HDU 1247 Hat’s Words 字典樹

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;//一個單詞可能有多個組成，或是一個單詞遍歷的時候，會有多個true在裏面，這裏都記錄下來
    }
    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;//注意這個地方，噹噹前從i開始不能找到，並不代表棧裏面所有的都找不到
                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;
}