Hdu Hat's Words (Dictionary tree)

Hat's WordsTime Limit: 2000/1000 ms (Java/Other) Memory Limit: 65536/32768 K (Java/Other) Total Submission (s): 7 Accepted Submission (s): 4 Font: times New Roman | Verdana | GeorgiaFont Size: Regular → Problem DescriptionA 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. inputStandard 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. OutputYour output shoshould contain all the hat's words, one per line, in alphabetical order. Sample Input

aahathathatwordhzieeword

Sample Output

ahathatword

Author determines whether a word can be composed of two other words. Note that other words may also consist of several words! Use the dictionary tree to query a word. When a word becomes obsolete, I search for the remaining part of the word. In this step, I use recursion to control the number of recursive layers with a count variable, but the code is messy and there is not much reference value.

# Include <stdio. h> # include <string. h> int cnt; int ant; char text [50005] [105]; struct Node {int next [26]; int flag;} tree [100000]; void Insert (char * s) {int p = 0; while (* s) {if (tree [p]. next [* s-'a'] =-1) {tree [p]. next [* s-'a'] = cnt; cnt ++;} p = tree [p]. next [* s-'a']; s ++;} tree [p]. flag = 1;} // use count as the flag to record the nth layer of int Query (char * s, int count) {int p = 0; while (* s) {if (tree [p]. next [* s-'a'] =-1 ){ Return 0;} p = tree [p]. next [* s-'a']; // when the dictionary word if (tree [p]. flag = 1 & * (s + 1) {// when the first word if (count = 1) {count ++; Query (s + 1, count); // search for the remaining words count = 1; // restore count to 1 and continue searching in the first recursion} // when searching for the remaining words, it is also possible that the remaining words are composed of more than two words // Therefore, You Need To separately determine else if (count = 2) {s ++; continue ;}} s ++ ;} if (tree [p]. flag = 1) {// a messy place... Do not tangle if (ant = 2) {return 0;} ant ++;} return 0;} int main () {int I, j; int len; cnt = 1; for (I = 0; I <100000; I ++) {memset (tree [I]. next,-1, sizeof (tree [I]. next); tree [I]. flag = 0;} I = 0; while (scanf ("% s", text [I])! = EOF) {Insert (text [I]); I ++;} len = I; for (j = 0; j <len; j ++) {ant = 0; // Query (text [j], 1); if (ant = 2) {printf ("% s \ n ", text [j]) ;}} return 0 ;}