2503题Trie树实现

1、题目描述：

Babelfish

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 17355		Accepted: 7484

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.

Source

Waterloo local 2001.09.22

二、题意分析

可以采用字典树，用Foreign words建立起trie树，English作为信息储存在结点里即可

三、代码：

#include <stdio.h> #include <stdlib.h> #define MAX 11 char result[MAX] = "eh"; typedef struct _Node{ char key; char *eng; struct _Node *parent; struct _Node *child; struct _Node *sibling; }Node; typedef struct _TrieTree{ Node * root; }TrieTree; void init(TrieTree *T){ T->root = (Node*)malloc(sizeof(Node)); if(T->root == NULL) return; T->root->key = 0; T->root->parent = NULL; T->root->child = NULL; T->root->sibling = NULL; } void insert(TrieTree *T, const char *s, const char source[MAX]){ Node *p = NULL; Node *ch = NULL; if(T == NULL || s == NULL) return; p = T->root; while(p != NULL && *s != 0){ ch = p->child; //查找子节点 while(ch != NULL && ch->key != *s){ ch = ch->sibling; } //子节点中不存在此字符，插入 if(ch == NULL){ Node *newNode = (Node*)malloc(sizeof(Node)); if(newNode == NULL) return; newNode->key = *s; newNode->eng = NULL; newNode->parent = p; newNode->child = NULL; newNode->sibling = p->child; p->child = newNode; p = newNode; s++; } /*子节点中存在此字符 *则此节点作为新的父节点继续查找 */ else if(ch->key == *s){ p = ch; s++; } } p->eng = (char*)malloc(sizeof(char)*MAX); if(p->eng == NULL) return; strcpy(p->eng, source); } char* search(const TrieTree T, const char *s){ Node *p = T.root; Node *ch = p->child; while(p != NULL && *s !=0){ ch = p->child; while(ch != NULL && ch->key !=*s){ ch = ch->sibling; } if(ch == NULL){ return result; } else{ p = ch; s++; } } if(p->eng == NULL) return result; else return p->eng; } int main(int argc, char **args){ char eng[MAX] = {0}; char dialect[MAX] = {0}; char buff[2*MAX] = {0}; TrieTree T; init(&T); while(gets(buff) && strcmp(buff, "") != 0){ sscanf(buff, "%s%s", eng, dialect); insert(&T, dialect, eng); memset(buff, 0 , sizeof(char)*2*MAX); memset(eng, 0, sizeof(char)*MAX); memset(dialect, 0, sizeof(char)*MAX); } while(scanf("%s", dialect) != EOF){ printf("%s/n", search(T, dialect)); memset(dialect, 0, sizeof(char)*MAX); } return 0; }