Hats’World

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
题目大意：
给定一组单词，让你找到一个单词，使得这个单词由这组单词中出现过的两个单词构成。
例：ahat由a和hat构成，而a和hat在这组单词中出现过。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=50010;
struct node
{
    int next[27];
    int b;
}tree[maxn];
int tot=1,sum,top,stack[maxn];
char s[maxn][27];
void build_tree(int l,char s[])
{
    int now=0;
    for(int i=0;i<l;i++)
    {
        int x=s[i]-96;
        if(tree[now].next[x])
        now=tree[now].next[x];
        else
        {
            tree[now].next[x]=++sum;
            now=sum;
        }
    }
    tree[now].b=true;
}
int can(int l,char s[])
{
    int i=0,now=0,top=0;
    for(int i=0;i<l;i++)
    {
        int x=s[i]-96;
        if(tree[now].next[x])
        now=tree[now].next[x];
        else return 0;
        if(tree[now].b&&s[i])
        stack[top++]=i+1;
    }
    while(top)
    {
        int now=0;
        bool flag=1;
        int x=stack[--top];
        while(s[x])
        {
            if(!tree[now].next[s[x]-96])
            {
                flag=0;
                break;
            }
            now=tree[now].next[s[x]-96];
            x++;
        }
        if(flag&&tree[now].b)
        return 1;
    }
    return 0;
}
int main()
{
    while(gets(s[tot])&&strlen(s[tot]))
    {
        int len=strlen(s[tot]);
        build_tree(len,s[tot]);
        tot++;
    }
    for(int i=1;i<=tot;i++)
    {
        int len=strlen(s[i]);
        if(can(len,s[i]))
        cout<<s[i]<<endl;
    }
    return 0;
}