标签：exception getchar process scanf lex code div ble init
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".
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 consists of Q lines, the i-th contains a string which is the answer to the i-th asked question.
Input:
aaa
2
2
3
Output: aa
aaa
Edited: Some input file contains garbage at the end. Do not process them.
题意：
给你一个串
Q个询问，在其去重子串中，字典序排名为K的字串是哪一个 ，并输出来
题解：
说一下后缀自动机的做法
设定F[i] 表示 以状态i为起点 所能 形成的不同字串的个数
求出来，再类似于贪心的找法求出答案串
后缀数组做更简单些
后缀自动机
#include <bits/stdc++.h> inline long long read(){long long x=0,f=1;char ch=getchar();while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}return x*f;} using namespace std; const int N = 3e5+7; const long long mod = 1000000007; int isPlus[N * 2],endpos[N * 2];int d[N * 2]; int tot,slink[2*N],trans[2*N][28],minlen[2*N],maxlen[2*N],pre; int newstate(int _maxlen,int _minlen,int* _trans,int _slink){ maxlen[++tot]=_maxlen;minlen[tot]=_minlen; slink[tot]=_slink; if(_trans)for(int i=0;i<26;i++)trans[tot][i]=_trans[i],d[_trans[i]]+=1; return tot; } int add_char(char ch,int u){ int c=ch-‘a‘,v=u; int z=newstate(maxlen[u]+1,-1,NULL,0); isPlus[z] = 1; while(v&&!trans[v][c]){trans[v][c]=z;d[z]+=1;v=slink[v];} if(!v){ minlen[z]=1;slink[z]=1;return z;} int x=trans[v][c]; if(maxlen[v]+1==maxlen[x]){slink[z]=x;minlen[z]=maxlen[x]+1;return z;} int y=newstate(maxlen[v]+1,-1,trans[x],slink[x]); slink[z]=slink[x]=y;minlen[x]=minlen[z]=maxlen[y]+1; while(v&&trans[v][c]==x){trans[v][c]=y;d[x]--,d[y]++;v=slink[v];} minlen[y]=maxlen[slink[y]]+1; return z; } void init_sam() { for(int i = 1; i <= tot; ++i) for(int j = 0; j < 26; ++j) trans[i][j] = 0; pre = tot = 1; } long long f[N],all[N]; char a[N]; int cnt[N],pos[N]; void query(long long k) { int p = 1; while(k) { long long now = 0; for(int i = 0; i < 26; ++i) { if(!trans[p][i]) continue; if(f[trans[p][i]] >= k) { k--; p = trans[p][i]; printf("%c",i+‘a‘); break; } else k -= f[trans[p][i]]; } } printf("\n"); } int main() { scanf("%s",a); int n = strlen(a); init_sam(); for(int i = 0; i < n; ++i) pre = add_char(a[i],pre); for(int i = 0; i <= n; ++i) cnt[i] = 0; for(int i = 1; i <= tot; ++i) cnt[maxlen[i]]++,all[i] = maxlen[i] - minlen[i] + 1; for(int i = 1; i <= n; ++i) cnt[i] += cnt[i-1]; for(int i = tot; i >= 1; --i) pos[cnt[maxlen[i]]--] = i; //for(int i = 2; i <= tot; ++i) cout<<all[i]<<" "<<slink[i]<<endl; for(int i = tot; i >= 2; --i) { int v = pos[i]; f[v] = 1; for(int j = 0; j < 26; ++j) { f[v] += f[trans[v][j]]; } } int Q; scanf("%d",&Q); while(Q--) { long long k; scanf("%lld",&k); query(k); } return 0; }
后缀数组
#include <bits/stdc++.h> inline long long read(){long long x=0,f=1;char ch=getchar();while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}return x*f;} using namespace std; const int N = 3e5+7; const long long mod = 1000000007; int *ran,r[N],sa[N],height[N],wa[N],wb[N],wm[N]; bool cmp(int *r,int a,int b,int l) { return r[a] == r[b] && r[a+l] == r[b+l]; } void SA(int *r,int *sa,int n,int m) { int *x=wa,*y=wb,*t; for(int i=0;i<m;++i)wm[i]=0; for(int i=0;i<n;++i)wm[x[i]=r[i]]++; for(int i=1;i<m;++i)wm[i]+=wm[i-1]; for(int i=n-1;i>=0;--i)sa[--wm[x[i]]]=i; for(int i=0,j=1,p=0;p<n;j=j*2,m=p){ for(p=0,i=n-j;i<n;++i)y[p++]=i; for(i=0;i<n;++i)if(sa[i]>=j)y[p++]=sa[i]-j; for(i=0;i<m;++i)wm[i]=0; for(i=0;i<n;++i)wm[x[y[i]]]++; for(i=1;i<m;++i)wm[i]+=wm[i-1]; for(i=n-1;i>=0;--i)sa[--wm[x[y[i]]]]=y[i]; for(t=x,x=y,y=t,i=p=1,x[sa[0]]=0;i<n;++i) { x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++; } } ran=x; } void Height(int *r,int *sa,int n) { for(int i=0,j=0,k=0;i<n;height[ran[i++]]=k) for(k?--k:0,j=sa[ran[i]-1];r[i+k] == r[j+k];++k); } int n; long long f[N]; char a[N]; void query(long long k) { int pos = lower_bound(f+1,f+n+1,k) - f; long long lll = k - f[pos-1] + height[pos]; for(int i = sa[pos]; i < sa[pos]+lll; ++i) printf("%c",a[i]); printf("\n"); } int main() { scanf("%s",a); n = strlen(a); for(int i = 0; i < n; ++i) r[i] = a[i] - ‘a‘+1; r[n] = 0; SA(r,sa,n+1,300); Height(r,sa,n); for(int i = 1; i <= n; ++i) f[i] = f[i-1] + n - sa[i] - height[i]; int Q; long long k; scanf("%d",&Q); while(Q--) { scanf("%lld",&k); query(k); } return 0; }
SPOJ SUBLEX - Lexicographical Substring Search 后缀自动机 / 后缀数组
标签：exception getchar process scanf lex code div ble init
原文地址：http://www.cnblogs.com/zxhl/p/7617446.html
