zoj1109-Language of FatMouse 字典树

Language of FatMouse

We all know that FatMouse doesn't speak English. But now he has to be prepared since our nation will join WTO soon. Thanks to Turing we have computers to help him.

Input Specification

Input consists of up to 100,005 dictionary entries, followed by a blank line, followed by a message of up to 100,005 words. Each dictionary entry is a line containing an English word, followed by a space and a FatMouse word. No FatMouse word appears more than once in the dictionary. The message is a sequence of words in the language of FatMouse, one word on each line. Each word in the input is a sequence of at most 10 lowercase letters.

Output Specification

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

Sample Input

dog ogdaycat atcaypig igpayfroot ootfrayloops oopslayatcayittenkayoopslay

Output for Sample Input

catehloops

题目大意：给你两个字符串a,b，询问字符串是不是和b相等，如果相等就输出字符串a,否则输出eh.

解题思路：

1、使用字典树，保存两个字符串，利用字典树判断是否存在第二个字符串，存在即输出第一个字符串。

2、使用map容器来存储字符串，如果存在该字符串即输出。

字典树专题：点击打开链接http://blog.csdn.net/wang_heng199/article/details/76448804

字典树ac代码：

#include <stdio.h>  #include <string.h>  #include <malloc.h>#include <iostream>  using namespace std;struct Tree{      int num;      char s[15];      struct Tree *next[27];  };  struct Tree *root;  void Initialize()  //初始化全部为空 {      root = (struct Tree*)malloc(sizeof(struct Tree));      root->num = 0;      for(int i = 0 ; i < 26 ; ++i)          root->next[i] = NULL;  }  void Insert(char *s1,char *s2)//以s1为基础建立字典树   {      int len = strlen(s1);      struct Tree *node,*p = root;            for(int i = 0 ; i < len ; ++i){          int pos = s1[i] - 'a';          if(p->next[pos] == NULL){               node = (struct Tree*)malloc(sizeof(struct Tree));              for(int j = 0 ; j < 26 ; ++j){                  node->next[i] = NULL;                  node->num = 0;              }              p->next[pos] = node;          }          p = p->next[pos];      }       p->num = 1;  //标记为1     strcpy(p->s,s2);  //存储字符串待输出 }  void Find(char *s)  {      int len = strlen(s);      int i;      struct Tree *p = root;            for(i = 0 ; i < len ; ++i){          int pos = s[i] - 'a';          if(p->next[pos] != NULL)              p = p->next[pos];          else              break;      }      if(p->num == 1) //为1即表示如果存在         printf("%s\n",p->s);      else          printf("eh\n");  }  int main()  {      char str[30];            Initialize();      while(gets(str)){          if(strlen(str) == 0)              break;          char str1[12],str2[12];          memset(str1,0,sizeof(str1));          memset(str2,0,sizeof(str2));          for(int i = 0 ; i < strlen(str) ; ++i){              if(str[i] == ' '){                  strncpy(str1,str,i);                  str1[i] = '\0';                  strncpy(str2,str+i+1,strlen(str)-i-1);                  str2[strlen(str) - i - 1] = '\0';              }          }          Insert(str2,str1);      }      while(scanf("%s",str) != EOF)          Find(str);      return 0;  }  

STLac代码：

#include <cstdio>#include <cstring>#include <map>#include <algorithm>#include <iostream>using namespace std;char s1[20], s2[20], s3[20];char s[100007][15];int main(){    map<string, int>ma;    int e=1;    while(scanf("%c", &s1[0]) && s1[0]!='\n' )    {        scanf("%s %s%*c", s1+1, s2 );         ma[s2]=e++ ;        strcpy(s[e-1], s1);    }    int flag;    while(scanf("%s", s3)!=EOF)    {        if(ma[s3]>=1)        {            printf("%s\n", s[ma[s3]] );        }        if(ma[s3]==0)        {            printf("eh\n");        }    }    return 0;}
题目链接：点击打开链接http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1109