暑假集训第三周周三赛 STL G - Intelligent IME

G - Intelligent IME

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 4287

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
      
      
      
      

分析：

这题用到map，最大的优点，他像字典一样是一一对应的关系，可以进行翻译转换

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#include<stdio.h>
#include<map>
#include<iostream>
#include<algorithm>
using namespace std;
char  v[5050]= {2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,9,9,9,9};
int main()
{
    int m,n,i,j,t,k,l;
    map<int,int> p;
    
    int  a[5050];
    scanf("%d",&t);
    while(t--)
    {
        p.clear();
        m=n=0;
        scanf("%d %d",&n,&m);
        for(i=0; i<n; i++)
            cin>>a[i];
        string te;
        for(i=0; i<m; i++)
        {
            cin>>te;
            l=te.length();
            j=0;
            for(k=0; k<l; k++)
                j=j*10+v[te[k]-'a'];
            p[j]++;
        }
        for(i=0; i<n; i++)
            cout<<p[a[i]]<<endl;
    }
return 0;

}