Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13798    Accepted Submission(s): 4950

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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/*
一开始想到将单词分割，但是怕超时，后来一想单词长度可能10^2差不多，10^7的运行时间应该不会超时
*/
#include<bits/stdc++.h>
#define N 30
using namespace std;
#define MAX 26
const int maxnode=4000*100+100;///预计字典树最大节点数目
const int sigma_size=26;///每个节点的最多儿子数

struct Trie
{
    ///这里ch用vector<26元素的数组> ch；实现的话，可以做到动态内存
    int ch[maxnode][sigma_size];///ch[i][j]==k表示第i个节点的第j个儿子是节点k
    int val[maxnode];///val[i]==x表示第i个节点的权值为x
    int sz;///字典树一共有sz个节点,从0到sz-1标号

    ///初始化
    void Clear()
    {
        sz=1;
        memset(ch[0],0,sizeof(ch[0]));///ch值为0表示没有儿子
    }

    ///返回字符c应该对应的儿子编号
    int idx(char c)
    {
        return c-‘a‘;
    }

    ///在字典树中插入单词s,但是如果已经存在s单词会重复插入且覆盖权值
    ///所以执行Insert前需要判断一下是否已经存在s单词了
    void Insert(char *s)
    {
        int u=0,n=strlen(s);
        for(int i=0;i<n;i++)
        {
            int id=idx(s[i]);
            if(ch[u][id]==0)///无该儿子
            {
                ch[u][id]=sz;
                memset(ch[sz],0,sizeof(ch[sz]));
                val[sz++]=0;
            }
            u=ch[u][id];
        }
        val[u]=n;
    }

    ///在字典树中查找单词s
    bool Search(char *s)
    {
        int n=strlen(s),u=0;
        for(int i=0;i<n;i++)
        {
            int id=idx(s[i]);
            if(ch[u][id]==0)
                return false;
            u=ch[u][id];
        }
        return val[u];
    }
};
Trie trie;///定义一个字典树
char op[50005][N];
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    int len=0;
    trie.Clear();
    while(gets(op[len++]))
        trie.Insert(op[len-1]);
    //cout<<len<<endl;
    for(int i=0;i<len-1;i++)
    {
        int n=strlen(op[i]);
        //cout<<"op[i]="<<op[i]<<" i="<<i<<endl;
        for(int k=1;k<n;k++)
        {
            char str1[N],str2[N];
            strncpy(str1,op[i],k);
            str1[k]=‘\0‘;
            strncpy(str2,op[i]+k,n);
            str2[n-k]=‘\0‘;
            if(trie.Search(str1)&&trie.Search(str2))
            {
                puts(op[i]);
                break;//这一句是必须加的因为不加就会输出两遍
            }
            //cout<<str1<<" "<<str2<<endl;
        }
        //puts(op[i]);
    }
    return 0;
}