hdu 4287 Intelligent IME

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

3 5

64448

night

might

Sample Output

Source

2012 ACM/ICPC Asia Regional Tianjin Online

题解：手机九宫格，字母与对应的数字.... 自己过了案例，但是还是WA，之后才知道自己写的代码不能判断相等的情况....srO

WA代码：

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <math.h>
  4 #include <algorithm>
  5 #include <iostream>
  6 #include <ctype.h>
  7 #include <iomanip>
  8 #include <queue>
  9 #include <stdlib.h>
 10 using namespace std;
 11 
 12 
 13 struct node
 14 {
 15     int count;     
 16     node *next[200];    
 17     node(){          //构造函数
 18         count=0;
 19         memset(next,0,sizeof(next));
 20     }
 21 };
 22 node *root;
 23 int k=0;
 24 void insert(char *a)
 25 {
 26     int l=strlen(a);
 27     node *p=root;
 28     int i;
 29     for(i=0;i<l;i++)
 30     {
 31         if(p->next[a[i]-'a']==0)
 32         {
 33             p->next[a[i]-'a']=new node;
 34             if(a[i]<'a'&&a[i]>'z')
 35                p->next[a[i]-'a']->count=p->count+0;
 36             if(a[i]>='a'&&a[i]<='c')
 37                p->next[a[i]-'a']->count+=p->count+2;
 38             if(a[i]>='d'&&a[i]<='f')
 39                p->next[a[i]-'a']->count+=p->count+3;
 40             if(a[i]>='g'&&a[i]<='i')
 41                p->next[a[i]-'a']->count+=p->count+4;
 42             if(a[i]>='j'&&a[i]<='l')
 43                p->next[a[i]-'a']->count+=p->count+5;
 44             if(a[i]>='m'&&a[i]<='o')
 45                p->next[a[i]-'a']->count+=p->count+6;
 46             if(a[i]>='p'&&a[i]<='s')
 47                p->next[a[i]-'a']->count+=p->count+7;
 48             if(a[i]>='t'&&a[i]<='v')
 49                p->next[a[i]-'a']->count+=p->count+8;
 50             if(a[i]>='w'&&a[i]<='z')
 51                p->next[a[i]-'a']->count+=p->count+9;
 52         }    
 53         // 已存在此前缀
 54         p=p->next[a[i]-'a'];
 55     }
 56 }
 57 int find(char *s)
 58 {
 59     struct node *p;
 60     int len=strlen(s);
 61     if(len==0) return 0;
 62     p=root;
 63     for(int i=0;i<len;i++){
 64         if(p->next[s[i]-'a']!=0)
 65             p=p->next[s[i]-'a'];
 66         else
 67             return 0;
 68     }
 69     return p->count;
 70 }
 71 void de(node *p)
 72 {
 73     if(p==0)
 74         return ;
 75     int i;
 76     for(i=0;i<26;i++)
 77     {
 78         de(p->next[i]);
 79     }
 80     delete p;
 81 
 82 }
 83 int main()
 84 {
 85     int t;
 86     scanf("%d",&t);
 87     char a[30];
 88     int b[5050],s[5050];
 89     while(t--)
 90     {
 91         root = new node;
 92         memset(b,0,sizeof(b));
 93         memset(s,0,sizeof(s));
 94         memset(a,'',sizeof(a));
 95         int n,m,p,x,j,sum=0;
 96         k=0;
 97         scanf("%d%d",&n,&m);
 98         int i;
 99         for(i=0;i<n;i++){
100             scanf("%d",&p);
101             while(p!=0){
102                 x=p%10;
103                 p/=10;
104                 k+=x;
105             }
106             b[i]=k;
107             k=0;
108         }
109         /*for (i = 0; i < n; ++i)
110         {
111             cout<<b[i]<<endl;
112         }*/
113         for(i=0;i<m;i++)
114         {
115             scanf("%s",a);
116             insert(a);
117             s[i]=find(a);
118         }
119         /*for (i = 0; i < m; ++i)
120         {
121             cout<<s[i]<<endl;
122         }*/
123         for (i = 0; i < n; ++i)
124         {
125             for(j=0;j<m;j++){
126                 if(b[i]==s[j]){
127                     sum++;
128                 }
129             }
130             printf("%d
",sum);
131             sum=0;
132         }
133         de(root);
134     }
135     return 0;
136 }

View Code

玉民代码：

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <string>
 4 #include <stdio.h>
 5 #include <cstring> 
 6 using namespace std;
 7 #define Max 13
 8 struct dot
 9 {
10     dot *next[Max];
11     int flag;
12 } ;
13 dot *newnode()
14 {
15     dot *temp=new dot;
16          temp->flag=0;
17     for(int i=0;i<Max;i++)
18        temp->next[i]=NULL;
19     return temp;
20 }
21 void tree(char *st,dot *root)
22 {
23     dot *p=root;
24     int id=0;
25     int len=strlen(st);
26     for(int i=0;i<len;i++)
27     {
28         if(st[i]>='a'&&st[i]<='c')
29           id=2;
30         else if(st[i]>='d'&&st[i]<='f')
31           id=3;
32         else if(st[i]>='g'&&st[i]<='i')
33           id=4;
34         else if(st[i]>='j'&&st[i]<='l')
35           id=5;
36         else if(st[i]>='m'&&st[i]<='o')
37           id=6;
38         else if(st[i]>='p'&&st[i]<='s')
39           id=7;
40         else if(st[i]>='t'&&st[i]<='v')
41           id=8;
42         else if(st[i]>='w'&&st[i]<='z')
43           id=9;
44         if(p->next[id]==NULL)
45             p->next[id]=newnode();
46         p=p->next[id];
47     }
48     p->flag++;
49 }
50 int find(char *st,dot *root)
51 {
52     int id;
53     dot *p=root;
54     for(int i=0;i<strlen(st);i++)
55     {
56         id=st[i]-'0';
57         if(p->next[id]==NULL)
58             return 0;
59         p=p->next[id];
60     }
61     return p->flag;
62 }
63 void del(dot *t)
64 {
65     if(t==NULL) return ;
66     for(int i=0;i<Max;i++)
67        if(t->next[i]==NULL)
68           del(t->next[i]);
69     delete t;
70 }
71 int main()
72 {
73     int n,m,i,j,k,t;
74     cin>>t;
75     char s[5005][20],st[30];
76     while(t--)
77     {
78         cin>>n>>m;
79         dot *root=new dot;
80            root=newnode();
81         for(i=0;i<n;i++)
82            cin>>s[i];
83         while(m--)
84         {
85             cin>>st;
86             tree(st,root);
87         }
88         for(i=0;i<n;i++)
89           cout<<find(s[i],root)<<endl;
90         del(root);
91     }
92     return 0;
93 }

View Code