HDU - 4287 Intelligent IME 字典树

HDU - 4287

Intelligent IME

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

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 
     
     
     
     

 

/*
Author: 2486
Memory: 3288 KB		Time: 93 MS
Language: G++		Result: Accepted
*/
//大家看懂题目这份代码基本就明白了
//很简单，先将数字进行字典树处理，然后通过一个s记录有多少个符合条件，最后用数字输出所有的结果
#include <cstdio>
#include <vector>
#include <string>
#include <iostream>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAXN = 5000 + 5;
const int MAXM = 1000000 + 5;
struct node {
    int v,s,next[30];
    void init() {
        v = -1;
        s = 0;
        memset(next, -1, sizeof(next));
    }
} L[MAXM];

int N, M, T, tot;
char str[MAXN][10],op[10];


int Get_Num(char ch) {
    if(ch >= 'a' && ch <= 'c')return 2;
    if(ch >= 'd' && ch <= 'f')return 3;
    if(ch >= 'g' && ch <= 'i')return 4;
    if(ch >= 'j' && ch <= 'l')return 5;
    if(ch >= 'm' && ch <= 'o')return 6;
    if(ch >= 'p' && ch <= 's')return 7;
    if(ch >= 't' && ch <= 'v')return 8;
    if(ch >= 'w' && ch <= 'z')return 9;
}

int add(char * a, int len) {
    int now = 0;
    for(int i = 0; i < len; i ++) {
        int tmp = a[i] - '0';
        int next = L[now].next[tmp];
        if(next == -1) {
            next = ++ tot;
            L[next].init();
            L[now].next[tmp] = next;
        }
        now = next;
    }
    L[now].v = 0;
    return L[now].s;
}


void query(char * a, int len) {
    int now = 0;
    for(int i = 0; i < len; i ++) {
        int tmp = Get_Num(a[i]);
        int next = L[now].next[tmp];
        if(next == -1)return;
        now = next;
    }
    if(L[now].v == 0)
        L[now].s ++;
}

int main() {
    scanf("%d", &T);
    while(T --) {
        tot = 0;
        L[0].init();
        scanf("%d%d", &N, &M);
        for(int i = 0; i < N; i ++) {
            scanf("%s", str[i]);
            add(str[i], strlen(str[i]));
        }
        for(int i = 0 ; i < M; i ++) {
            scanf("%s", op);
            query(op, strlen(op));
        }
        for(int i = 0; i < N ; i ++) {
            printf("%d\n", add(str[i], strlen(str[i])));
        }
    }
    return 0;
}