hdu 1247 Hat’s Words(dfs+trie)

最新推荐文章于 2018-11-18 17:28:18 发布

黎辰

最新推荐文章于 2018-11-18 17:28:18 发布

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CC 4.0 BY-SA版权

文章标签： C++ trie 字符串

本文链接：https://blog.youkuaiyun.com/qq_24451605/article/details/44903323

本文介绍了一种使用字典树和深搜算法来识别字典中可以由两个其他单词组成的特殊单词（hat's words）。通过算法实现，可以高效地在给定的字典中找到所有符合定义的组合单词，并按字母顺序输出。

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Hat’s Words

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9037 Accepted Submission(s): 3231

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

题目分析:
用字典树先把字符串存下,然后利用深搜trie,利用指针跳转判断当前串能否被严格分成两个已有的串

#include <algorithm>
#include <cstring>
#include <cstdio>
#include <iostream>

using namespace std;

char s[50007][100];

struct Node
{
    int cnt;
    Node * b[27];
    Node ( )
        :cnt(0)
    {
        memset ( b , 0 , sizeof ( b ) );
    }
}*root;

bool flag;

void dfs ( int i , Node * p , int state = 0, int j = 0 )
{
    if ( flag ) return;
    if ( !s[i][j] )
    {
        if ( state == 1 && p->cnt == 1 ) flag = true;
        return;
    }
    int x = s[i][j]-'a';
    if ( p->cnt )
        dfs ( i , root , state+1 , j );
    if ( p->b[x] )
        dfs ( i , p->b[x] , state , j+1 );
}

int main ( )
{
    int num = 0;
   // freopen ( "a.txt" , "r" , stdin );
    root = new Node ( );
    while ( ~scanf ( "%s" , s[num] ) )
    {
        int i = 0;
        Node *p = root;
        while ( s[num][i] )
        {
            int x = s[num][i]-'a';
            if ( !(p->b[x]) ) 
                p->b[x] = new Node();
            p = p->b[x];
            i++;
        }
        p->cnt = 1;
        num++;
    }
    for ( int i = 0 ; i < num ; i++ )
    {
        flag = false;
        dfs ( i , root );
        if ( flag ) 
            printf ( "%s\n" , s[i] );
    }
}