題干：

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

題目大意：

? ? 輸入N個單詞，輸出N個單詞本身以及在該單詞在字典中不會與其他單詞混淆的最短前綴縮寫。

解題報告：

? 裸的字典樹。。不解釋。注意字符串的讀入方法和rt每次查詢的初始化就可以了。。。但是這題好像不用初始化也可以AC不知道為什么、、、

AC代碼：

#include<cstdio> #include<iostream> #include<cstring> using namespace std; char a[10001][31]; int tot,size; int sz,trie[300001][26],s[300001]; void insert(char ch[]) {int len=strlen(ch),rt=0;for(int i=0; i<len; i++) {int cur=ch[i]-'a';if(trie[rt][cur]==0) trie[rt][cur]=++sz;rt=trie[rt][cur];s[rt]++;} } void ask(char ch[]) {int now=0,rt=0,len=strlen(ch);for(int i=0; i<len; i++) {int cur = ch[i]-'a';printf("%c",ch[i]);rt=trie[rt][cur];if(s[rt]==1) break; //也可以放到int cur = ch[i] - 'a'上面} } int main() {char str[1005];while(~scanf("%s",str)) {tot++;strcpy(a[tot],str);insert(a[tot]);}for(int i=1; i<=tot; i++) {printf("%s ",a[i]);ask(a[i]);printf("\n");}return 0; }

?

總結

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