hdu-Danganronpa(AC自动机)

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
   
   
   
   

 

wange2014   |   We have carefully selected several similar problems for you:   5421  5420  5419  5418  5417


最简单的AC自动机题型，关于AC自动机详解见： http://blog.csdn.net/niushuai666/article/details/7002823  http://www.cppblog.com/menjitianya/archive/2014/07/10/207604.html

#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<map>
#include<set>
#include<time.h>
#include<vector>
#include<queue>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long
#define rd(a) scanf("%d",&a)
#define rdLL(a) scanf("%I64d",&a)
#define rdd(a,b) scanf("%d%d",&a,&b)
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
typedef pair<int , int> P;
#define MOD 1000000007
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,1,sizeof(a))
#define Max 500010

struct Trie{
    int next[Max][26],fail[Max],end[Max];
    int root,L;

    int newnode()
    {
        memset(next[L],-1,sizeof(next[L]));  ///初始化根节点没有一个子节点
        end[L++]=0;
        return L-1;
    }

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

    void insert(char buf[])  ///建树
    {
        int len = strlen(buf) , now = root;
        for(int i= 0; i<len; i++)
        {
            if(next[now][  buf[i]-'a'  ] == -1)
                next[now][buf[i]-'a'] = newnode();
            now = next[now][buf[i]-'a'];
        }
      end[now]++;
    }

    void build()  ///bfs建立fail指针
    {
        queue<int>Q;
        fail[root] = root;
        for(int i=0 ; i<26 ; i++)  ///初始化根节点子节点信息
            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++)  ///遍历子节点
                if(next[now][i] == -1)
                    next[now][i] = next[   fail[now]   ][i];
                else
                {
                    fail[   next[now][i]  ] = next[   fail[now]  ][i];  ///失败的指针指向该子节点父亲节点fail位置的第i个孩子
                    Q.push(  next[now][i]  );
                }
        }
    }
    int query (char buf[])  ///当前字符串中有哪些部分在树中
    {
        int len = strlen(buf) , now = root , sum = 0;
        for(int i=0 ; i<len ; i++)
        {
            now = next[now][buf[i]-'a'];

            int temp = now;
            while(  temp   !=  root  )
            {
                sum += end[temp];
                ///end[temp] = 0 ; ///查看出现过读多少串
                temp = fail[temp];
            }
        }
        return sum;
    }
};


char buf[Max*2];
Trie ac;

int main()
{
    int T;
    int n,m;
    scanf("%d",&T);
    while(T--)
    {
        string str[100005];
        scanf("%d%d",&m,&n);
        for(int i=0;i<m;i++){
            cin>>str[i];
        }
        ac.init();
        for(int i=0 ; i<n ; i++)
        {
            scanf("%s",buf);
            ac.insert(buf);
        }
        ac.build();
        for(int i = 0 ; i<m ; i++){
          strcpy(buf, str[i].c_str() );
          printf("%d\n",ac.query(buf));
        }
    }
    return 0;
}