                    }                    if(p==NULL)                        temp->next[i]->fail=root;                }                q[head++]=temp->next[i];            }        }    }}int query(node *root){    int i=0,cnt=0,index,len=strlen(str);    node *p=root;    while(str[i])    {        index = str[i]-'A';        while(p->next[index]==NULL&&p!=root)            p = p->fail;        p=p->next[index];        p=(p==NULL)?root:p;        node *temp = p;        while(temp!=root&&temp->count!=-1)//沿失配边走，走过的不走{            cnt+=temp->count;            temp->count=-1;            temp=temp->fail;        }        i++;    }    return cnt;}int value(int p,int q)  {      int i,ans=0,w=1;      for (i=q;i>=p;i--)      {          ans+=(str1[i]-'0')*w;          w*=10;      }        return ans;  }  int main(){    int n,t;    scanf("%d",&t);    while(t--)    {        head=tail=0;        node *root = new node();        scanf("%d",&n);        while(n--)        {            scanf("%s",keyword);            insert(keyword,root);        }        build_ac(root);        scanf("%s",str1);int l=strlen(str1),i,j,k;            j=0;          for (i=0;i<l;)          {              if (str1[i]!='[') str[j++]=str1[i++];              else              {                  /*for (k=i+1;str1[k]!=']';k++);  int v=0,w=1;for (int l=k-2;l>=i+1;l--)  {  v+=(str1[l]-'0')*w;  w*=10;  }  */int v=0;i++;while(str1[i]>='0'&&str1[i]<='9'){v=v*10+str1[i]-'0';i++;}                for (int k1=1;k1<=v;k1++) str[j++]=str1[i];                    i+=2;             }          }          str[j]='\0';          //printf("%s\n",str);            int h=query(root);          char chh;            l=strlen(str);          for (i=0;i<=(l-1)/2;i++)          {              chh=str[l-i-1];              str[l-i-1]=str[i];              str[i]=chh;          }          //printf("%s",str);          h+=query(root);          printf("%d\n",h);      }    return 0;}/*32ABDCBDACB3ABCCDEGHIABCCDEFIHG4ABBACDEEBBBFEEEA[2B]CD[4E]F*/