Intelligent IME

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2681 Accepted Submission(s): 1322


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

Source 2012 ACM/ICPC Asia Regional Tianjin Online
題意：手機英文輸入法九宮格鍵盤中，2~9數字各自代表了自己的字母，求給出的幾個數字串分別能夠打出下列幾個英文單詞。
分析：一開始我是建立給出的單詞的字典樹，然後dfs數字串，枚舉數字串能夠組成的單詞數，這樣毫無疑問地TLE了。後來換了種方法，建立給出數字串的字典樹，然後把單
詞串hash為對應的的數字串，再進行查詢。
題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=4287
AC代碼：

#include
#include
#include
#include
using namespace std;
const int MAX = 10 ;
const int maxn = 5005 ;
struct trie{
    int first;
    trie *next[MAX];
};

trie *root;

int T,N,M;
char str[7],ss;
int ans[maxn],slen;
void createTrie(char *s,int n){
    trie *p=root,*q;
    int len=strlen(s),pos;
    for(int i=0;inext[pos]==NULL){
            q=new trie;
            q->first=0;
            for(int j=0;jnext[j]=NULL;
            p->next[pos]=q;
            p=p->next[pos];
        }
        else{
            p=p->next[pos];
        }
    }
    p->first=n;
}

int findTrie(char *s){
    trie *p=root;
    int len=strlen(s),pos;
    for(int i=0;inext[pos]==NULL)
            return 0;
        p=p->next[pos];
    }
    return p->first;
}

void delTrie(trie *Root){
    for(int i=0;inext[i]!=NULL)
            delTrie(Root->next[i]);
    }
    free(Root);
}

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

int main(){
    scanf("%d",&T);
    while(T--){
        root=new trie;
        for(int i=0;inext[i]=NULL;
        memset(ans,0,sizeof(ans));
        memset(str,'\0',sizeof(str));
        scanf("%d%d",&N,&M);
        for(int i=1;i<=N;i++){
            scanf("%s",str);
            createTrie(str,i);
        }
        for(int i=0;i

TLE代碼：（trie+dfs）
#include
#include
#include
#include
using namespace std;
const int MAX = 28;
const int maxn = 5005;
char str[10][4]={{'a','b','c'},{'d','e','f'},{'g','h','i'},{'j','k','l'},
                {'m','n','o'},{'p','q','r','s'},{'t','u','v'},{'w','x','y','z'}};
int ank[8]={3,3,3,3,3,4,3,4};

struct trie{
    bool point;
    trie *next[MAX];
};

struct edge{
    int n;
    int x[6];
}num[maxn];

trie *root;
char s[7];
int T,N,M,slen,nn,sum;
int vis[10][7];
void createTrie(char *s){
    trie *p=root,*q;
    int len=strlen(s),pos;
    for(int i=0;inext[pos]==NULL){
            q=new trie;
            q->point=false;
            for(int j=0;jnext[j]=NULL;
            p->next[pos]=q;
            p=p->next[pos];
        }
        else{
            p=p->next[pos];
        }
    }
    p->point=true;
}

bool findTrie(char *s){
    trie *p=root;
    int len=strlen(s),pos;
    for(int i=0;inext[pos]==NULL)
            return false;
        p=p->next[pos];
    }
    return p->point;
}

void delTrie(trie *Root){
    for(int i=0;inext[i]!=NULL)
            delTrie(Root->next[i]);
    }
    free(Root);
}

void findAns(int a[],int k,int n,char *s){
    if(k==n){
        //printf("%s  ",s);
        if(findTrie(s)){
            sum++;
        }return ;
    }
    int kk=a[k]-2;
    for(int i=0;inext[i]=NULL;
        scanf("%d%d",&N,&M);
        for(int i=0;i