二分查找 - bailian 2503:Babelfish

题目描述

描述
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.
输入
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.
输出
Output is the message translated to English, one word per line. Foreign words not in the dictionary should be translated as "eh".
样例输入
dog ogday
cat atcay
pig igpay
froot ootfray
loops oopslay

atcay
ittenkay
oopslay
样例输出
cat
eh
loops
提示
Huge input and output,scanf and printf are recommended.
来源
Waterloo local 2001.09.22

解题分析

题目的意思就是查字典，但是由于字典的比较大，考虑使用二分查找。二分查找需要注意两点：1.查找过程中区间一致性，最初的查找区间是闭区间，子区间也要是闭区间，左闭右开也是类似的2.查找序列一定需要排好序。

解题代码

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

struct Entry{
    char en[11];
    char fr[11];
}dict[100010];

bool operator < (const Entry & a, const Entry & b){
    return (strcmp(a.fr, b.fr) < 0);
}

int main(){
    char word[11];
    int k = 0;
    while(true){
        scanf("%s%s", dict[k].en, dict[k].fr);
        k++;
        cin.get();
        if(cin.peek() == '\n') break;
    }
    sort(dict, dict + k);
    while ((scanf("%s", word)) != EOF) {
        int l = 0, r = k - 1;
        int mid = l;
        while(l <= r){
            mid = l + (r - l)/2;
            int t = strcmp(word, dict[mid].fr);
            if( t == 0)
            {
                printf("%s\n", dict[mid].en);
                break;
            }
            else if(t < 0)
                r = mid - 1;
            else
                l = mid + 1;
        }
        if(r < l)
            printf("eh\n");
    }
    return 0;
}