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 ogday cat atcay pig igpay froot ootfray loops oopslay atcay ittenkay oopslay
Output for Sample Input
cat eh loops
題目大意：給你兩個字符串a,b，詢問字符串是不是和b相等，如果相等就輸出字符串a,否則輸出eh.
解題思路：
1、使用字典樹，保存兩個字符串，利用字典樹判斷是否存在第二個字符串，存在即輸出第一個字符串。
2、使用map容器來存儲字符串，如果存在該字符串即輸出。
字典樹專題：點擊打開鏈接http://blog.csdn.net/wang_heng199/article/details/76448804
字典樹ac代碼：
#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() //初始化全部爲空
{
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爲基礎建立字典樹
{
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; //標記爲1
strcpy(p->s,s2); //存儲字符串待輸出
}
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) //爲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代碼：
#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;
}
