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

题意是给出一些字符串，找出每个字符串对应的唯一能表示该字符串的最短前缀（不会和其他字符串冲突）
用字典树节点维护一个经过该节点的字符串个数，然后find的时候从开始输出字符，直到cnt==1，即经过这个节点只有一个字符串，那就是唯一标识了，return

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
using namespace std;
char s[30];
char t[1005][30];
int k;
struct Node{
    int cnt;
    Node* Next[26];
    Node(int x):cnt(x){
        for(int i=0;i<26;i++){
            Next[i]=NULL;
        }
    }
};
void insert(Node* &rt,char* s){
    int l=strlen(s);
    int val;
    Node* p=rt;
    for(int i=0;i<l;i++){
        val=s[i]-'a';
        if(p->Next[val]==NULL){
            p->Next[val]=new Node(0);
        }
        p=p->Next[val];
        p->cnt++;
    }
}
void find(Node* &rt,char* s){
    int l=strlen(s);
    Node* p=rt;
    int val;
    for(int i=0;i<l;i++){
        val=s[i]-'a';
        if(p->Next[val]==NULL){
            return;
        }
        p=p->Next[val];
        printf("%c",s[i]);
        if(p->cnt==1){
            return;
        }
    }
}
void clear(Node* a){
    if(a==NULL){
        return;
    }
    else{
        for(int i=0;i<26;i++){
            clear(a->Next[i]);
        }
    }
    delete(a);
}
int main(void){
    //freopen("data.txt","r",stdin);
    Node* root=new Node(0);
    while(~scanf("%s",s)){
        insert(root,s);
        strcpy(t[k++],s);
    }
    for(int i=0;i<k;i++){
        printf("%s ",t[i]);
        find(root,t[i]);
        printf("\n");
    }
    return 0;
}