HDU1247 Hat’s Words 【trie树】

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7502    Accepted Submission(s): 2705

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

这题開始总想着走捷径，想依据trie树内部节点将单词分成前后缀，前缀一定在树里，所以仅仅需推断后缀是否在树里，可是非常多细节非常棘手，比方如何确定后缀。后缀如何检索。如何在一个单词内改变前后缀等，折腾了非常久，实在没法解决，然后就用了一開始就非常鄙夷的方法，将每一个字符串都存到数组并插入到树里。然后将每一个字符串遍历拆分成前后缀，再检索前后缀是否都在树中。

这题再次验证了一个道理，就是在没有想出更好的方法之前。最笨的方法就是最好的方法。

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

struct Node{
    struct Node *next[26];
    int wordCover;
};
Node *root = (Node *)malloc(sizeof(Node));
char suffix[50], prefix[50], strArr[50000][50];

void cleanStruct(Node *p)
{
    memset(p->next, 0, sizeof(p->next));
    p->wordCover = 0;
}

void insert(char *str)
{
    int id;
    Node *p = root;
    while(*str)
    {
        id = *str - 'a';
        if(p->next[id] == NULL){
            p->next[id] = (Node *)malloc(sizeof(Node));
            cleanStruct(p->next[id]);
        }
        p = p->next[id];
        ++str;
    }
    ++p->wordCover;
}

int isExist(char *str)
{
    Node *p = root;
    int id;
    while(*str){
        id = *str - 'a';
        if(p->next[id] == NULL) return 0;
        p = p->next[id];
        ++str;
    }
    return p->wordCover;
}

void deleteTrie(Node *p)
{
    for(int i = 0; i < 26; ++i)
        if(p->next[i]) deleteTrie(p->next[i]);
    free(p);
}

int main()
{
    //freopen("stdin.txt", "r", stdin);
    int id = 0, i, j, len;
    cleanStruct(root);
    while(gets(strArr[id])) insert(strArr[id++]);
    for(i = 0; i < id; ++i){
        len = strlen(strArr[i]);
        for(j = 1; j < len; ++j){
            strcpy(prefix, strArr[i]);
            prefix[j] = '\0';
            strcpy(suffix, strArr[i] + j);
            if(isExist(prefix) && isExist(suffix)){
                puts(strArr[i]); break;
            }
        }
    }
    deleteTrie(root);
    return 0;
}

2014.12.16更新

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

#define maxn 1000000

char str[50], str1[50], str2[50], dic[50002][50];
struct Trie {
    int ch[maxn][26];
    int val[maxn], sz;

    Trie() {
        memset(ch[0], 0, sizeof(ch[0]));
        sz = 1;
    }
    int idx(char ch) { return ch - 'a'; };
    void insert(const char *str) {
        int u = 0, i, id, len = strlen(str);
        for (i = 0; i < len; ++i) {
            id = idx(str[i]);
            if (!ch[u][id]) {
                memset(ch[sz], 0, sizeof(ch[sz]));
                ch[u][id] = sz;
                val[sz++] = 0;
            }
            u = ch[u][id];
        }
        val[u] = 1;
    }
    bool find(const char *str) {
        int u = 0, i, id, len = strlen(str);
        for (i = 0; i < len; ++i) {
            id = idx(str[i]);
            if(!ch[u][id]) return false;
            u = ch[u][id];
        }
        return val[u];
    }
} T;

int main() {
    // freopen("stdin.txt", "r", stdin);
    int id = 0, i, j, len;
    while (scanf("%s", str) == 1) {
        T.insert(str);
        strcpy(dic[id++], str);
    }
    for (i = 0; i < id; ++i) {
        len = strlen(dic[i]);
        for (j = 1; j < len; ++j) {
            strncpy(str1, dic[i], j);
            strncpy(str2, dic[i] + j, len - j);
            str1[j] = '\0';
            str2[len-j] = '\0';
            if (T.find(str1) && T.find(str2)) {
                puts(dic[i]); break;
            }
        }
    }
    return 0;
}