Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17088    Accepted Submission(s): 6141

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
 

Recommend
Ignatius.L

思路:给一个字典,输出这些词,满足条件: 该词可以拆分成两个字典中的词.

注意判断不是前缀而是字典中的词

复杂度:O(n*len)


#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define bug1 cout <<"bug1"<<endl
#define bug2 cout <<"bug2"<<endl
#define bug3 cout <<"bug3"<<endl
using namespace std;
typedef long long ll;

struct node{
    node *next[26];
    bool flag;
    node(){
        flag=false;
        for(int i=0;i<26;i++)   next[i]=NULL;
    }
};

char str[51000][200];
node *root;

void buildtrie(char *s){
    int len=(int)strlen(s);
    node *p=root;
    for(int i=0;i<len;++i){
        int num=s[i]-'a';
        if(!p->next[num]){
            node *temp=new node;
            p->next[num]=temp;
        }
        p=p->next[num];
    }
    p->flag=true;
    return ;
}

bool findtrie(char *s){
    int len=strlen(s);
    node *p=root;
    for(int i=0;i<len;++i){
        int num=s[i]-'a';
        if(!p->next[num])   return false ;
        else{
            p=p->next[num];
        }
    }
    return p->flag;
}



int main(void){
    int num=0;  
    root=new node();
    while(scanf("%s",str[num++])!=EOF){
        buildtrie(str[num-1]);
    }
    for(int i=0;i<num;++i){
        int len=strlen(str[i]);
        for(int j=0,k;j<len-1;++j){
            char a[5000],b[5000];
            for(k=0;k<=j;++k)   a[k]=str[i][k];
            a[k]='\0';
            for(k=0;k<=len-1-j;++k)   b[k]=str[i][k+j+1];
            b[k]='\0';
//            printf("a=%s b=%s\n",a,b);
            if(findtrie(a) && findtrie(b)){
                printf("%s\n",str[i]);
                break;
            }
        }
    }
    return 0;
}