POJ-2001-Shortest Prefixes（字典树入门题）

Shortest Prefixes
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 20382 Accepted: 8819
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
解题思路： 这是一道简单的字典树题，在字典树的模板上修改了如何存储字符串，但在插入时结构没有改变，在查找时加入了判断条件：如果在当前节点sum[当前节点]==1即在该节点之前的前缀需要用于代表其他与此字符串有最大相同前缀的字符串；
AC代码

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int trie[40001][26];//trie[i][j]=k，表示第i个节点的第j个子节点相对于树的编号，j用字母的ASCLL码表示
char ch[1005][25];
int total=0;//表示该节点相对与数的编号，即上述提到的k
int sum[40001];//用sum数组表示每个节点的访问次数
int len;
int counti=0;
void insert()
{
    int rt=0;
    for(int i=0;i<len;i++)
    {
        int x=ch[counti][i]-'a';
        if(!trie[rt][x])
            trie[rt][x]=++total;
        sum[trie[rt][x]]++;
        rt=trie[rt][x];
    }
}
void research(int countx)
{
    int rt=0;
    for(int i=0;i<len;i++)
    {
        int x=ch[countx][i]-'a';
        if(sum[trie[rt][x]]==1)//指当前节点与其他与此节点有相同前缀的最大相通前缀的下标+1
        {
            for(int j=0;j<=i;j++)
                printf("%c",ch[countx][j]);
            printf("\n");
             return;
        }
        if(!trie[rt][x])
            return;
        rt=trie[rt][x];
    }
    for(int j=0;j<len;j++)
    {
        printf("%c",ch[countx][j]);
    }
    printf("\n");
    return;
}
int main()
{
    memset(trie,0,sizeof(trie));
    memset(sum,0,sizeof(sum));
    memset(ch,'\0',sizeof(ch));
    while(gets(ch[++counti]))//因为scanf（）自动忽略空格，所以只能用gets()，之前因为忽略了这个问题，调试了半天
    {
        len=strlen(ch[counti]);
        insert();
    }
    for(int i=1;i<=counti;i++)
    {
        len=strlen(ch[i]);
        for(int j=0;j<len;j++)
            printf("%c",ch[i][j]);
        cout<<" ";
        research(i);
    }
    return 0;
}