Hat’s Words Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19500 Accepted Submission(s): 6867Problem 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 wordSample Output ahat hatwordAuthor 戴帽子的Recommend Ignatius.L | We have carefully selected several similar problems for you: 1075 1671 1298 1800 2846 T

這道題也是比較簡單的trie樹的題，開始覺得很懵逼，最后才發現其實暴力就可以

對于每個單詞枚舉斷點，然后查前后是否都存在

我們來計算下時間復雜度，假設n個單詞，每個單詞長度最大為m

那么插入O（nm）；

查詢時候O（nmm）；

總復雜度O（nm+m2n）；

這道題的n是50000，m假設是30，那么很明顯可以過

上代碼

1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #define idx(i) (i-'a') 5 #define N 50011 6 using namespace std; 7 char in[N][300]; 8 int cnt=1,pp; 9 struct TRIE{int nxt[30],cnt;}tree[N*300]; 10 inline int regist(){return cnt++;} 11 void insert(char *now) 12 { 13 int c,rt=0,len=strlen(now); 14 for(int i=0;i<len;i++) 15 { 16 c=idx(now[i]); 17 if(!tree[rt].nxt[c]) 18 tree[rt].nxt[c]=regist(); 19 rt=tree[rt].nxt[c]; 20 } 21 tree[rt].cnt=1; 22 } 23 bool find(char *now,int st,int ed) 24 { 25 int rt=0; 26 for(int i=st-1;i<ed;i++) 27 { 28 if(!tree[rt].nxt[idx(now[i])])return 0; 29 rt=tree[rt].nxt[idx(now[i])]; 30 } 31 return tree[rt].cnt; 32 } 33 int main() 34 { 35 while(scanf("%s",in[++pp]+1)!=EOF)insert(in[pp]+1); 36 for(int i=1;i<pp;i++) 37 { 38 int len=strlen(in[i]+1); 39 for(int j=1;j<=len;j++) 40 if(find(in[i]+1,1,j)&&find(in[i]+1,j+1,len)){printf("%s\n",in[i]+1);break;} 41 } 42 return 0; 43 }

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轉載于:https://www.cnblogs.com/Qin-Wei-Kai/p/10224182.html

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