• 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] != ''; ++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] != ''; ++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] != ''; ++i){
            int id = str[i]-'a';
            if(p->pt[id] == NULL){
                return;
            }
            else{
                if(p->pt[id]->isEnd && str[i+1] != ''){
                    bool ok = trie_find(str+i+1);
                    if(ok){
                        printf("%s
    ", 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;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/inmoonlight/p/5921071.html
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