hdu 1247 Hat’s Words(map)

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8527 Accepted Submission(s): 3069



Problem Description

A hat’s word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.

You are to find all the hat’s words in a dictionary.



Input

Standard input consists of a number of lowercase words, one per line, in alphabetical order. There will be no more than 50,000 words.

Only one case.



Output

Your output should contain all the hat’s words, one per line, in alphabetical order.



Sample Input

a
ahat
hat
hatword
hziee
word

Sample Output

ahat
hatword

Author

戴帽子的



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题 意：找出一个单词，前半部是一个出现过的单词，后半部也是，记住，要严格满足这个条件

所以，其实也就是先查找一个单词的是否有前缀，再用这个单词除去前缀的部分查找是否存在一个这样的单词

CODE:

#include <iostream>
#include <string>
#include <map>
using namespace std;
map <string, int> mp;
string str[50005];
int main()
{
    int n = 0, j, len;
    while ( cin >> str
 ) //输入所有的单词
        mp[str[n++]] = 1; //构成映射
    for ( int i = 0; i < n; ++i )
    {
        len = str[i].size();
        for ( j = 1; j < len; ++j )
        {
            string s1 ( str[i], 0, j ); //分割单词
            string s2 ( str[i], j, len - j );
            if ( mp[s1] == 1 && mp[s2] == 1 ) //在单词中查找存在
            {
                cout << str[i] << endl; //输出该单词
                break;
            }
        }
    }
    return 0;
}