spoj7258:Lexicographical Substring Search（后缀自动机+拓扑排序）

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

题意：给一个字符串，多次询问字典序大小为kth的本质不同字符串。

题解：后缀自动机。

对串构建SAM并拓扑排序统计每个状态能走到的字符串个数

DP转移方程：f[i]=sigma(f[to[i][j])

询问时贪心即可。

代码：

#include <iostream>#include <cstring>using namespace std;const int Maxn=2e5+50;char ch[Maxn];int n,Q,c[Maxn],que[Maxn];inline int read(){char ch=getchar();int i=0,f=1;while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){i=(i<<3)+(i<<1)+ch-'0';ch=getchar();}return i*f;}struct sam{int to[Maxn][26],par[Maxn],l[Maxn],f[Maxn];int p,np,q,nq,last,num;sam(){last=++num;}inline void extend(int c){p=last;last=np=++num;l[np]=l[p]+1;for(;p&&!to[p][c];p=par[p])to[p][c]=np;if(!p)par[np]=1;else{q=to[p][c];if(l[q]==l[p]+1)par[np]=q;else{nq=++num,l[nq]=l[p]+1;memcpy(to[nq],to[q],sizeof(to[q]));par[nq]=par[q],par[q]=par[np]=nq;for(;p&&to[p][c]==q;p=par[p])to[p][c]=nq;}}}inline void build(){scanf("%s",ch+1);n=strlen(ch+1);for(int i=1;i<=n;i++)extend(ch[i]-'a');}inline void calc(){for(int i=1;i<=num;i++)c[l[i]]++;for(int i=1;i<=n;i++)c[i]+=c[i-1];for(int i=num;i>=1;i--)que[c[l[i]]--]=i;for(int i=num;i>=1;i--){int q=que[i];f[q]=1;for(int j=0;j<26;j++)if(to[q][j])f[q]+=f[to[q][j]];}}inline void query(int kth){int pos=1;while(kth>0){for(int j=0;j<26;j++)if(to[pos][j]){if(f[to[pos][j]]>=kth){pos=to[pos][j];putchar('a'+j);--kth;break;}else kth-=f[to[pos][j]];}}puts("");}}sam;int main(){sam.build();sam.calc();Q=read();while(Q--){int kth=read();sam.query(kth);}}