Shortest Prefixes
Time Limit:1000MSMemory Limit:30000K
Total Submissions:14336Accepted:6184
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 car
carbonate carbona
字典树入门题！
题意：
输出在字典中有相同的前缀+1个字符来唯一识别这个单词，该单词在字典中是前缀，，则输出它。
代码：
/*1、对每个结点开一个字母集大小的数组，对应的下标是儿子所表示的字母，内容则是这个儿子对应在大数组上的位置，即标号；*/#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int Max=30;const int maxn=11000;typedef struct node{int count;// 该节点前缀 出现的次数struct node *next[Max];//该节点的后续节点}node;char s[1020][25];node a[maxn];int cur=0;int n=0;node *root;//初始化一个节点。nCount计数为1， next都为nullnode *creat(){node *tmp=&a[cur++];tmp->count=1;for(int i=0;i<Max;i++)tmp->next[i]=NULL;return tmp;}void insert(char *str){int temp,len;node *p;p=root;len=strlen(str);for(int i=0;i<len;i++){temp=str[i]-'a';if(p->next[temp]!=NULL){p=p->next[temp];p->count++;}else{p->next[temp]=creat();p=p->next[temp];}}}void find(char *str){int len,temp;node *p;p=root;len=strlen(str);for(int i=0;i<len;i++){temp=str[i]-'a';p=p->next[temp];if(p->count>1)printf("%c",str[i]);else{printf("%c",str[i]);//多输出了一个return;}}}int main(){root=creat();//根节点while(~scanf("%s",s[n])){insert(s[n]);//建树n++;}for(int i=0;i<n;i++){printf("%s ",s[i]);find(s[i]);//查询printf("\n");}return 0;}
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