zoj 1825 Compound Words

ZOJ Problem Set - 1825

Compound Words

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Time Limit: 5 Seconds      Memory Limit: 32768 KB

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You are to find all the two-word compound words in a dictionary. A two-word compound word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.

Input

Standard input consists of a number of lowercase words, one per line, in alphabetical order. There will be no more than 120,000 words.

Output

Your output should contain all the compound words, one per line, in alphabetical order.

Sample Input

a
alien
born
less
lien
never
nevertheless
new
newborn
the
zebra

Sample Output

alien
newborn

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Source: University of Waterloo Local Contest 1996.09.28
Submit    Status

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//1846335 2009-04-28 15:41:20 Accepted  1825 C++ 270 1240 Wpl 
#include <iostream>
#include <string>
#include <set>
using namespace std;
set<string>S;
set<string>::iterator p;
int main()
{
    string str,str1,str2;
    S.clear();
    int len,i;
    while(cin>>str)
    {
        S.insert(str);
    }
    for(p=S.begin();p!=S.end();p++)
    {
        str=*p;
        len=str.length();
        for(i=1;i<len;i++)
        {
            str1=str.substr(0,i);
            str2=str.substr(i,len-i);
            if(S.find(str1)!=S.end()&&S.find(str2)!=S.end())
            {
                cout<<str<<endl;
                break;
            }
        }
    }
    return 0;
}

//用map

//1846363 2009-04-28 16:02:33 Wrong Answer  1825 C++ 410 3616 Wpl 
//1846374 2009-04-28 16:11:40 Accepted  1825 C++ 240 1372 Wpl 
#include <iostream>
#include <map>
#include <string>
using namespace std;
map<string,int>M;
map<string,int>::iterator p;
int main()
{
    string str,str1,str2;
    int len,i;
    M.clear();
    while(cin>>str)
        M[str]=1;
    for(p=M.begin();p!=M.end();p++)
    {
        str=p->first;
        len=str.length();
        for(i=1;i<len;i++)
        {
            str1=str.substr(0,i);
            str2=str.substr(i,len-i);
        //    if(M[str1]==1&&M[str2]==1)  //找数不用这样找的,因为这样会把那个数放进M里
            if(M.find(str1) != M.end() && M.find(str2) != M.end())
            {
                cout<<str<<endl;
                break;
            }
        }
    }
    return 0;
}