字典树trie树 模板题 POJ2001 Shortest Prefixes

Shortest Prefixes

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 19565 Accepted: 8486

Description

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

carbohydratecartcarburetorcaramelcariboucarboniccartilagecarboncarriagecartoncarcarbonate

Sample Output

carbohydrate carbohcart cartcarburetor carbucaramel caracaribou caricarbonic carbonicartilage carticarbon carboncarriage carrcarton cartocar carcarbonate carbona

Source

Rocky Mountain 2004

题目大意：给定不超过1000个长度不超过20的子串，找出能唯一表示每个子串的最短前缀，如果找不出，就输出该字符串。

那当然是trie树的裸题啦，睡前写一道喔。

trie树，就是每一个节点向下发出26条边(可以不是，但这里是小写字母所以26条)，这是一种存储子串的方法，也叫字典树。

这棵树中的节点有些是危险节点，即从根节点走到当前节点所经过的边上字母组成的字符串在子串中出现过。

对于此题，显而易见，如果在某一个节点往下再走只有一个危险节点了，那么这一定是最短前缀啦~

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cstdlib>#include <cmath>using namespace std;char a[2000][100];int cnt,n,t[300300][26],sum[300300];void insert(char ch[]){int len=strlen(ch+1),now=0;for(int j=1;j<=len;j++){int k=ch[j]-'a';if(t[now][k]==0) t[now][k]=++cnt;now=t[now][k];sum[now]++;}}void solve(char ch[]){int len=strlen(ch+1),now=0;for(int j=1;j<=len;j++){if(sum[now]==1) break;int k=ch[j]-'a';printf("%c",ch[j]);now=t[now][k];}}int main(){while(scanf("%s",a[++n]+1) != EOF) insert(a[n]);for(int i=1;i<=n;i++){printf("%s ",a[i]+1);solve(a[i]);printf("\n");}return 0;}