[HDOJ1247]Hat’s Words
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1247
Hat’s Words
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10982 Accepted Submission(s): 3937
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
在所有单词中找被其他两个单词表示的单词,全部丢到字典树里维护,暴力枚举每个单词,再将每个单词「芙兰达」判断是否在字典树中维护即可。注意如果找到符合情况的要及时跳出内层的枚举。
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <algorithm> 5 #include <iostream> 6 #include <cmath> 7 #include <queue> 8 #include <map> 9 #include <stack> 10 #include <list> 11 #include <vector> 12 13 using namespace std; 14 15 char str[66666][81]; 16 17 typedef struct Node { 18 Node *next[26]; 19 int cnt; 20 Node() { 21 cnt = 0; 22 for(int i = 0; i < 26; i++) { 23 next[i] = NULL; 24 } 25 } 26 }Node; 27 28 void insert(Node *p, char *str) { 29 for(int i = 0; str[i]; i++) { 30 int t = str[i] - 'a'; 31 if(p->next[t] == NULL) { 32 p->next[t] = new Node(); 33 } 34 p = p->next[t]; 35 } 36 p->cnt++; //is a word 37 } 38 39 int find(Node *p, char *str) { 40 for(int i = 0; str[i]; i++) { 41 int t = str[i] - 'a'; 42 p = p->next[t]; 43 if(!p) { 44 return 0; 45 } 46 } 47 if(p->cnt >= 1) { 48 return 1; 49 } 50 return 0; 51 // return p->cnt; 52 } 53 54 int main() { 55 int n = 0; 56 Node *root = new Node(); 57 // freopen("in", "r", stdin); 58 while(gets(str[n]) && strlen(str[n])) { 59 insert(root, str[n]); 60 n++; 61 } 62 char a[81]; 63 char b[81]; 64 for(int i = 0; i < n; i++) { 65 int len = strlen(str[i]); 66 for(int j = 1; j < len; j++) { 67 memset(a, 0, sizeof(a)); 68 memset(b, 0, sizeof(b)); 69 strncpy(a, str[i], j); 70 strncpy(b, str[i] + j, len - j); 71 if(find(root, a) && find(root, b)) { 72 printf("%s\n", str[i]); 73 break; 74 } 75 } 76 } 77 return 0; 78 }
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】博客园社区专享云产品让利特惠,阿里云新客6.5折上折
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步