Lexicographical Substring Search
| Time Limit: 1000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
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:
aaa
2
2
3
Output:
aa
aaa
题目大意:
给出一个字符串以及若干个询问.
求出这个字符串中第K大的字符串.....
思路:
这个思路非常的明了..
先统计出字串的个数,然后按照BST的寻找第K大的方法去寻找就ok了....
常数卡得紧,就下次再优化吧....
1 #include<cstdlib>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5 using namespace std;
6 typedef unsigned int uint;
7 const int maxn = (int)1e5,sigma = 26;
8 char str[maxn];
9 int q;
10 uint k;
11 int cmp(int,int);
12 struct Sam{
13 int ch[maxn * 2][sigma],par[maxn * 2],stp[maxn * 2],id[maxn * 2],sub[maxn * 2];
14 int sz,last;
15 void init(){
16 memset(ch,0,sizeof(ch)); memset(par,0,sizeof(par)); memset(stp,0,sizeof(stp));
17 sz = last = 1;
18 }
19 void ext(int c){
20 stp[++sz] = stp[last] + 1;
21 int p = last, np = sz;
22 for(; !ch[p][c]; p = par[p]) ch[p][c] = np;
23 if(p == 0) par[np] = 1;
24 else{
25 int q = ch[p][c];
26 if(stp[q] != stp[p] + 1){
27 stp[++sz] = stp[p] + 1;
28 int nq = sz;
29 memcpy(ch[nq],ch[q],sizeof(ch[q]));
30 par[nq] = par[q];
31 par[q] = par[np] = nq;
32 for(; ch[p][c] == q; p = par[p]) ch[p][c] = nq;
33 }
34 else
35 par[np] = q;
36 }
37 last = np;
38 }
39 void ins(char *pt){
40 int i;
41 init();
42 for(i = 0; pt[i]; ++i) ext(pt[i] - ‘a‘);
43 }
44 void prep(){
45 int i,j;
46 for(i = 1; i <= sz; ++i) id[i] = i;
47 sort(id + 1, id + sz + 1, cmp);
48 for(i = 1; i <= sz; ++i) sub[i] = 1;
49 for(i = 1; i <= sz; ++i)
50 for(j = 0; j < sigma; ++j)
51 if(ch[id[i]][j]) sub[id[i]] += sub[ch[id[i]][j]];
52 }
53 void solve(uint k){
54 int i,x = 1;
55 while(k){
56 for(i = 0; i < sigma; ++i)
57 if(ch[x][i] && k <= sub[ch[x][i]]){
58 printf("%c",i + ‘a‘);
59 x = ch[x][i];
60 k--;
61 break;
62 }
63 else k -= sub[ch[x][i]];
64 }
65 printf("\n");
66 }
67 }sam;
68 int cmp(int x,int y){
69 return sam.stp[x] > sam.stp[y];
70 }
71 int main()
72 {
73 freopen("sublex.in","r",stdin);
74 freopen("sublex.out","w",stdout);
75 scanf("%s",str);
76 sam.ins(str);
77 sam.prep();
78 scanf("%d",&q);
79 while(q--){
80 scanf("%u",&k);
81 sam.solve(k);
82 }
83 return 0;
84 }
85
86 int cmp(int x,int y){
87 return sam.stp[x] > sam.stp[y];
88 }
89 int main()
90 {
91 scanf("%s",str);
92 sam.ins(str);
93 sam.prep();
94 scanf("%d",&q);
95 while(q--){
96 scanf("%u",&k);
97 sam.solve(k);
98 }
99 return 0;
100 }