poj 2001 Shortest Prefixes(字典树)

题目链接:http://poj.org/problem?id=2001

思路分析：

在Trie结点中添加数据域childNum，表示以该字符串为前缀的字符数目；

在创建结点时,路径上的所有除叶子节点以外的结点的childNum增加1,叶子结点的childNum设置为1；

在查询某个结点的最短前缀时：

(1)若查找路径上的所有结点的childNum大于1,表明该字符串的最短路径为其自身;

(2)若查找路径上存在结点的childNum等于1,表明查找的字符串是唯一以该字符串为前缀的字符串;

 

代码如下：

#include <iostream>
#include <cstdlib>

const int MAXN = 26;
const int N = 1000 + 10;
const int M = 20 + 1;
char dic[N][M];
struct Trie
{
    int childNum;
    Trie *child[MAXN];
    Trie()
    {
        for (int i = 0; i < MAXN; ++ i)
            child[i] = NULL;
        childNum = 1;
    }
};

Trie *root = NULL;
void insert(char *word)
{
    Trie *cur = root;
    int len = strlen(word);

    for (int i = 0; i < len; ++ i)
    {
        int id = word[i] - 'a';
     
        if (cur->child[id] == NULL)
            cur->child[id] = new Trie;
        else
            cur->child[id]->childNum++;
        cur = cur->child[id];
    }
}

int findShortestPrefixes(char *word)
{
    int i, len = strlen(word);
    Trie *cur = root;

    for (i = 0; i < len; ++ i)
    {
        int id = word[i] - 'a';

        cur = cur->child[id];
        if (cur->childNum <= 1)
            return i;
    }
    return -1;
}

int main()
{
    int count = 0;

    root = new Trie;
    while (scanf("%s", dic[count]) != EOF)
        insert(dic[count++]);

    for (int j = 0; j < count; ++ j)
    {
        int ans = findShortestPrefixes(dic[j]);

        if (ans == -1)
            printf("%s %s\n", dic[j], dic[j]);
        else
        {
            printf("%s ", dic[j]);
            dic[j][ans + 1] = '\0';
            printf("%s\n", dic[j]);
        }
    }

    return 0;
}