hdu 2222 Keywords Search //ac自动机

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7476    Accepted Submission(s): 2541

Problem Description

In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.

 

Input

First line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.

 

Output

Print how many keywords are contained in the description.

 

Sample Input

15shehesayshrheryasherhs

 

Sample Output

3

 

Author

Wiskey

 

Recommend

lcy

 

//模板题#include<cstdio>#include<cstring>const int NODE=10000*26;const int CH=26;int tree[NODE][CH],cnt;int word[NODE];int fail[NODE];int Que[NODE];int num[128];char str[100];//题目要求char s[1000001];void Ins(char *a,int val){ int p=0; for(;*a;a++) { int c=num[*a]; if(!tree[p][c]) { memset(tree[cnt],0,CH*sizeof(int));//初始化新的节点 word[cnt]=0; tree[p][c]=cnt++; } p=tree[p][c]; } word[p]+=val;//这一点要看题目要求所定}void AC(){ int *s=Que,*e=Que; for(int i=0;i<CH;++i) if(tree[0][i]) { fail[tree[0][i]]=0; *e++=tree[0][i]; } while(s!=e) { int p=*s++; for(int i=0;i<CH;++i) { if(tree[p][i]) { int v=tree[p][i]; *e++=v; fail[v]=tree[fail[p]][i]; } else { tree[p][i]=tree[fail[p]][i]; } } }}int query(char *a){ int ret=0,p=0; for(int c=num[*a];*a;a++,c=num[*a]) { p=tree[p][c];//因为有虚拟儿子这一更改，所以p就不用再怕找不到fail指针所指的节点了 int pp=p; while(pp&&word[pp]!=-1) { ret+=word[pp]; word[pp]=-1; pp=fail[pp]; } } return ret;}int main(){ int n,t; scanf("%d",&t); for(int i=0;i<26;i++) num[i+'a']=i; while(t--) { scanf("%d",&n); fail[0]=0; memset(tree[0],0,CH*sizeof(int));//初始化根节点 cnt=1;//已经有根节点 for(int i=0;i<n;++i) { scanf("%s",str); Ins(str,1); } AC(); scanf("%s",s); printf("%d/n",query(s)); } return 0;}