Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9620    Accepted Submission(s): 3438

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

昨天晚上用STL 1a渺杀 ，今天改用trie树 wa一上午 也是醉了！指针真的好烦人！

ＳＴＬ：

#include<iostream>
#include<sstream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include<cctype>
#include<string>
#include<cmath>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
using namespace std;

int main()
{
    string str;
    map<string,int >cnt;
    set<string>dict;
    vector<string>buf;
    while(cin>>str)
    {
        cnt[str]=1;
        buf.push_back(str);
    }
    for(int i=0; i<buf.size(); i++)
    {
        string str2;
        for(int j=0; j<buf[i].size()-1; j++)
        {
            str2+=buf[i][j];
            if(cnt.count(str2)&&cnt.count(buf[i].substr(j+1,buf[i].size()-j-1)))
            {
                dict.insert(buf[i]);
                break;
            }
        }
    }

    for(set<string>::iterator it=dict.begin(); it!=dict.end(); it++)
        cout<<*it<<endl;
    return 0;
}

ＴＲＩＥ树：

#include<iostream>
#include<sstream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include<cctype>
#include<string>
#include<cmath>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
using namespace std;
char cnt[50031][153];

struct Node
{
    struct Node *next[26];
    bool isword;
    Node()
    {
        memset(next,NULL,sizeof(next));
        isword=0;
    }
}*root;

void insertWord(Node * node ,char *co)
{
    int id;
    node = root ;
    while(*co)
    {
        id=*co-'a';
        if(node->next[id]==NULL)
            node->next[id]=new Node;
        node = node->next[id];
        co++;
    }
    node->isword=1;
}

bool searchWord(Node *node ,char *co)
{

    int flag=0;
    node = root;
    while(*co)
    {
        int id=*co-'a';
        if(node->next[id])
        {
            node=node->next[id];
            if(node->isword)
            {
                flag=1;
                Node *p=root;
                char *pt=co;
                pt++;
                while(*pt)
                {

                    int id=*pt-'a';
                    if(p->next[id])
                        p=p->next[id];
                    else
                    {
                        flag=0;
                        break;
                    }
                    pt++;
                }
                if(flag)
                {
                    if(p->isword)
                        return 1;
                }

            }

        }
        co++;
    }
    return 0;
}

int main()
{
    int t=0;
    root=new Node;
    while(~scanf("%s",cnt[t++]))
    {
        insertWord(root,cnt[t-1]);
    }
    for(int i=0; i<t; i++)
    {
        if(searchWord(root,cnt[i]))
        {
            printf("%s\n",cnt[i]);
        }
    }
    return 0;
}