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): 12123    Accepted Submission(s): 4329

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

aahathathatwordhzieeword

 

Sample Output

ahathatword

 

Author

戴帽子的

 

字典树的具体用法可看：http://www.cnblogs.com/tanky_woo/archive/2010/09/24/1833717.html

贴个模板学习一下。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdlib>#include <string>#include <vector>#include <cstdio>#include <cmath>#include <queue>#include <stack>#include <map>#include <set>using namespace std;#define INF 0x3f3f3f3const int N=50005;const int mod=1e9+7;typedef struct Trie{    Trie *next[26];    bool v;}Trie;Trie *root=(Trie *)malloc(sizeof(Trie));void insert(char *str){    int len=strlen(str);    Trie *p = root,*q;    for (int i=0; i<len; i++) {        int id=str[i]-'a';        if (p->next[id]==NULL) {            q=(Trie *)malloc(sizeof(Trie));            q->v=0;            for (int j=0; j<26; j++)                q->next[j]=NULL;            p->next[id]=q;        }        p=p->next[id];    }    p->v=true;}int find(char *str){    int len=strlen(str);    Trie *p = root;    for (int i=0; i<len; i++) {        int id=str[i]-'a';        p=p->next[id];        if (p==NULL) return 0;    }    return p->v;}void del(Trie *root){    for (int i=0; i<26; i++) {        if (root->next[i]!=NULL) {            del(root->next[i]);        }    }    free(root);}char s[50005][50];int main() {    int count=0;    char str[50];    for (int i=0; i<26; i++) {        root->next[i]=NULL;    }    root->v=0;    while (scanf("%s",str)!=EOF) {        strcpy(s[count++], str);        insert(str);    }    for (int i=0; i<count; i++) {        for (int j=1; j<strlen(s[i]); j++) {            char temp1[50]={},temp2[50]={};            strncpy(temp1, s[i], j);            strncpy(temp2, s[i]+j, strlen(s[i])-j);            if (find(temp1)&&find(temp2)) {                printf("%s\n",s[i]);                break;            }        }    }    del(root);    return 0;}