Keywords Search
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 67950 Accepted Submission(s): 22882
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 | We have carefully selected several similar problems for you: 2896 3065 2243 2825 3341 题意是给出几个模式串，给一个匹配串，问能匹配多少个模式串。 代码如下：
//======================
// HDU 2222
// 求目标串中出现了几个模式串
//====================
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;
struct Trie
{
int Next[500010][26];//26是这里讨论26个小写字母的情况，根据情况修改
int fail[500010],end[500010];//end数组表示以该节点结尾的字符串的数量
int root,L;//L用来标记节点序号，以广度优先展开的字典树的序号
int newnode() //建立新节点
{
for(int i = 0;i < 26;i++)
Next[L][i] = -1; //将该节点的后继节点域初始化
end[L++] = 0;
return L-1; //返回当前节点编号
}
void init() //初始化操作
{
L = 0;
root = newnode();
}
void insert(char buf[])
{
int len = strlen(buf);
int now = root;
for(int i = 0;i < len;i++)
{
if(Next[now][buf[i]-'a'] == -1) //如果未建立当前的后继节点，建立新的节点
Next[now][buf[i]-'a'] = newnode();
now = Next[now][buf[i]-'a'];
}
end[now]++;//以该节点结尾的字符串数量增加1
}
void build()
{
queue<int>Q; //用广度优先的方式，将树层层展开
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];//该段的最后一个节点匹配后，跳到拥有最大公共后缀的fail节点继续匹配
else
{
fail[Next[now][i]]=Next[fail[now]][i];//当前节点的fail节点等于它前驱节点的fail节点的后继节点
Q.push(Next[now][i]);
}
}
}
int query(char buf[])
{
int len = strlen(buf);
int now = root;
int res = 0;
for(int i = 0;i < len;i++)
{
now = Next[now][buf[i]-'a'];
int temp = now;
while( temp != root )
{
res += end[temp];//加上以当前节点结尾的字符串数
end[temp] = 0;//该题是防止计算重复的字符串
temp = fail[temp];//每次找最大公共后缀对应的fail节点
}
}
return res;
}
void debug()
{
for(int i = 0;i < L;i++)
{
printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
for(int j = 0;j < 26;j++)
printf("%2d",Next[i][j]);
printf("]\n");
}
}
};
char buf[1000010];
Trie ac;
int main()
{
int t,n,ans;
scanf("%d",&t);
while(t--)
{
ac.init();
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%s",buf);
ac.insert(buf);
}
ac.build();
scanf("%s",buf);
printf("%d\n",ac.query(buf));
}
return 0;
}