Keywords Search
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
PS: 很久之前就想开AC自动机，一直没狠下心来去搞，今天晚上虽说昏昏沉沉的度过，不过还是慢慢的啃懂了AC自动机的最简单用法。用一道AC自动机比较经典的例题来当模板吧，还有如果你想很快的学会的话推荐一个博客【
AC自动机】去认真的看，相信你也很快就会懂哒！不过要提前学会字典树和KMP噢。
回归正题：
【例题题意】
题意很简单，首先给定n个长度不超过50的要查找字符串，最后一行给出一个模式串。问在模式串中出现了多少个待查找字符串，有坑点是这n个字符串中可能会有重复的。
【解决方法】
盲目的跑n遍KMP不知道可行不可行，不过算了一下极限复杂度是绝对要炸的。
那么这时候就用到AC自动机了，AC自动机用通俗的话来说就是多个数据结构放在了一起，这次是KMP+字典树而已，如果你已经学会了KMP，相信你理解AC自动机中使用BFS建立每个节点的fail指针和最后查找字符串的时候就不会那么难了。
具体的做法，我们先把给定的n个待查字符串以字典树的形式存储下来，以下是建立字典树的过程。
//建立字典树
char str[1000100];
struct node{
int count;
node *next[26];
node *fail;
void init(){ //初始化函数，这样写会节省大量代码量
count = 0; //因为n个待查字符串会有重复的，所以我们设立一个count来统计一下数量
fail = NULL;
for(int i=0;i<26;i++)
next[i] = NULL;
}
}*root;
void insert(char *str){ //传进来的str是每个待查找字符串
int len = strlen(str);
node *p = root;
for(int i=0;i<len;i++){
int pos = str[i] - 'a';
if(p->next[pos] == NULL){
p->next[pos] = new node;
p->next[pos]->init();
p = p->next[pos];
}
else
p = p->next[pos];
}
p->count++;
}//构造失败指针(fail指针)
//其实这个过程很简单，我们归结为一句话
//设这个节点上的字母为C，沿着他父亲的失败指针走，直到走到一个特殊的节点，
//恰好这个特殊的节点的儿子中也有字母为C的节点。然后把当前节点的失败指针指向那个字母也
//为C的儿子。如果一直走到了root都没找到，那就把失败指针指向root
void Getfail(){
node *p = root,*son,*temp;
queue<node*>Q; //节点队列
Q.push(p);
while(!Q.empty()){
temp = Q.front();Q.pop();
for(int i=0;i<26;i++){
son = temp->next[i];
if(son != NULL){
if(temp == root)
son->fail = root;
else{
p = temp->fail;
while(p){
if(p->next[i]){
son->fail = p->next[i];
break;
}
p = p->fail;
}
if(p == NULL) //一直找到最后为NULL了还没找到，就直接指向root
son->fail = root;
}
Q.push(son); //每更新了一个节点，把这个节点存到队列里，方便去更新它的子孙节点
}
}
}
}//查找模式串
int cal_ans(){
//Getfail();
int len = strlen(str),num = 0;
node *p,*temp;
p = root;
for(int i=0;i<len;i++){ //遍历模式串
int pos = str[i] - 'a';
while(!p->next[pos] && p!=root) //将fail指针指到最后有特殊的节点的节点
p = p->fail;
p = p->next[pos];
if(p == NULL)
p = root;
temp = p;
while(temp!=root){
if(temp->count>0){ //如果count大于0，更新答案，并删掉count值
num += temp->count;
temp->count = 0;
}
else
break;
temp = temp->fail; //遍历他所有的失败节点，即以当前字母作为结尾的所有答案
}
}
return num;
}
AC Code
#include <algorithm>
#include <iostream>
#include <numeric>
#include <cstring>
#include <iomanip>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const LL M = 100099;
const double esp = 1e-6;
const double PI = 3.14159265359;
const LL INF = 0x3f3f3f3f;
using namespace std;
//建立字典树
char str[1000100];
struct node{
int count;
node *next[26];
node *fail;
void init(){ //初始化函数，这样写会节省大量代码量
count = 0; //因为n个待查字符串会有重复的，所以我们设立一个count来统计一下数量
fail = NULL;
for(int i=0;i<26;i++)
next[i] = NULL;
}
}*root;
void insert(char *str){ //传进来的str是每个待查找字符串
int len = strlen(str);
node *p = root;
for(int i=0;i<len;i++){
int pos = str[i] - 'a';
if(p->next[pos] == NULL){
p->next[pos] = new node;
p->next[pos]->init();
p = p->next[pos];
}
else
p = p->next[pos];
}
p->count++;
}
//构造失败指针(fail指针)
//其实这个过程很简单，我们归结为一句话
//设这个节点上的字母为C，沿着他父亲的失败指针走，直到走到一个特殊的节点，
//恰好这个特殊的节点的儿子中也有字母为C的节点。然后把当前节点的失败指针指向那个字母也
//为C的儿子。如果一直走到了root都没找到，那就把失败指针指向root
void Getfail(){
node *p = root,*son,*temp;
queue<node*>Q; //节点队列
Q.push(p);
while(!Q.empty()){
temp = Q.front();Q.pop();
for(int i=0;i<26;i++){
son = temp->next[i];
if(son != NULL){
if(temp == root)
son->fail = root;
else{
p = temp->fail;
while(p){
if(p->next[i]){
son->fail = p->next[i];
break;
}
p = p->fail;
}
if(p == NULL) //一直找到最后为NULL了还没找到，就直接指向root
son->fail = root;
}
Q.push(son); //每更新了一个节点，把这个节点存到队列里，方便去更新它的子孙节点
}
}
}
}
//查找模式串
int cal_ans(){
Getfail();
int len = strlen(str),num = 0;
node *p,*temp;
p = root;
for(int i=0;i<len;i++){ //遍历模式串
int pos = str[i] - 'a';
while(!p->next[pos] && p!=root) //将fail指针指到最后有特殊的节点的节点
p = p->fail;
p = p->next[pos];
if(p == NULL)
p = root;
temp = p;
while(temp!=root){
if(temp->count>0){ //如果count大于0，更新答案，并删掉count值
num += temp->count;
temp->count = 0;
}
else
break;
temp = temp->fail; //遍历他所有的失败节点，即以当前字母作为结尾的所有答案
}
}
return num;
}
int main(){
int t,n;
scanf("%d",&t);
while(t--){
root = new node;
root->init();
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%s",str);
insert(str);
}
scanf("%s",str);
printf("%d\n",cal_ans());
}
return 0;
}
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