POJ 2503 Babelfish

Babelfish

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 28103		Accepted: 12128

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 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

 

题意：

输入一系列单词 输出对应单词

 

代码：

#include<iostream>
#include<string>
#include<map>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#define MAX 100001
using namespace std;
 
typedef struct 
{
	char tran[20];
	char str[20];
}node;

node s[MAX];
int i=0;

int cmp(const void *a,const void *b)
{
	node *c=(node *)a,*d=(node *)b;
	return strcmp(c->str,d->str);
}

int bin_search(char *t)
{
	int l=0,r=i-1;
	int mid;
	while(l<=r)
	{
		mid=(l+r)/2;
		if(strcmp(s[mid].str,t)==0) return mid;
		else if(strcmp(s[mid].str,t)<0) l=mid+1;
		else r=mid-1;
	}
	return -1;
}

int main()
{
	char tmp[20];

	while(true)
	{
		char t;
		if((t=getchar())=='\n') break;
		s[i].tran[0]=t;
		int j=1;
		while(true)
		{
			t=getchar();
			if(t==' ') { s[i].tran[j++]='\0'; break; }
			else s[i].tran[j++]=t;
		}
		
		scanf("%s",s[i].str);
		getchar();
		i++;
	}

	qsort(s,i,sizeof(node),cmp);

	while(gets(tmp)!=NULL)
	{
		int j=bin_search(tmp);
		if(j==-1) printf("eh\n");
		else printf("%s\n",s[j].tran);
	}
	return 0;
}

思路：

二分查找法

处理空行的时候比较麻烦