hdu 1247 Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19109    Accepted Submission(s): 6748

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

 

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给出一个字典里的单词，找出这种单词：由另外两个单词组成。字典树记录所有单词，对于可能的单词进行查询。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#define MAX 600005
using namespace std;
int trie[MAX][26];
bool flag[MAX];
int pos,num;
bool may[50000];
void Insert(char *s,int snum) {
    int i = 0,c = 0;
    while(s[i]) {
        int d = s[i] - 'a';
        if(!trie[c][d]) trie[c][d] = ++ pos;
        c = trie[c][d];
        if(flag[c]) {
            may[snum] = true;
        }
        i ++;
    }
    flag[c] = true;
}
bool Query(char *s,int k,int t) {
    if(t == 2) {
        if(!s[k])return true;
        return false;
    }
    int i = k,c = 0;
    bool ok = false;
    while(s[i]) {
        int d = s[i] - 'a';
        if(!trie[c][d]) trie[c][d] = ++ pos;
        c = trie[c][d];
        i ++;
        if(flag[c]) {
            ok |= Query(s,i,t + 1);
        }
    }
    return ok;
}
char str[50000][30];
int main() {
    while(~scanf("%s",str[num])) {
        Insert(str[num],num);
        num ++;
    }
    for(int i = 0;i < num;i ++) {
        if(may[i] && Query(str[i],0,0)) {
            printf("%s
",str[i]);
        }
    }
}