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最新推荐文章于 2025-08-24 09:29:29 发布

转载最新推荐文章于 2025-08-24 09:29:29 发布 · 81 阅读

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原文链接：http://www.cnblogs.com/songorz/p/9386575.html

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#数据结构与算法

本文介绍了一道关于AC自动机的经典模板题，通过详细解释题目的要求和给出完整的代码实现，帮助读者理解AC自动机的工作原理及其实现细节。

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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

   1 5 she he say shr her yasherhs
  

Sample Output

3

题解：AC自动机的模版题~~；

代码为：

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#define MAXNODE 500005
using namespace std;
int n,T;
char s[1000005];
struct Aho-Corasick
{
int ch[MAXNODE][26];
int f[MAXNODE]; // fail函数
int val[MAXNODE]; // 是否为单词结尾
int last[MAXNODE]; // 后缀链接
int cnt[10005]; // 每个单词出现次数
int tot; // trie 单词总数
int num; // 单词出现了几个
int time[10005];

void init()
{
num = 0;
tot = 1;
memset(ch[0],0,sizeof(ch[0]));
memset(cnt,0,sizeof(cnt));
memset(time,0,sizeof(time));
}

int idx(char c) //获取编号
{
return c - 'a';
}

void insert(char *s,int v) //插入
{
int u = 0,n = strlen(s);
for(int i = 0;i < n;i++)
{
int c = idx(s[i]);
if(!ch[u][c])
{
memset(ch[tot],0,sizeof(ch[tot]));
val[tot] = 0;
ch[u][c] = tot++;
}
u = ch[u][c];
}
if(val[u])
time[val[u]]++;
else
val[u] = v,time[v] = 1;
}

void print(int i,int j)
{
if(j)
{
if(!cnt[val[j]])
num += time[val[j]];
cnt[val[j]]++;
print(i,last[j]);
}
}

void find(char *T)
{
int n = strlen(T);
int j = 0;
for(int i = 0;i < n;i++)
{
int c = idx(T[i]);
j = ch[j][c];
if(val[j]) print(i,j);
else if(last[j]) print(i,last[j]);
}
}

void getFail()
{
queue<int> q;
f[0] = 0;
for(int c = 0;c < 26;c++)
{
int u = ch[0][c];
if(u)
{
f[u] = 0;
q.push(u);
last[u] = 0;
}
}

while(!q.empty())
{
int r = q.front();
q.pop();
for(int c = 0;c < 26;c++)
{
int u = ch[r][c];
if(!u)
{
ch[r][c] = ch[f[r]][c];
continue;
}
q.push(u);
int v = f[r];
f[u] = ch[v][c];
last[u] = val[f[u]] ? f[u] : last[f[u]];
}
}
}
} tree;

int main()
{
cin>>T;
while(T--)
{
tree.init();
cin>>n;
for(int i = 1;i <= n;i++)
{
scanf("%s",s);
tree.insert(s,i);
}
tree.getFail();
scanf("%s",s);
tree.find(s);
cout<<tree.num<<endl;
}
}

Problem Description

Input

Output

Print how many keywords are contained in the description.

Sample Input