POJ 2001 Shortest Prefixes(字典树)

最新推荐文章于 2021-08-05 18:41:49 发布

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本文介绍了一道关于字典树的经典算法题，通过构建字典树来找到每个输入单词的最短唯一前缀，并提供了完整的代码实现。

Shortest Prefixes

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 0		Accepted: 0

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

/************************************************************************/

题意：有多个字符串，现要用各自的前缀唯一表示各自的字符

对于唯一的说法，当有两个字符串car和carriage时，若用前缀car来表示carriage是不可行的，这并不唯一，因为前缀car也有可能是表示car的，所以我们不得不用前缀carr来表示carriage

解题思路：一道字典树的入门题，也算是模板题吧

将给定的字符串建立一棵字典树，每个字符出现的次数都要记录下来，而询问阶段只需遍历字典树，当某个字符出现次数为1或者当前字符串遍历完，都可以将其作为前缀表示

样例可以表示成如下这棵树：

圈内表示字符以及其出现的次数

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 25;
const int inf = 1000000000;
const int mod = 2009;
struct tree
{
    int t;
    int a[26];
}s[100000];
int p=0;
void init(int z)
{
    s[z].t=0;
    for(int i=0;i<26;i++)
        s[z].a[i]=-1;
}
void buildtree(char *c)
{
    int i,k;
    for(i=k=0;c[i]!='\0';i++)
    {

        if(s[k].a[c[i]-'a']==-1)
        {
            s[k].a[c[i]-'a']=++p;
            init(k=p);
        }
        else
            k=s[k].a[c[i]-'a'];
        s[k].t++;
    }
}
void find(char *c)
{
    int i,k;
    for(i=k=0;c[i]!='\0';i++)
    {
        k=s[k].a[c[i]-'a'];
        if(s[k].t==1)
        {
            c[i+1]='\0';
            break;
        }
    }
    puts(c);
}
char ch[50*N][N];
int main()
{
    int i,j;
	init(0);
	for(i=0;~scanf("%s",ch[i]);i++)
        buildtree(ch[i]);
    for(j=0;j<i;j++)
    {
        printf("%s ",ch[j]);
        find(ch[j]);
    }
	return 0;
}

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