Little Daniel loves to play with strings! He always finds different ways to have fun with strings! Knowing that, his friend Kinan decided to test his skills so he gave him a string S and asked him Q questions of the form:
If all distinct substrings of string S were sorted lexicographically, which one will be the K-th smallest?
After knowing the huge number of questions Kinan will ask, Daniel figured out that he can‘t do this alone. Daniel, of course, knows your exceptional programming skills, so he asked you to write him a program which given S will answer Kinan‘s questions.
Example:
S = "aaa" (without quotes)
substrings of S are "a" , "a" , "a" , "aa" , "aa" , "aaa". The sorted list of substrings will be:
"a", "aa", "aaa".
Input
In the first line there is Kinan‘s string S (with length no more than 90000 characters). It contains only small letters of English alphabet. The second line contains a single integer Q (Q <= 500) , the number of questions Daniel will be asked. In the next Q lines a single integer K is given (0 < K < 2^31).
Output
Output consists of Q lines, the i-th contains a string which is the answer to the i-th asked question.
Example
Input:aaa223 Output: aaaaa
Edited: Some input file contains garbage at the end. Do not process them.
让你求按字典序排名第k名的字符串是什么
dp处理出每个点往下走能够走出多少个串。
f[i]=sigma(f[ch[i][c])+1
这个可以按Max排序之后倒着推就好了。
询问的时候看一下走下去个数是否<=k，是的话就走下去，然后–k；否则就找下一条边。
1 #include"bits/stdc++.h"
2 using namespace std;
3 const int N = 101000;
4 #define int long long
5 struct node
6 {
7 int fa;
8 int ch[26];
9 int len;
10 }dian[N<<1];
11 int cnt[N<<1];
12 int f[N<<1];
13 int last=1;int tot=1;
14
15 inline void add(int c)
16 {
17 int p=last; int np=last=++tot;
18 dian[np].len=dian[p].len+1;
19 for(;p&&!dian[p].ch[c];p=dian[p].fa)dian[p].ch[c]=np;
20 if(!p) dian[np].fa=1,cnt[1]++;
21 else
22 {
23 int q=dian[p].ch[c];
24 if(dian[q].len == dian[p].len+1)dian[np].fa=q,cnt[q]++;
25 else
26 {
27 int nq=++tot;
28 dian[nq]=dian[q];
29 dian[nq].len=dian[p].len+1;
30 dian[q].fa=dian[np].fa=nq;
31 ; cnt[nq]+=2;
32 for(;p&&dian[p].ch[c] == q;p=dian[p].fa)dian[p].ch[c]=nq;
33 }
34 }
35 }
36
37
38 void dfs(int x)
39 {
40 if(f[x]) return ;
41 f[x]=1;
42 for(int i=0;i<26;i++)
43 {
44 if(dian[x].ch[i])
45 {dfs(dian[x].ch[i]);
46 f[x]+= f[dian[x].ch[i]];
47
48 }
49 }
50 }
51 void que()
52 {
53 int x;cin>>x; int now=1;
54 while(x)
55 {
56 for(int i=0;i<26;i++)if(dian[now].ch[i])
57 {
58 if(f[dian[now].ch[i]]>=x)
59 {
60 putchar(‘a‘+i); --x;now=dian[now].ch[i];
61 break;
62 }
63 else x-=f[dian[now].ch[i]];
64 }
65 }
66 cout<<‘\n‘;
67 }
68 signed main()
69 {
70 // freopen("datte.txt","r",stdin);
71 //freopen("my.out","w",stdout);
72 string s;cin>>s; int len =s.length();
73 for(int i=0;i<len;i++)add(s[i]-‘a‘);
74
75 dfs(1); int m;cin>>m;
76 while(m--)que();
77 }
原文地址：https://www.cnblogs.com/zhangbuang/p/10872357.html