POJ 2503 Babelfish map做法

Babelfish

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 31965 Accepted: 13732

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

Hint

Huge input and output,scanf and printf are recommended.

Source

Waterloo local 2001.09.22

map的基础题。比较大的输入输出，所以重载new，delete运算符。但其实这题输入输出也不是很大，没什么必要。

代码如下：

#include <iostream>
#include<cstdio>
#include<map>
#include<string>
using namespace std;

//手动分配内存空间
char buf[50000000];
int sizet =0;


void* operator new(size_t size){
    void *p = buf + sizet;
    sizet  += size;
    return p;


}
void operator delete(void* p){
}

//
int main()
{
char e[11],f[11],str[25];
map<string,string>t;
for(;;)
{
   gets(str);
   //cin>>str;
   if(str[0]=='\0') break;//str[0]=='\n'无效,应该是gets和cin等的运行机制问题。若要保险，再加一句：str[0]=='\n'.
   sscanf(str,"%s%s",e,f);//从str读取
   t[f]=e;
}
while(scanf("%s",str)!=EOF)
{
   string s=t[str];
   if (s.length()==0)//string类对象s的函数length（）
   printf("eh\n");
   else
   printf("%s\n",s.c_str());//转换为cstring类型的数据，用s.c_str()函数
}
    return 0;
}