POJ 2001 Shortest Prefixes 【 trie树（别名字典树)】

Shortest Prefixes

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 15574		Accepted: 6719

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

看完题后，感慨：简直不能忍，这么裸的字典树。接下来就是无脑式敲 Trie 树了。这个题目能够拿来练练手速~。另外。一时没到比較合适的函数名，望 勿喷

题解就略了，容我贴下代码~

/****************************>>>>HEADFILES<<<<****************************/
#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <sstream>
#include <algorithm>
using namespace std;
/****************************>>>>>DEFINE<<<<<*****************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define FIN             freopen("input.txt","r",stdin)
#define FOUT            freopen("output.txt","w",stdout)
#define rep(i,a,b)      for(int i = a;i <= b;i++)
#define rep1(i,a)       for(int i = 1;i <= a;i++)
#define rep0(i,a)       for(int i = 0;i < a;i++)
#define MP(a,b)         make_pair(a,b)
#define PB(a)           push_back(a)
#define fst             first
#define snd             second
#define lson            l,mid,rt<<1
#define rson            mid+1,r,rt<<1|1
/****************************>>>>>>DEBUG<<<<<<****************************/
#define out(x)          cout<<x<<""
/****************************>>>>SEPARATOR<<<<****************************/
const int maxk = 26;
const int maxl = 20+5;
int N,M;
struct Node
{
    int cnt;
    Node* pNext[maxk];
    Node() : cnt(0)
    {
        rep0(i,maxk) pNext[i] = NULL;
    }
};
struct Trie
{
    Node* const pRoot;
    Trie() : pRoot(new Node()) {}
    void AddWord(const char str[],int len);
   // int FindPredix(const char str[],int len);
    void Query(const char str[],int len);
    void Release(const Node *p);
}dic;
void Trie::AddWord(const char str[],int len)
{
    Node* ptr = pRoot;
    for(int i = 0;i < len;i++)
    {
        int nPos = str[i] - 'a';
        if(ptr->pNext[nPos] == NULL) ptr->pNext[nPos] = new Node();
        ptr->cnt++;
        ptr = ptr->pNext[nPos];
    }
    ptr->cnt++;
}
void Trie::Release(const Node* p)
{
    for(int i = 0;i < maxk;i++) if(p->pNext[i] != NULL) Release(p->pNext[i]);
    delete p;
}
void Trie::Query(const char str[],int len)
{
    Node* ptr = pRoot;
    for(int i = 0;i < len;i++)
    {
        int nPos = str[i] - 'a';
        if(ptr->pNext[nPos] == NULL) return;
        if(ptr->cnt == 1) return;
        putchar(str[i]);
        ptr = ptr->pNext[nPos];
    }
}
char buf[1005][maxl];
int main()
{
    //FIN;
    int cnt = 0;
    while(~scanf("%s",buf[cnt]))
        dic.AddWord(buf[cnt],strlen(buf[cnt])),cnt++;
    for(int i = 0;i < cnt;i++)
    {
        printf("%s ",buf[i]);
        dic.Query(buf[i],strlen(buf[i]));
        puts("");
    }
    dic.Release(dic.pRoot);
    return 0;
}