HDU 1247 Hat’s Words

题目链接 HDU 1247

Hat’s Words

                                                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                         Total Submission(s): 9872    Accepted Submission(s): 3536

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

aahathathatwordhzieeword

 

Sample Output

ahathatword

 

Author

戴帽子的

 

题意：

给一堆单词，找出其中有多少个由另外两个单词组成的单词。

分析：

Trie的应用，先将所有的单词保存到树中，再将每个单词分为两段，如果这两段都在树中，那么这个单词满足条件。

代码：

#include <stdio.h>#include <iostream>#include <algorithm>#include <cstring>using namespace std;char str[50000][50];struct Trie         //字典树{    int IsWord;     //判断该节点是否为单词    Trie* next[26];    Trie() {        //构造函数，每创建一棵字典树时自动调用        IsWord = 0;        for(int i = 0; i < 26; i++) next[i] = NULL;    }} *Root;void Insert(char ch[])      //插入一个单词{    Trie* p = Root;    for(int i = 0; ch[i]; i++) {        int k = ch[i] - 'a';        if(p->next[k] == NULL) {            Trie* temp = new Trie;            p->next[k] = temp;        }        p = p->next[k];    }    p->IsWord = 1;}bool Search(char ch[])      //查找单词是否在树中{    Trie* p = Root;    for(int i = 0; ch[i]; i++) {        int k = ch[i] - 'a';        if(p == NULL || p->next[k] == NULL) return false;        p = p->next[k];    }    return p->IsWord;}void Delete(Trie* root)     //递归释放内存空间{    for(int i = 0; i < 26; i++) {        if(root->next[i] != NULL)            Delete(root->next[i]);    }    delete root;}int main(){    Root = new Trie;    int s = 0;    while(~scanf("%s", str[s])) {        Insert(str[s]);        s++;    }    for(int i = 0; i < s; i++) {        int len = strlen(str[i]);        for(int j = 1; j < len; j++) {            char t1[50] = "\0", t2[50] = "\0";            strncpy(t1, str[i], j);         //将单词截断为两部分            strncpy(t2, str[i] + j, len - j);            if(Search(t1) && Search(t2)) {                printf("%s\n", str[i]);                break;            }        }    }    Delete(Root);    return 0;}