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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> POJ2503:Babelfish(二分)

POJ2503:Babelfish(二分)

編輯:C++入門知識

POJ2503:Babelfish(二分)


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

咋看是字典樹,但是我在搜二分專題的時候看到的這道題,那麼這道題肯定能用二分解決
思路很好想,運用到了sscanf函數

#include 
#include 
#include 
using namespace std;

struct node
{
    char s1[20],s2[20];
} a[100005];
int len;
int cmp(node a,node b)
{
    return strcmp(a.s2,b.s2)<0;
}

int main()
{
    len = 0;
    int i,j;
    char str[50];
    while(gets(str))
    {
        if(str[0] == '\0')
        break;
        sscanf(str,"%s%s",a[len].s1,a[len].s2);
        len++;
    }
    sort(a,a+len,cmp);
    while(gets(str))
    {
        int l = 0,r= len-1,mid,flag = 1;
        while(l<=r)
        {
            int mid = (l+r)>>1;
            if(strcmp(str,a[mid].s2)==0)
            {
                printf("%s\n",a[mid].s1);
                flag = 0;
                break;
            }
            else if(strcmp(str,a[mid].s2)<0)
                r = mid-1;
            else
                l = mid+1;
        }
        if(flag)
            printf("eh\n");
    }

    return 0;
}


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