HDU - 2222 Keywords Search

关键词搜索与匹配算法实现

最新推荐文章于 2023-07-09 22:02:16 发布

原创最新推荐文章于 2023-07-09 22:02:16 发布 · 398 阅读

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本文探讨了如何通过关键词搜索算法实现图像检索系统的关键词匹配功能。详细解释了输入格式、输出要求以及具体实现过程，包括使用AC自动机进行关键词匹配，并提供了示例输入输出解析。

Keywords Search

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

[Submit] [Go Back] [Status]

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

    
      1
5
she
he
say
shr
her
yasherhs

Sample Output

Source

简单的AC自动机（待解决）

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 500010;
char str[1000010];
struct ACAutomaton{
int ch[maxn][26],fail[maxn],val[maxn],last[maxn],sz,root;
int newnode(){
memset(ch[sz],0,sizeof(ch[sz]));
val[sz] = 0;
return sz++;
}

void init(){
sz = 0;
root = newnode();
}

void insert(char *str){
int len = strlen(str);
int now = root;
for(int i = 0;i < len;i++){
int &tmp = ch[now][str[i]-'a'];
if(!tmp) tmp = newnode();
now = tmp;
}
val[now]++;
}

void getfail(){
queue<int> q;
fail[root] = root;
for(int i = 0;i < 26;i++){
int u = ch[root][i];
if(u){
fail[u] = last[u] = 0;
q.push(u);
}
}
while(!q.empty()){
int now = q.front();q.pop();
for(int i = 0;i < 26;i++){
int u = ch[now][i];
if(!u) ch[now][i] = ch[fail[now]][i];
else{
fail[u] = ch[fail[now]][i];
last[u] = val[fail[u]] ? fail[u]:last[fail[u]];
q.push(u);
}
}
}
}

int query(char *str){
int len = strlen(str);
int now = root;
int ret = 0;
for(int i = 0;i < len;i++){
now = ch[now][str[i]-'a'];
int tmp = now;
while(tmp != root && val[tmp]){
ret += val[tmp];
val[tmp] = 0;
tmp = last[tmp];
}
}
return ret;
}
}ac;

int main(){
int kase,n;
scanf("%d",&kase);
while(kase--){
ac.init();
scanf("%d",&n);
for(int i = 0;i < n;i++){
scanf("%s",str);
ac.insert(str);
}
ac.getfail();
scanf("%s",str);
printf("%d\n",ac.query(str));
}
return 0;
}