hdu5384 AC自动机模板题，统计模式串在给定串中出现的个数

http://acm.hdu.edu.cn/showproblem.php?pid=5384

Problem Description

Danganronpa is a video game franchise created and developed by Spike Chunsoft, the series' name is compounded from the Japanese words for "bullet" (dangan) and "refutation" (ronpa).

Now, Stilwell is playing this game. There are  n  verbal evidences, and Stilwell has  m  "bullets". Stilwell will use these bullets to shoot every verbal evidence.

Verbal evidences will be described as some strings  Ai , and bullets are some strings  Bj . The damage to verbal evidence  Ai  from the bullet  Bj  is  f(Ai,Bj) .

f(A,B)=∑i=1|A|−|B|+1[ A[i...i+|B|−1]=B ]

In other words,  f(A,B)  is equal to the times that string  B  appears as a substring in string  A .
For example:  f(ababa,ab)=2 ,  f(ccccc,cc)=4

Stilwell wants to calculate the total damage of each verbal evidence  Ai  after shooting all  m  bullets  Bj , in other words is  ∑mj=1f(Ai,Bj) .

 

Input

The first line of the input contains a single number  T , the number of test cases.
For each test case, the first line contains two integers  n ,  m .
Next  n  lines, each line contains a string  Ai , describing a verbal evidence.
Next  m  lines, each line contains a string  Bj , describing a bullet.

T≤10
For each test case,  n,m≤105 ,  1≤|Ai|,|Bj|≤104 ,  ∑|Ai|≤105 ,  ∑|Bj|≤105
For all test case,  ∑|Ai|≤6∗105 ,  ∑|Bj|≤6∗105 ,  Ai  and  Bj  consist of only lowercase English letters

 

Output

For each test case, output  n  lines, each line contains a integer describing the total damage of  Ai  from all  m  bullets,  ∑mj=1f(Ai,Bj) .

 

Sample Input

   
   
   
   

    
    
    
    1
5 6
orz
sto
kirigiri
danganronpa
ooooo
o
kyouko
dangan
ronpa
ooooo
ooooo
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
1
0
3
7
   
   
   
   
   
   
   
   


   
   
   
   

/**
hdu5384 AC自动机模板题，统计模式串在给定串中出现的个数
*/
#include<string.h>
#include<algorithm>
#include<iostream>
#include<stdio.h>
#include<queue>
using namespace std;
char str[100010][10010];
int num[100010],n,m;
struct Trie
{
    int next[10010*50][28],fail[10010*50],end[10010*50];
    int root,L;
    int newnode()
    {
        for(int i=0; i<26; i++)
        {
            next[L][i]=-1;
        }
        end[L++]=-1;
        return L-1;
    }
    void init()
    {
        L=0;
        root=newnode();
    }
    void insert(char *s)
    {
        int len=strlen(s);
        int now=root;
        for(int i=0; i<len; i++)
        {
            if(next[now][s[i]-'a']==-1)
            {
                next[now][s[i]-'a']=newnode();
            }
            now=next[now][s[i]-'a'];
        }
        if(end[now]==-1)///标记模式串出现的次数
        {
            end[now]=1;
        }
        else
        {
            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())
        {
           // printf("**\n");
            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]);
                }
            }
        }
    }
    void query(char* s)
    {
        memset(num,0,sizeof(num));
        int len=strlen(s);
        int now=root;
        for(int i=0; i<len; i++)
        {
            now=next[now][s[i]-'a'];
            int temp=now;
            while(temp!=root)
            {
                if(end[temp]!=-1)///统计所有模式串出现的次数，num数组在0~m之间定能取到所有end[temp]必不大于m
                {
                    num[end[temp]]+=end[temp];
                }
                temp=fail[temp];
            }
        }
        int ans=0;
        for(int i=0; i<=m; i++)
        {
            if(num[i]>0)
                ans+=num[i];
        }
        printf("%d\n",ans);
    }
} ac;
char s[10005];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0; i<n; i++)
        {
            scanf("%s",str[i]);
        }
        ac.init();
        for(int i=0; i<m; i++)
        {
            scanf("%s",s);
            ac.insert(s);
        }
        ac.build();
        //printf("**\n");
        for(int i=0; i<n; i++)
        {
            ac.query(str[i]);
        }
    }
    return 0;
}