HDU 1247 Hat’s Words

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13752    Accepted Submission(s): 4919

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

 

 

 

解析：字典树。

 

 

 

#include <cstdio>
#include <cstring>

char s[50005][50];

struct Node{
    Node* pt[26];
    bool isEnd;
};

Node* root;
Node memory[100000];
int allo;

void trie_init()
{
    memset(memory, 0, sizeof(memory));
    allo = 0;
    root = &memory[allo++];
}

void trie_insert(char str[])
{
    Node *p = root;
    for(int i = 0; str[i] != '\0'; ++i){
        int id = str[i]-'a';
        if(p->pt[id] == NULL){
            p->pt[id] = &memory[allo++];
        }
        p = p->pt[id];
    }
    p->isEnd = true;
}

bool trie_find(char str[])
{
    Node *p = root;
    for(int i = 0; str[i] != '\0'; ++i){
        int id = str[i]-'a';
        if(p->pt[id] == NULL){
            return false;
        }
        p = p->pt[id];
    }
    return p->isEnd;
}

void getans(char str[])
{
    Node* p = root;
    for(int i = 0; str[i] != '\0'; ++i){
        int id = str[i]-'a';
        if(p->pt[id] == NULL){
            return;
        }
        else{
            if(p->pt[id]->isEnd && str[i+1] != '\0'){
                bool ok = trie_find(str+i+1);
                if(ok){
                    printf("%s\n", str);
                    return;
                }
            }
        }
        p = p->pt[id];
    }
}

void solve(int n)
{
    trie_init();
    for(int i = 0; i < n; ++i){
        trie_insert(s[i]);
    }
    for(int i = 0; i < n; ++i){
        getans(s[i]);
    }
}

int main()
{
    int n = 0;
    while(gets(s[n])){
        ++n;
    }
    solve(n);
    return 0;
}