Keywords Search
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29818 Accepted Submission(s): 9724
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.
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.
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
1 5 she he say shr her yasherhs
Sample Output
3
ac自动机模板题
代码：
/* ***********************************************
Author :xianxingwuguan
Created Time :2014-2-2 16:23:08
File Name :2.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
struct Trie {
int next[600100][26],fail[600020],L,end[600100],root;
int newnode(){
for(int i=0;i<26;i++)
next[L][i]=-1;
end[L++]=0;
return L-1;
}
void init(){
L=0;
root=newnode();
}
void insert(char *str){
int len=strlen(str);
int now=root;
for(int i=0;i<len;i++){
if(next[now][str[i]-‘a‘]==-1)
next[now][str[i]-‘a‘]=newnode();
now=next[now][str[i]-‘a‘];
}
end[now]++;
}
void build(){
queue<int> q;
fail[root]=root;
for(int i=0;i<26;i++)
if(next[root][i]==-1)
next[root][i]=root;
else {
fail[next[root][i]]=root;
q.push(next[root][i]);
}
while(!q.empty()){
int now=q.front();
q.pop();
for(int i=0;i<26;i++)
if(next[now][i]==-1)
next[now][i]=next[fail[now]][i];
else {
fail[next[now][i]]=next[fail[now]][i];
q.push(next[now][i]);
}
}
}
int solve(char *str){
int len=strlen(str);
int now=root;
int res=0;
for(int i=0;i<len;i++){
now=next[now][str[i]-‘a‘];
int temp=now;
while(temp!=root&&end[temp]!=-1){
res+=end[temp];
end[temp]=-1;
temp=fail[temp];
}
}
return res;
}
}ac;
char str[1000100];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
scanf("%d",&T);
while(T--){
int n;
scanf("%d",&n);
ac.init();
while(n--){
scanf("%s",str);
ac.insert(str);
}
ac.build();
scanf("%s",str);
printf("%d\n",ac.solve(str));
}
return 0;
}
/*
5 she he say shr her yasherhshersher
*/