POJ_2503 Babelfish 字典树

题目链接:http://poj.org/problem?id=2503

依旧字典树

注意新建节点初始化值,读取数据的时候可以用gets判空行,最后删除分配的空间防RE

评论区还有用map过的,比较耗时。

代码:

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <iostream>
 4 #include <cstring>
 5 #include <string>
 6 using namespace std;
 7 #define maxnode 27
 8 #define maxn 100005
 9 struct trie{
10     trie *next[maxnode];
11     int index;
12 };
13 trie * root;
14 char st[maxn][maxnode];
15 
16 trie * newnode(){
17     trie *p = (trie *)malloc(sizeof(trie));
18     p->index = 0;
19     for(int i = 0; i < maxnode; i++)
20         p->next[i] = 0;
21     return p;
22 }
23 void init(){
24     root = newnode();
25     root->index = 0;
26     for(int i = 0; i < maxnode; i++)
27         root->next[i] = 0;
28 }
29 int insert_node(char tms[], int sindex){
30     trie * p = root, *q;
31     int len = strlen(tms), i = 0;
32     for(; i < len; i++){
33         if(p->next[tms[i] - 'a'])
34             p = p->next[tms[i] - 'a'];
35         else{
36             q = newnode();
37             p->next[tms[i] - 'a'] = q;
38             p = q;
39         }
40     }
41     p->index = sindex;
42     return 1;
43 }
44 // -1 == No Found
45 int find_node(char tms[]){
46     trie *p = root;
47     int len = strlen(tms);
48     for(int i = 0; i < len; i++){
49         if(p->next[tms[i] - 'a'])
50             p = p->next[tms[i] - 'a'];
51         else
52             return -1;
53     }
54     if(p->index){
55         return p->index;
56     }
57     return -1;
58 }
59 void deltrie(trie *node){
60     if(node){
61         for(int i = 0; i < maxnode; i++){
62             deltrie(node->next[i]);
63         }
64         free(node);
65     }
66 }
67 
68 int main(){
69 
70     char tms[maxnode];
71     int totid = 1;
72     init();
73     while(gets(tms) && strcmp(tms, "")){
74         char tps[maxnode];
75         sscanf(tms, "%s %s", st[totid], tps);
76         //printf("
-->totid: %d<--
", totid);
77         insert_node(tps, totid);
78         totid++;
79     }
80     while(gets(tms) && strcmp(tms, "")){
81         int tp = find_node(tms);
82         if(tp == -1){
83             printf("eh
");
84         }
85         else{
86             //printf("-->tp: %d-->tms: %s
", tp, tms);
87             printf("%s
", st[tp]);
88         }
89     }
90     deltrie(root);
91 }

题目:

Babelfish
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 43616   Accepted: 18403

Description

You have just moved from Waterloo to a big city. The people here speak an incomprehensible dialect of a foreign language. Fortunately, you have a dictionary to help you understand them.

Input

Input consists of up to 100,000 dictionary entries, followed by a blank line, followed by a message of up to 100,000 words. Each dictionary entry is a line containing an English word, followed by a space and a foreign language word. No foreign word appears more than once in the dictionary. The message is a sequence of words in the foreign language, one word on each line. Each word in the input is a sequence of at most 10 lowercase letters.

Output

Output is the message translated to English, one word per line. Foreign words not in the dictionary should be translated as "eh".

Sample Input

dog ogday
cat atcay
pig igpay
froot ootfray
loops oopslay

atcay
ittenkay
oopslay

Sample Output

cat
eh
loops

Hint

Huge input and output,scanf and printf are recommended.
原文地址:https://www.cnblogs.com/bolderic/p/6800736.html