HDU_1247_Hat’s Words（字典树）

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Hat’s Words

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8886 Accepted Submission(s): 3194

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

题意：问给定的单词中，输出可以由另外两个单词拼成的单词，按照字典序输出。
分析：先用字典树处理所有单词，然后对单词进行枚举，每个单词分成两部分来查找。
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1247
代码清单：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int MAX = 30;struct trie{    bool point;        //单词的结尾标记    trie *next[MAX];};trie *root=new trie();int k=0,slen,k1,k2;char s1[MAX],s2[MAX];char str[50005][MAX],s[MAX];void createTrie(char *s){    trie *p=root, *q;    int len=strlen(s),pos;    for(int i=0;i<len;i++){        pos=s[i]-'a';        if(p->next[pos]==NULL){            q=new trie();            q->point=false;            for(int j=0;j<MAX;j++)                q->next[j]=NULL;            p->next[pos]=q;            p=p->next[pos];        }        else{            p=p->next[pos];        }    }    p->point=true;}bool findTrie(char *s){    trie *p=root;    int len=strlen(s),pos;    for(int i=0;i<len;i++){        pos=s[i]-'a';        if(p->next[pos]==NULL)            return false;        p=p->next[pos];    }    return p->point;}void delTrie(trie *Root){    for(int i=0;i<MAX;i++){        if(Root->next[i]!=NULL)            delTrie(Root->next[i]);    }    free(Root);}int main(){    //freopen("liuchu.txt","r",stdin);    for(int i=0;i<MAX;i++)        root->next[i]=NULL;    while(scanf("%s",s)!=EOF){        slen=strlen(s);        for(int i=0;i<slen;i++)            str[k][i]=s[i];        k++;        createTrie(s);    }    for(int i=0;i<k;i++){        slen=strlen(str[i]);        k1=0;        for(int j=0;j<slen-1;j++){            k2=0;            s1[k1++]=str[i][j];            for(int h=j+1;h<slen;h++)                s2[k2++]=str[i][h];            s2[k2]='\0';            if(findTrie(s1)&&findTrie(s2)){                printf("%s\n",str[i]);                break;            }        }        memset(s1,'\0',sizeof(s1));        memset(s2,'\0',sizeof(s2));    }    delTrie(root);    return 0;}

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