POJ2001 Shortest Prefixes(字典树)

最新推荐文章于 2021-08-05 18:41:49 发布

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最新推荐文章于 2021-08-05 18:41:49 发布

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分类专栏：数据结构-字典树

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本文介绍了一种用于从一组字符串中找出每个字符串的最短唯一前缀的算法实现。通过构建字典树（Trie），该算法可以高效地解决字符串集合中最短前缀的问题，适用于如词汇缩写或关键字快速识别等应用场景。

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题解：

查找并输出字符串在串集里面唯一确定的最短前缀

代码

#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 10000
#define LL long long
int cas=1,T;
struct Tree
{
    int L,root;
    int Next[maxn][27];
    int word[maxn];

    void init()
    {
       L = 0;
       root = newnode();
    }
    int newnode()
    {
        for (int i = 0;i<27;i++)
            Next[L][i] = -1;
        word[L++] = 0;
        return L-1;
    }
    void insert(char *s)
    {
        int l = strlen(s);
        int u = root,v;
        for (int i = 0;i<l;i++)
        {
            int c = s[i] - 'a';
            if (Next[u][c] == -1)
                Next[u][c] = newnode();
            u = Next[u][c];
            word[u]++;
        }
    }
    void findprefix(char *s)
    {
        int l = strlen(s);
        int u = root;
        for (int i = 0;i<l;i++)
        {
            int c = s[i]-'a';
            u = Next[u][c];
            printf("%c",s[i]);
            if (word[u] == 1)
                break;
        }
        printf("\n");
    }

};
char list[1005][25];
int main()
{
    int t = 0;
    Tree tree;
    tree.init();
    while (scanf("%s",list[t])!=EOF)
    {
        tree.insert(list[t]);
        t++;
    }
    for (int i = 0;i<t;i++)
    {
        printf("%s ",list[i]);
        tree.findprefix(list[i]);
    }
    return 0;
}

题目

POJ2001Shortest Prefixes

Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u

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