HDU 1247 Hat’s Words 字典树

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8778    Accepted Submission(s): 3151

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

#include <stdio.h>
#include <string.h>

struct node
{
    bool flag;
    node *next[26];
    node()
    {
        flag = false;
        memset(next,0,sizeof(next));
    }
};

char mp[50001][110];
int mplen[50001];
int stk[50001];

void build(node *&head,char str[],int len)
{
    node *p = head,*q;
    for(int i = 0; i < len; i++)
    {
        int zh = str[i] - 'a';
        if(p->next[zh])
        {
            p = p->next[zh];
        }
        else
        {
            q = new node;
            p->next[zh] = q;
            p = q;
        }
    }
    p->flag = true;
}

bool nfind(node *&head,char str[],int len)
{
    node *p = head;
    int tp = 0;
    for(int i = 0; i < len; i++)
    {
        int zh = str[i] - 'a';
        if(p->next[zh])
            p = p->next[zh];
        if(p->flag && i + 1 < len)
            stk[tp++] = i + 1;//一个单词可能有多个组成，或是一个单词遍历的时候，会有多个true在里面，这里都记录下来
    }
    while(tp)
    {
        bool flag = true;
        int i = stk[tp - 1];
        tp--;
        p = head;
        for(; i < len; i++)
        {
            int zh = str[i] - 'a';
            if(p->next[zh])
            {
                p = p->next[zh];
            }
            else
            {
                flag = false;//注意这个地方，当当前从i开始不能找到，并不代表栈里面所有的都找不到
                break;
            }
        }
        if(p->flag && flag)
            return true;
    }
    return false;
}

int main()
{
    node *head = new node;
    int k = 0;
    while(~scanf("%s",mp[k]))
    {
        mplen[k] = strlen(mp[k]);
        build(head,mp[k],mplen[k]);
        k++;
    }
    for(int i = 0; i < k; i++)
    {
        if(nfind(head,mp[i],mplen[i]))
            printf("%s\n",mp[i]);
    }
    return 0;
}