hdu 1247 字典树以及map+string 2种做法

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3847    Accepted Submission(s): 1456

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

 

Recommend

Ignatius.L

 

#include<stdio.h>
#include<string.h>
#include<malloc.h>
struct haha
{
    int cnt;
    struct haha *next[26];
}q,temp;
char str[50010][100];
struct haha *root;

struct haha *creat()
{
    int i;
    struct haha *p;
    p=(struct haha *)malloc(sizeof(struct haha));
    p->cnt=0;
    for(i=0;i<26;i++)
    {
        p->next[i]=NULL;
    }
    return p;
}
void insert(char *s)
{
    int d,i,pos;
        struct haha *p=root;
        d=strlen(s);
        for(i=0;i<d;i++)
        {
            pos=s[i]-'a';
            if(p->next[pos]==NULL)
            {
                p->next[pos]=creat();
                p=p->next[pos];
            }
            else
                p=p->next[pos];
        }
        p->cnt=1;
}
int find2(char *s)
{
    int i,pos,d;
    struct haha *p;
    p=root;
    d=strlen(s);
//    printf("d=%d \n",d);
    for(i=0;i<d;i++)
    {
        pos=s[i]-'a';
        if(p->next[pos]!=NULL)
        {
               p=p->next[pos];
             
        }
        else return 0;
    }
  if(p->cnt==1)
          return 1;
    return 0;

}
int find1(char *s)
{
    int i,pos,d;
    struct haha *p;
    p=root;
    d=strlen(s);
//    printf("d1=%d\n",d);
    for(i=0;i<d;i++)
    {
        pos=s[i]-'a';
        if(p->next[pos]!=NULL)
        {
            if(p->cnt==1) {if(find2(s+i)) return 1;}
            p=p->next[pos];
        }
        else return 0;
    }
    return 0;

}

int main()
{
    int i,count=0,t=6;
    root=creat();
    while(scanf("%s",str[count])!=EOF)
//    while(t--)
    {
    //    scanf("%s",str[count]);
        insert(str[count]);
        count++;
    }
    for(i=0;i<count;i++)
    {
        if(find1(str[i]))
            puts(str[i]);
    }
    return 0;
}

map+string

#include <iostream>
#include <string>
#include <map>

using namespace std;

map <string, int> m_v;

string str[50006];

int main() {
    int k(-1);
    while(cin >> str[++k]) {
        m_v[str[k]] = 1;
    }
    for(int i = 0; i <= k; i++) {
        int e = str[i].size()-1;
        for(int j = 1; j < e; j++) {
            string s1(str[i], 0, j);
            string s2(str[i], j);
            if(m_v[s1] == 1 && m_v[s2] == 1) {
                cout << str[i] << endl;
                break;
            }
        }
    }
    return 0;
}