Hat’s Words

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 #include #include #include #include char a[50001][50]; struct tire { int n; struct tire *next[26]; } *root; void jianshu(char a[50]) { tire *p1,*p2; p1=root; int n,i,id; n=strlen(a); for(i=0; i<n; i++) { id=a[i]-'a'; if(p1->next[id]==NULL) { p2=(tire*)malloc(sizeof(tire)); for(int j=0; j<26; j++) p2->next[j]=NULL; p2->n=0; p1->next[id]=p2; p1=p2; } else p1=p1->next[id]; } p1->n=1; } int find1(char b[50]) { tire *p1; p1=root; int n,i,id; n=strlen(b); for(i=0; i<n; i++) { id=b[i]-'a'; if(p1==NULL||p1->next[id]==NULL)return 0; p1=p1->next[id]; } return p1->n; } void xiao(tire *p1) { if(p1==NULL)return ; int i; for(i=0; i<26; i++) if(p1->next[i]!=NULL)xiao(p1->next[i]); free(p1); } int main() { int i=0,n; root=(tire*)malloc(sizeof(tire)); for(i=0; i<26; i++) root->next[i]=NULL; root->n=0; i=0; while(scanf("%s",a[i])!=EOF) jianshu(a[i]),i++; for(int j=0; j<i; j++) { n=strlen(a[j]); for(int k=1; k<n; k++) { char b[50]= {''},c[50]= {''}; strncpy(b,a[j],k); strncpy(c,a[j]+k,n-k); if(find1(b)&&find1(c)) { printf("%s ",a[j]); break; } } } xiao(root); }