HDU - 1247 Hat’s Words 字典树

Posted 2020-09-30 gavanwanggw

HDU - 1247 Hat’s Words Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u Submit Status 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: 2486 Memory: 9096 KB Time: 31 MS Language: G++ Result: Accepted */ //这道题目思路非常easy //就是找两个单词组成还有一个单词就可 #include <cstdio> #include <vector> #include <string> #include <iostream> #include <queue> #include <cstring> #include <algorithm> using namespace std; const int MAXN = 1000000 + 5; const int MAXM = 50000 + 5; const int INF = 0x3f3f3f3f; struct node { int v,next[30]; void init() { v = -1; memset(next, -1, sizeof(next)); } } L[MAXN]; int tot; char str[MAXM][100]; priority_queue<string , vector<string>, greater<string> >P; void add(char * a, int len) { int now = 0; for(int i = 0; i < len ; i ++) { int tmp = a[i] - 'a'; int next = L[now].next[tmp]; if(next == -1) { next = ++ tot; L[next].init(); L[now].next[tmp] = next; } now = next; } L[now].v = 0; } bool querys(char * a) { int len = strlen(a); int now = 0; for(int i = 0; i < len; i ++) { int tmp = a[i] - 'a'; int next = L[now].next[tmp]; if(next == -1) return false; now = next; } return L[now].v == 0; } bool query(char * a, int len) { int now = 0; for(int i = 0; i < len; i ++) { int tmp = a[i] - 'a'; int next = L[now].next[tmp]; if(L[now].v == 0) { if(querys(a + i)) return true;//是否存在还有一个单词能够组成剩下的字符串 } now = next; } return false; } int main() { int cnt = 0, Min = INF; L[0].init(); tot = 0; //freopen("D://imput.txt", "r", stdin); while(~scanf("%s", str[cnt ++])) { add(str[cnt - 1], strlen(str[cnt - 1])); Min = min(Min, int(strlen(str[cnt - 1]))); } for(int i = 0; i < cnt ; i ++) { if(Min * 2 > strlen(str[i]))continue; if(query(str[i], strlen(str[i]))) P.push(str[i]); } string s; while(!P.empty()) {//优先队列自己主动排序也可使用set或者map s = P.top(); P.pop(); cout<<s<<endl; } return 0; }