POJ 2001：Shortest Prefixes

POJ 2001：Shortest Prefixes

Shortest Prefixes

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

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

题意：给出若干字符串，求把他们缩写所能形成的最小程度。

如：

abdef

abcdefg

ab

其缩写最小程度为

abd

abc

ab

可以使用字典树求出衍生边为1的各个单词，然后输出！

#include#include#includestruct tree{ char c; struct tree *next[27]; int num;};void init(char *c,tree *T){ tree *p; for(int i=0; i<(int)strlen(c); i++) { if(T->next[c[i]-'a']==NULL) { p=(tree*)malloc(sizeof(tree)); T->next[c[i]-'a']=p; p->c=c[i]; T=T->next[c[i]-'a']; memset(T->next,0,sizeof(T->next)); T->num=1; } else { T=T->next[c[i]-'a']; T->num++; } }}void PR(char *c,tree *t){ t=t->next[c[0]-'a']; for(int i=0; i<(int)strlen(c); i++) { if(t->num>1) { printf("%c",t->c); t=t->next[c[i+1]-'a']; } else { printf("%c",t->c); break; } }}int main(){ char c[1005][35]={{0}}; int i; tree *T=(tree*)malloc(sizeof(tree)); T->num=0; memset(T->next,0,sizeof(T->next)); for(i=0; gets(c[i]); i++) { init(c[i],T); } for(int k=0; k

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