描述
http://poj.org/problem?id=2001
给出一组单词,求每个单词的最小唯一前缀.
最小唯一前缀:该前缀不能是其他单词的前缀,并且最小,如果不存在,则为该单词本身.
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 16921 | Accepted: 7338 |
Description
In the sample input below, "carbohydrate" can be abbreviated to
"carboh", but it cannot be abbreviated to "carbo" (or anything shorter)
because there are other words in the list that begin with "carbo".
An exact match will override a prefix match. For example, the prefix
"car" matches the given word "car" exactly. Therefore, it is understood
without ambiguity that "car" is an abbreviation for "car" , not for
"carriage" or any of the other words in the list that begins with "car".
Input
input contains at least two, but no more than 1000 lines. Each line
contains one word consisting of 1 to 20 lower case letters.
Output
output contains the same number of lines as the input. Each line of the
output contains the word from the corresponding line of the input,
followed by one blank space, and the shortest prefix that uniquely
(without ambiguity) identifies this word.
Sample Input
carbohydrate
cart
carburetor
caramel
caribou
carbonic
cartilage
carbon
carriage
carton
car
carbonate
Sample Output
carbohydrate carboh
cart cart
carburetor carbu
caramel cara
caribou cari
carbonic carboni
cartilage carti
carbon carbon
carriage carr
carton carto
car car
carbonate carbona
Source
分析
唯一前缀就要求前缀的末位置只出现过一次,所以在Trie中insert的时候顺便记下每个点经过的次数.查找时一直找到第一个只出现了一次的点,或者查完整个单词.
p.s.打打Trie模板,已经忘干净了.
#include <cstdio>
#include <cstring>
using namespace std; const int type=; struct Trie{
struct node{
node* next[type];
int num;
node(){
num=;
for(int i=;i<type;i++) next[i]=NULL;
}
}*root;
Trie(){ root=new node; }
inline int id(char c) { return c-'a'; }
void insert(char *c){
node* o=root;
while(*c){
int t=id(*c);
if(o->next[t]==NULL) o->next[t]=new node;
o=o->next[t];
o->num++;
c++;
}
}
void search(char *c){
node* o=root;
while(*c){
int t=id(*c);
o=o->next[t];
printf("%c",*c);
if(o->num==) return;
c++;
}
}
}tree;
char c[][]; int main(){
int cnt=;
while(scanf("%s",c[++cnt])!=EOF) tree.insert(c[cnt]);
for(int i=;i<=cnt;i++){
printf("%s ",c[i]);
tree.search(c[i]);
printf("\n");
}
return ;
}