算法：字典树

题意：手机9键拼音：2：abc  3：def 4：ghi 5：jkl 6：mno 7：pqrs 8：tuv 9：wxyz；

第一行读入一个T，代表测试组数；

之后输入两个整数n和m，

有n个数字串和m个字符串

让你判断给定的数字串可以得到几种字符组合，在给定的字符串中；并输出相应的个数；

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

思路：用字符串在数字2-9之间建树，查找就方便多了；

代码：

#include <iostream>
#include <algorithm>
#include <string>
#include <stdio.h>
#include <cstring>
using namespace std;
#define Max 13
struct dot
{dot *next[Max];int flag;
} ;
dot *newnode()
{dot *temp=new dot;temp->flag=0;for(int i=0;i<Max;i++)temp->next[i]=NULL;return temp;
}
void tree(char *st,dot *root)
{dot *p=root;int id=0;int len=strlen(st);for(int i=0;i<len;i++){if(st[i]>='a'&&st[i]<='c')id=2;else if(st[i]>='d'&&st[i]<='f')id=3;else if(st[i]>='g'&&st[i]<='i')id=4;else if(st[i]>='j'&&st[i]<='l')id=5;else if(st[i]>='m'&&st[i]<='o')id=6;else if(st[i]>='p'&&st[i]<='s')id=7;else if(st[i]>='t'&&st[i]<='v')id=8;else if(st[i]>='w'&&st[i]<='z')id=9;if(p->next[id]==NULL)p->next[id]=newnode();p=p->next[id];}p->flag++;
}
int find(char *st,dot *root)
{int id;dot *p=root;for(int i=0;i<strlen(st);i++){id=st[i]-'0';if(p->next[id]==NULL)return 0;p=p->next[id];}return p->flag;
}
void del(dot *t)
{if(t==NULL) return ;for(int i=0;i<Max;i++)if(t->next[i]!=NULL)del(t->next[i]);delete t;
}
int main()
{int n,m,i,j,k,t;cin>>t;char s[5005][20],st[30];while(t--){cin>>n>>m;dot *root=new dot;root=newnode();for(i=0;i<n;i++)cin>>s[i];while(m--){cin>>st;tree(st,root);}for(i=0;i<n;i++)cout<<find(s[i],root)<<endl;del(root);}return 0;
}