HDU 1247 Hat’s Words （字典树 && map）

分析：一开始是用递归做的，没做出来，于是就换了现在的数组。即，把每一个输入的字符串都存入二维数组中，然后创建字典树。输入和创建完毕后，开始查找。

其实一开始就读错题目了，题目要求字符串是由其他两个输入的字符串组成的前后缀，自己根本没有判断前缀是否满足，就直接判断后缀，一直想不通自己哪里错了，很惭愧，水平还是不行。

查找分为前缀查找和后缀查找，其实两个函数基本差不多的。下面放代码。

#include <cstdio>#include <cstring>#include <iostream>using namespace std;struct trie{     trie *next[26]; int v;    //字符相同的个数 trie() { memset(next,NULL,sizeof(next)); v=1; }};trie *root=new trie();char s[50001][101];void creat_trie(char *str){    int i,id;trie *p;    for(p = root,i=0;str[i]; ++i)    {        id = str[i]-'a';        if(p->next[id] == NULL)        {            p->next[id] = new trie();        }p = p->next[id];}p->v = -1;}int find_trie(char *str){int i=0,j,id;trie *p = root;    for(;*str != '\0';)    {        id= *str - 'a' ;if (p->next[id] != NULL){p = p->next[id];if(p->v == -1 && *(str+1) == '\0')return 1;str++;}elsereturn 0;    }return 0;}int find(char *str){trie *p = root;int m;for (;*str != '\0';){m = *str - 'a';p = p->next[m];if(p != NULL){if (p->v == -1 && find_trie(str+1)){return 1;}str++;}else return 0;} return 0;}int main(){int N,n,i=0,j,t,m,flag=0;int a,b,c,d,k;int sum;trie *p;while (gets(s[i]),s[i][0] != '\0')//,s[i][0] != '\0'{creat_trie(s[i++]);}for (j=0;j<i;j++){   if (find(s[j]))   {   puts(s[j]);   }}return 0;}

代码2：map容器

#include <iostream>#include <map>#include <cstring>using namespace std;map <string,int> m;string s[50005];int main() {int k=-1;while(cin>>s[++k]){m[s[k]] = 1;}for(int i=0;i<=k;i++){int len = s[i].length();for(int j=1;j<len;j++){string s1(s[i],0,j);  //从0开始的j个数string s2(s[i],j);    //从j开始(不包括)一直到结尾if(m[s1] == 1 && m[s2] == 1){cout<<s[i]<<endl;break;}   }}return 0;}

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9436    Accepted Submission(s): 3369

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