Intelligent IME

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2479 Accepted Submission(s): 1212

Problem Description

　　We all use cell phone today. And we must be familiar with the intelligent English input method on the cell phone. To be specific, the number buttons may correspond to some English letters respectively, as shown below:

　　2 : a, b, c 3 : d, e, f 4 : g, h, i 5 : j, k, l 6 : m, n, o

　　7 : p, q, r, s 8 : t, u, v 9 : w, x, y, z

　　When we want to input the word “wing”, we press the button 9, 4, 6, 4, then the input method will choose from an embedded dictionary, all words matching the input number sequence, such as “wing”, “whoi”, “zhog”. Here comes our question, given a dictionary, how many words in it match some input number sequences?

Input

　　First is an integer T, indicating the number of test cases. Then T block follows, each of which is formatted like this:

　　Two integer N (1 <= N <= 5000), M (1 <= M <= 5000), indicating the number of input number sequences and the number of words in the dictionary, respectively. Then comes N lines, each line contains a number sequence, consisting of no more than 6 digits. Then comes M lines, each line contains a letter string, consisting of no more than 6 lower letters. It is guaranteed that there are neither duplicated number sequences nor duplicated words.

Output

　　For each input block, output N integers, indicating how many words in the dictionary match the corresponding number sequence, each integer per line.

Sample Input

       1 3 5 46 64448 74 go in night might gn 

Sample Output

       3 2 0 

       题意 ：在手机键盘的背景下，给出一串数字，然后由这些数字能够枚举出一系列的字符串，然后给出一个字典，问这些字符串一共同拥有几个在字典中，逆向思考一下。一開始我是考虑着由给出的数字枚举出全部的可能的字符串然后在字典中一一查找，但时间复杂度太高了，后来看到题解才明确，字符串是能够转换成数字的。并且唯一相应，然后哈希一下就能够了。 

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                   #define L long longusing namespace std;const int INF=0x3f3f3f3f;const int maxn=1000020;int n,m,num[maxn],h[maxn];char phone[]="22233344455566677778889999";char word[5010][7];int Binary_search(int x){ int mid,low=0,high=n-1; while(low<=high) { mid=(low+high)/2; if(num[mid]==x) return mid; else if(num[mid]>x) high=mid-1; else if(num[mid]

pos(num,num+n); sort(num,num+n); for(int i=0;i

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