POJ2503Babelfish(MAP+字典树)



Babelfish

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 37423 Accepted: 15967

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 ogdaycat atcaypig igpayfroot ootfrayloops oopslayatcayittenkayoopslay

Sample Output

catehloops

题意：输入的每一单词跟每一个单词对应，查找第二个单词对应的第一个单词

直接map<string,string>建立关系，后来输出的时候scanf是不可以的，因为map里面的是string

然后看到有说用字典树写的，然后又写了个字典树，没调试，直接交，水过

//map 2000ms，原因可能是后面用的cout#include <iostream>#include <cstring>#include <string.h>#include <cstdio>#include <map>#include <ctype.h>#include <stdlib.h>#include <algorithm>using namespace std;int main(){    map<string,string> g;    char str[30],s1[15],s2[15];    while(gets(str) && strcmp(str,"") != 0)    {        sscanf(str,"%s %s",s1,s2);        g[s2] = s1;    }    while(gets(s2))    {        if(g.count(s2) == 0)            printf("eh\n");        else            cout << g[s2]<<endl;    }    return 0;}

字典树 735ms

#include <iostream>#include <cstring>#include <string.h>#include <cstdio>#include <map>#include <ctype.h>#include <stdlib.h>#include <algorithm>using namespace std;struct node{    int flag;    node *next[26];    char s[15];};node *root;void build(char *s2,node *root,char *s1){    node *p,*q;    p = root;    int len = strlen(s2);    for(int i = 0; i < len; i++)    {        int id = s2[i] - 'a';        if(p -> next[id] == NULL)        {            q = (node *) malloc(sizeof(node));            for(int i = 0; i < 26; i++)                q -> next[i] = NULL;            q -> flag = 0;            p -> next [id] = q;            p = p -> next[id];        }        else            p = p -> next[id];    }    p -> flag = 1;    strcpy(p -> s,s1);}void Find(node *root,char *s2){    node *p;    p = root;    int len = strlen(s2);    for(int i = 0; i < len; i++)    {        int id = s2[i] - 'a';        if(p -> next[id] == NULL)        {            printf("eh\n");            return ;        }        else            p = p -> next[id];    }    if(p -> flag)        printf("%s\n",p -> s);    else        printf("eh\n");}int main(){    char str[30],s1[15],s2[15];    root = (node *)malloc(sizeof(node));    for(int i = 0; i < 26; i++)        root->next[i] = NULL;    while(gets(str) && strcmp(str,"") != 0)    {        sscanf(str,"%s %s",s1,s2);        build(s2,root,s1);    }    while(gets(s2))    {        Find(root,s2);    }    return 0;}