poj Babelfish（二分||map）

 Babelfish

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.

方法一：map记录并查找

代码：

#include<stdio.h>#include<iostream>#include<map>using namespace std;int main(){    map<string,string>translate;    char a[22],b[11],c[11];    while(gets(a)&&a[0]!='\0')    {        sscanf(a,"%s %s",b,c);        translate[c]=b;    }    while(gets(a))    {        if(translate.find(a)!=translate.end())            cout<<translate[a]<<endl;        else            cout<<"eh"<<endl;    }    return 0;}

方法二：二分查找

代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node{    char s1[15],s2[15];}a[100010];int len;bool cmp(node a,node b){    return strcmp(a.s2,b.s2)<0;}int main(){    char s[22];    len=0;    while(gets(s)&&s[0]!='\0')    {        sscanf(s,"%s %s",a[len].s1,a[len].s2);        len++;    }    sort(a,a+len,cmp);    while(gets(s))    {        int x=0,y=len-1,flag=1;        while(x<=y)        {            int mid=(x+y)>>1;            if(!strcmp(s,a[mid].s2))            {                printf("%s\n",a[mid].s1);                flag=0;                break;            }            else if(strcmp(s,a[mid].s2)<0)                y=mid-1;            else                x=mid+1;        }        if(flag)            printf("eh\n");    }    return 0;}