构造字符串（hdu2970Suffix reconstruction）

最新推荐文章于 2024-08-13 10:26:25 发布

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最新推荐文章于 2024-08-13 10:26:25 发布

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本文介绍了一个算法问题，即根据给定的后缀数组排名构造原始字符串。探讨了如何通过线性时间复杂度来检查是否存在符合条件的字符串，并给出了具体的实现代码。

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Suffix reconstruction

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 278 Accepted Submission(s): 150

Problem Description

Given a text s[1..n] of length n, we create its suffix array by taking all its suffixes: s[1..n], s[2..n],...., s[n..n] and sorting them lexicographically. As a result we get a sorted list of suffixes: s[p(1)..n], s[p(2)..n],..., s[p(n)..n] and call the sequence p(1),p(2),...,p(n) the suffix array of s[1..n].

For example, if s = abbaabab, the sorted list of all suffixes becomes: aabab, ab, abab, abbaabab, b, baabab, bab, bbaabab and the suffix array is 4, 7, 5, 1, 8, 3,6, 2.

It turns out that it is possible to construct this array in a linear time. Your task will be completely different, though: given p(1), p(2), p(3),... , p(n) you should check if there exist at least one text consisting of lowercase letters of the English alphabet for which this sequence is the suffix array. If so, output any such text. Otherwise output -1.

Input

The input contains several descriptions of suffix arrays. The first line contains the number of descriptions t (t <= 100). Each description begins with a line containing the length of both the text and the array n (1 <= n <= 500000). Next line contains integers p(1), p(2), ... ,p(n). You may assume that 1 <= p(i) <= n and no value of p(i) occurs twice. Total size of the input will not exceed 50MB.

Output

For each test case
If there are multiple answers, output the smallest dictionary order in the given suffix array. In case there is no such text consisting of lowercase letters of the English alphabet, output -1.

Sample Input

6
2
1 2
2
2 1
3
2 3 1
6
3 4 5 1 2 6
14
3 10 2 12 14 5 13 4 1 8 6 11 7 9
7
5 1 7 4 3 2 6

Sample Output

ab
aa
bab
bcaaad
ebadcfgehagbdc
bcccadc

题意：给出排名第i的是从第几个位置开始的后缀，让你构造出满足条件的字符串

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=600010;
int sa[maxn],Rank[maxn];
char s[maxn];
int N;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&N);
        for(int i=1;i<=N;i++)
        {
            scanf("%d",&sa[i]);
            Rank[sa[i]]=i;
        }
        Rank[N+1]=-1;
        int k=0;
        s[sa[1]]=k+'a';
        for(int i=2;i<=N;i++)
        {
            int x=sa[i-1],y=sa[i];
            if(Rank[x+1]>Rank[y+1])k++;
            s[sa[i]]=k+'a';
        }
        if(k>=26)printf("-1\n");
        else
        {
            s[N+1]=0;
            printf("%s\n",s+1);
        }
    }
    return 0;
}