HDU 4287 Intelligent IME hash

Intelligent IME

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

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

HINT

题意

用9宫格敲了N次，给m个字符串，问你每次敲击，能够敲出多少个单词

题解：

hash或者map搞一搞

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
const int maxn=;
#define mod 1000000009
#define eps 1e-7
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//**************************************************************************************

//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
map<int,int> HH;
string s[];
map<string,int> H;
int ans[];
int main()
{
    int k=;
    for(int i=;i<=;i++)
    {
        if(i==||i==||i==||i==||i==||i==||i==)
            k++;
        HH[i]=k;
    }
    int t=read();
    while(t--)
    {
        H.clear();
        memset(ans,,sizeof(ans));
        int n=read(),m=read();
        for(int i=;i<n;i++)
            cin>>s[i];
        string s1;
        for(int i=;i<m;i++)
        {
            cin>>s1;
            for(int j=;j<s1.size();j++)
                s1[j]=char(HH[(s1[j]-'a')]+'');
            H[s1]++;
        }
        for(int i=;i<n;i++)
            printf("%d\n",H[s[i]]);
    }
}