ZOJ 1109 ————————字典树基础
We all know that FatMouse doesn't speak English. But now he has to be prepared since our nation will join WTO soon. Thanks to Turing we have computers to help him.
Input Specification
Input consists of up to 100,005 dictionary entries, followed by a blank line, followed by a message of up to 100,005 words. Each dictionary entry is a line containing an English word, followed by a space and a FatMouse word. No FatMouse word appears more than once in the dictionary. The message is a sequence of words in the language of FatMouse, one word on each line. Each word in the input is a sequence of at most 10 lowercase letters.
Output Specification
Output is the message translated to English, one word per line. FatMouse words not in the dictionary should be translated as "eh".
Sample Input
dog ogday cat atcay pig igpay froot ootfray loops oopslay atcay ittenkay oopslay
Output for Sample Input
cat eh loops
struct Trie{ int find; char s[15]; struct Trie *next[26]; };当且仅当,构造完一个单词的时候,例如acm这个单词:
#include <stdio.h> #include <string.h> #include <malloc.h> struct Tree{ int num; char s[15]; struct Tree *next[27]; }; struct Tree *root; void Initialize() { root = (struct Tree*)malloc(sizeof(struct Tree)); root->num = 0; for(int i = 0 ; i < 26 ; ++i) root->next[i] = NULL; } void Insert(char *s1,char *s2)//以s1为基础建立字典树 { int len = strlen(s1); struct Tree *node,*p = root; for(int i = 0 ; i < len ; ++i){ int pos = s1[i] - 'a'; if(p->next[pos] == NULL){ //printf("%c %d %c\n",s1[i],pos,pos + 'a'); node = (struct Tree*)malloc(sizeof(struct Tree)); for(int j = 0 ; j < 26 ; ++j){ node->next[i] = NULL; node->num = 0; } p->next[pos] = node; } p = p->next[pos]; } p->num = 1; strcpy(p->s,s2); } void Find(char *s) { int len = strlen(s); int i; struct Tree *p = root; for(i = 0 ; i < len ; ++i){ int pos = s[i] - 'a'; if(p->next[pos] != NULL) p = p->next[pos]; else break; } if(p->num == 1) printf("%s\n",p->s); else printf("eh\n"); } int main() { char str[30]; Initialize(); while(gets(str)){ if(strlen(str) == 0) break; char str1[12],str2[12]; memset(str1,0,sizeof(str1)); memset(str2,0,sizeof(str2)); for(int i = 0 ; i < strlen(str) ; ++i){ if(str[i] == ' '){ strncpy(str1,str,i); str1[i] = '\0'; strncpy(str2,str+i+1,strlen(str)-i-1); str2[strlen(str) - i - 1] = '\0'; //printf("%s %s\n",str1,str2); } } Insert(str2,str1); } while(scanf("%s",str) != EOF) Find(str); return 0; }
还有一种非常坑爹的做法...就是C++的STL的map,代码量很短也不用思考,就是速度很慢
#include <iostream> #include <map> #include <cstdio> #include <string> #include <cstring> using namespace std; map<string,string> M; int main() { char a[12],b[12],str[30]; while(gets(str)) { if(strcmp(str,"")==0)break; int i,l=strlen(str); for(i=0;i<l;++i) { if(str[i]==' ')break; } strncpy(a,str,i); a[i]='\0'; strncpy(b,str+i+1,l-i); b[l-i]='\0'; // cout<<a<<endl<<b<<endl; M[b]=a; } while(cin>>b) { if(M[b]=="") { cout<<"eh"<<endl; } else { cout<<M[b]<<endl; } } return 0; }