hdoj 1247 Hat’s Words 【字典树】【输出字符串集里面 可以由任意两个字符串所构成的字符串】

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9380    Accepted Submission(s): 3352

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
   
   
   
   

 

插入字符串时标记一下单词节点就ok了。

 

#include <cstdio>
#include <cstring>
#define MAX 50001
using namespace std;
char str[MAX][101];
int ch[MAX][101];
int word[MAX];//这里 记录单词数目这个数组没用上 
int val[MAX];//标记是否为单词节点 
int sz, n;
void init()
{
    sz = 1;
    memset(ch[0], 0, sizeof(ch[0]));
    memset(word, 0, sizeof(word));
}
int idx(char x)
{
    return x - 'a';
}
void insert(char *s)
{
    int i, j, l = strlen(s);
    int u = 0;
    for(i = 0; i < l; i++)
    {
        int c = idx(s[i]);
        if(!ch[u][c])
        {
            memset(ch[sz], 0, sizeof(ch[sz]));
            val[sz] = 0;//不是单词节点 
            ch[u][c] = sz;
            sz++;
        }
        u = ch[u][c];
        word[u]++;
    }
    val[u] = 1;
}
bool find(char *s)//判断该字符串 是不是 字符串集里面的一个字符串  
{
    int i, j, l = strlen(s);
    int u = 0;
    for(i = 0; i < l; i++)
    {
        int c = idx(s[i]);
        if(!ch[u][c]) return false;
        u = ch[u][c];
    } 
    return val[u];//最后一个是不是单词节点 
}
int judge(char *s)//判断该字符串 能否由另外两个字符串构成 
{
    int i, j, l = strlen(s);
    if(l == 1)
    return 0;
    char front[MAX], after[MAX];
    char convert[MAX];
    strcpy(convert, s);
    strrev(convert);
    int k = 0;
    for(i = 0; i < l; i++)
    {
        if(i == l-1)
        break;
        front[k++] = s[i];
        front[k] = '\0';
        convert[l-k] = '\0'; 
        if(find(front))//前面一部分可以找到 
        {
            strcpy(after, convert);
            strrev(after);
            if(find(after))//后面一部分可以找到 
            return 1;
        }
    }
    return 0;
}
int main()
{
    n = 0;
    init();
    while(scanf("%s", str[n]) != EOF)
    {
        insert(str[n]);
        n++;
    }
    for(int i = 0; i < n; i++)
    {
        if(judge(str[i]))//查找成功 
        printf("%s\n", str[i]);
    }
    return 0;
}