Hat’s Words
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14083 Accepted Submission(s): 5049
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
Recommend
Ignatius.L
/*
这题询问Trie树上是否存在一个单词,
满足拆成两部分,使得两部分单词都在树上.
然后其实直接从树上找一个单词
把是单词结尾的位置压栈
然后暴力枚举断点检验即可.
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#define MAXN 300001
#define MAXM 50001
using namespace std;
int tot;
char s[MAXM][27];
struct data{int next[27];bool b;bool w;}
tree[MAXN];
void Add_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 tot++,tree[now].next[x]=tot,now=tot;
}
tree[now].b=true;
}
bool jd(int l,char s[])
{
int i=0,top=0,stack[1001],now=0;
while(s[i])
{
if(tree[now].next[s[i]-96]) now=tree[now].next[s[i]-96];
else return 0;
if(tree[now].b&&s[i])//所谓割点
stack[top++]=i+1;
i++;
}
while(top)//检验
{
int now=0;
bool flag=1;
int x=stack[--top];
while(s[x])
{
if(!tree[now].next[s[x]-96]){flag=false;break;}
now=tree[now].next[s[x]-96];
x++;
}
if(tree[now].b&&flag)//该结点是单词的结尾
return 1;
}
return 0;
}
int main()
{
int i=1;
while(gets(s[i])&&strlen(s[i]))
{
int l=strlen(s[i]);
Add_tree(l,s[i]);
i++;
}
for(int j=1;j<=i;j++)
{
int l=strlen(s[j]);
if(jd(l,s[j]))
cout<<s[j]<<endl;
}
return 0;
}