FOJ topic 2075 Substring (suffix array to find the smallest dictionary order substring with k times)

problem 2075 SubstringAccept:70 submit:236
Time limit:1000 mSec Memory limit:65536 KB problem Descriptiongiven A string, find a substring of it which the Origi NAL string contains exactly n such substrings. Inputthere is several cases. The first line of each case contains an integer n.the second line contains a string, no longer than 100000. Outputif the such substring doesn ' t exist, output "impossible", else output the substring that appeared n times in the Ori Ginal string. If There is multiple solutions, output lexicographic smallest substring. Sample Input2ababba Sample OUTPUTAC code

<pre name= "code" class= "CPP" >,

<pre name= "code" class= "CPP" > #include <stdio.h> #include <string.h> #include <algori        thm> #include <iostream> #define MIN (A, b) (A&GT;B?B:A) #define MAX (A, B) (A&GT;B?A:B)          using namespace Std;        Char str[103030];      int sa[103030],rank[103030],rank2[103030],height[103030],c[103030],*x,*y;      int n;          void cmp (int n,int sz) {int i;          Memset (C,0,sizeof (c));          for (i=0;i<n;i++) c[x[y[i]]]++;          for (i=1;i<sz;i++) c[i]+=c[i-1];      for (i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];          } void Build_sa (char *s,int n,int sz) {x=rank,y=rank2;          int i,j;          for (i=0;i<n;i++) x[i]=s[i],y[i]=i;          CMP (N,SZ);          int Len;              for (len=1;len<n;len<<=1) {int yid=0; for (i=n-len;i<n;i++) {Y[yid++]=i;                  } for (i=0;i<n;i++) if (Sa[i]>=len) Y[yid++]=sa[i]-len;              CMP (N,SZ);              Swap (x, y);              X[sa[0]]=yid=0; for (i=1;i<n;i++) {if (Y[SA[I-1]]==Y[SA[I]]&AMP;&AMP;SA[I-1]+LEN&LT;N&AMP;&AMP;SA[I]+LEN&L                  T;n&&y[sa[i-1]+len]==y[sa[i]+len]) X[sa[i]]=yid;              else X[sa[i]]=++yid;              } sz=yid+1;          if (sz>=n) break;      } for (i=0;i<n;i++) rank[i]=x[i];          } void GetHeight (char *s,int n) {int k=0;              for (int i=0;i<n;i++) {if (rank[i]==0) continue;              K=max (0,k-1);              int j=sa[rank[i]-1];              while (S[i+k]==s[j+k]) k++;          Height[rank[i]]=k; }} int minv[103010][20],lg[103030];        void Init_lg () {int i;        lg[1]=0;        for (i=2;i<102020;i++) {lg[i]=lg[i>>1]+1;        }} void Init_rmq (int n) {int i,j,k;        for (i=1;i<=n;i++) {minv[i][0]=height[i];                  } for (j=1;j<=lg[n];j++) {for (k=0;k+ (1<<j) -1<=n;k++) {               Minv[k][j]=min (minv[k][j-1],minv[k+ (1<< (j-1))][j-1]);        }}} int LCP (int l,int R) {L=rank[l];        R=RANK[R];        if (l>r) swap (L,R);        l++;        int k=lg[r-l+1];      return min (minv[l][k],minv[r-(1<<k) +1][k]); } void Solve (int n,int k) {int i,j;for (i=1;i<=n;i++) {if (i+k-1>n) Break;int T=LCP (sa[i],sa[i+k-1]); if (t>height [i]&& (i+k>n| | T+k<=n&&t>height[i+k]) {for (j=0;j<t;j++) {printf ("%c", Str[j+sa[i]]);} printf ("\ n"); return;}} printf ("impossible\n");} int main () {int N;while (scanf("%d", &n)!=eof) {scanf ("%s", str), int len=strlen (str); Build_sa (str,len+1,128); getheight (Str,len); Init_lg (); INIT_RMQ (len); solve (len,n);}}

SOURCEFOJ Prize Month race-March 2012

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FOJ topic 2075 Substring (suffix array to find the smallest dictionary order substring with k times)