Hat’s Words
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20441 Accepted Submission(s): 7162
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.
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.
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
题意：只有一个测试样例，输入到文件结束。从输入的单词中，找到一个单词是由两个输入的其他单词组成的单词，找出所有符合条件的单词。
#include<iostream> #include<string.h> #include<string> using namespace std; char s[400005][26]; int vis[400005],tree[400005][26]; int n=0,num=0,root,id,len; void insert() { len=strlen(s[n]); root=0; for(int i=0;i<len;i++) { id=s[n][i]-'a'; if(!tree[root][id]) tree[root][id]=++num;//注意是++num root=tree[root][id]; } vis[root]=1;//每个单词结束标记一下 } int search(char *ss) { len=strlen(ss); root=0; for(int i=0;i<len;i++) { id=ss[i]-'a'; if(tree[root][id]) root=tree[root][id]; if(vis[root]==1) { int root2=0;//根节点继续从0开始 for(int j=i+1;j<len;j++) { int id2=ss[j]-'a'; if(!tree[root2][id2]) break; root2=tree[root2][id2]; if(vis[root2]==1&&j==len-1) return 1; } } } return 0; } int main() { while(~scanf("%s",&s[++n])) { insert(); } for(int i=1;i<=n;i++) { if(search(s[i])) printf("%s\n",s[i]); } return 0; }