Shortest Prefixes POJ - 2001（字典树）

    

Shortest Prefixes

POJ - 2001

A prefix of a string is a substring starting at the beginning of the given string. The prefixes of "carbon" are: "c", "ca", "car", "carb", "carbo", and "carbon". Note that the empty string is not considered
a prefix in this problem, but every non-empty string is considered to be a prefix of itself. In everyday language, we tend to abbreviate words by prefixes. For example, "carbohydrate" is commonly abbreviated by "carb". In this problem, given a set of words,
you will find for each word the shortest prefix that uniquely identifies the word it represents.

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

The 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

The 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

思路：把所给的字符串都放进字典树，记录每个前缀出现的次数，然后对于每个字符串进行一个字符一个字符从头查找，当查找到这个前缀记录次数为1的时候，说明这个前缀只有

这个字符串有，这个前缀即为所求前缀

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
using namespace std;
char s[1009][35];//存放字符串
struct TNode{
int time;
struct TNode* Next[26];//字典树结构体
}tre;
typedef struct TNode* Tree;
void Insert(char a[]){//字典树插入
int i,j;
Tree p = &tre;
Tree q;
for(i = 0; a[i]; i++){
if(p->Next[a[i]-'a']==NULL){
q = (Tree)malloc(sizeof(struct TNode));
q->time = 0;
for(j = 0; j < 26; j++)q->Next[j] = NULL;
p->Next[a[i]-'a'] = q;
}
p = p->Next[a[i]-'a'];
p->time++;
}
}
void Find(char a[]){
int i,j;
Tree p = &tre;
for(i = 0; a[i]; i++){
printf("%c",a[i]);//每次输出一个字符
p = p->Next[a[i]-'a'];
if(p->time==1){//判断这个字符，以这个字符为末尾的前缀出现次数是否为1，是1就终止，当前输出的即所求前缀
return;
}
}
}
int main(){
int i;
int cnt = 0;
for(i = 0; i < 26; i++)tre.Next[i] = NULL;//初始化头节点
while(~scanf("%s",s[cnt++]));
for(i = 0; i < cnt; i++){
Insert(s[i]);//输入
}
for(i = 0; i < cnt; i++){
printf("%s ",s[i]);
Find(s[i]);//查找
puts("");
}
return 0;
}