POJ 2503 Babelfish (Trie tree or map), pojtrie

POJ 2503 Babelfish (Trie tree or map), pojtrie
Babelfish
Time Limit: 3000 MS Memory Limit: 65536 K
Total Submissions: 34278 Accepted: 14706

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 shocould 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

Link: poj.org/problem? Id = 2503

Returns the correspondence between strings. The input is the first for the second request, and the output eh does not exist.

Question Analysis: The dictionary tree or map is very raw in the meaning of the question, and the corresponding relationship can be established. If map is used, the dictionary tree is over 2000 ms, and the dictionary tree is over 700 ms.

Trie tree:

#include <cstdio>#include <cstring>struct Dic{    char s1[15], s2[15];    int id;}d[100005];struct node{    node *next[26];    bool end;    int id;    node()    {        memset(next, NULL, sizeof(next));        end = false;        id = -1;    }};void Insert(node *p, char *s, int id){    for(int i = 0; s[i] != '\0'; i++)    {        int idx = s[i] - 'a';        if(p -> next[idx] == NULL)            p -> next[idx] = new node();        p = p -> next[idx];    }    p -> end = true;    p -> id = id;}int Search(node *p, char *s){    for(int i = 0; s[i] != '\0'; i++)    {        int idx = s[i] - 'a';        if(p -> next[idx] == NULL)            return -1;        p = p -> next[idx];    }    if(p -> end)        return p -> id;    return -1;}int main(){    char s[30], get[15];    int cnt = 0;    node *root = new node();    while(gets(s) && strlen(s))    {        d[cnt].id = cnt;        sscanf(s, "%s %s", d[cnt].s1, d[cnt].s2);        Insert(root, d[cnt].s2, cnt);        cnt++;    }    while(scanf("%s", get) != EOF)    {        int id = Search(root, get);        if(id == -1)            printf("eh\n");        else            printf("%s\n", d[id].s1);    }}

Map:

#include <iostream>#include <map>#include <cstring>#include <string>#include <cstdio>using namespace std;char s[30], s1[15], s2[15];string s3, tmp;map <string, string> mp;map <string, string> :: iterator it;int main(){    while(gets(s) && strlen(s))    {        sscanf(s, "%s %s", s1, s2);        mp[s2] = s1;    }    while(cin >> s3)    {        it = mp.find(s3);        if(it == mp.end())            cout << "eh" << endl;        else            cout << it -> second << endl;    }}