HDU 1247 Hat’s Words

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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

题目大意：找出一些单词满足，在输入的所有单词中，这些单词恰好可以是两个单词合成而成，并按字典许输出。

思路：建字典树，然后美剧每个单词的断点，找分裂的两个单词是否在字典树中，若存在则输出这个单词。

#include<stdio.h>#include<string.h>int ch[7000010][26];int val[7000010] ,sz=1;char s[50005][50];char a[50] ,b[50];int id(char c) {return c-'a';}void Inser(char *s) {    int u=0;    for(int i=0;s[i];i++) {        int c=id(s[i]);        if(!ch[u][c]) ch[u][c]=sz++;        u=ch[u][c];    }    val[u]=1;}int Find(char *s ){    int u=0;    for(int i=0;s[i];i++) {        int c=id(s[i]);        if(!ch[u][c]) return 0;        u=ch[u][c];    }    if(val[u] == 1) return 1;    else return 0;}int main(){    int k=0;    while(~scanf("%s",s[k])) {        Inser(s[k]);        k++;    }    for(int i=0;i<k;i++) {        int n=strlen(s[i]);        for(int j=0; j<n-1 ;j++) {            memset(a,0,sizeof(a));            memset(b,0,sizeof(b));            int kk=0 ,aa ,bb;            for(aa=0;aa<=j;aa++) a[aa]=s[i][aa];            for(bb=j+1;bb<n;bb++) b[kk++]=s[i][bb];            if(Find(a) && Find(b)) {                puts(s[i]);                break;            }        }    }    return 0;}

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