Description
Input
Output
Sample Input
abcd aaaa ababab .
Sample Output
1 4 3
Hint
#include <iostream> #include <stdio.h> #include <string.h> #include <stack> #include <queue> #include <map> #include <set> #include <vector> #include <math.h> #include <bitset> #include <algorithm> #include <climits> using namespace std; #define LS 2*i #define RS 2*i+1 #define UP(i,x,y) for(i=x;i<=y;i++) #define DOWN(i,x,y) for(i=x;i>=y;i--) #define MEM(a,x) memset(a,x,sizeof(a)) #define W(a) while(a) #define gcd(a,b) __gcd(a,b) #define LL long long #define N 1000005 #define MOD 1000000007 #define INF 0x3f3f3f3f #define EXP 1e-8 #define F(x) ((x)/3+((x)%3==1?0:tb)) #define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2) int wsf[N],wa[N],wb[N],wv[N],sa[N],rank[N],height[N],f[N]; int s[N],a[N]; char str[N],str1[N],str2[N]; //sa:字典序中排第i位的起始位置在str中第sa[i] //rank:就是str第i个位置的后缀是在字典序排第几 //height:字典序排i和i-1的后缀的最长公共前缀 int c0(int *r,int a,int b) { return r[a]==r[b]&&r[a+1]==r[b+1]&&r[a+2]==r[b+2]; } int c12(int k,int *r,int a,int b) { if(k==2) return r[a]<r[b]||r[a]==r[b]&&c12(1,r,a+1,b+1); else return r[a]<r[b]||r[a]==r[b]&&wv[a+1]<wv[b+1]; } void sort(int *r,int *a,int *b,int n,int m) { int i; for(i=0; i<n; i++) wv[i]=r[a[i]]; for(i=0; i<m; i++) wsf[i]=0; for(i=0; i<n; i++) wsf[wv[i]]++; for(i=1; i<m; i++) wsf[i]+=wsf[i-1]; for(i=n-1; i>=0; i--) b[--wsf[wv[i]]]=a[i]; return; } void dc3(int *r,int *sa,int n,int m) { int i,j,*rn=r+n,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p; r[n]=r[n+1]=0; for(i=0; i<n; i++) if(i%3!=0) wa[tbc++]=i; sort(r+2,wa,wb,tbc,m); sort(r+1,wb,wa,tbc,m); sort(r,wa,wb,tbc,m); for(p=1,rn[F(wb[0])]=0,i=1; i<tbc; i++) rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++; if(p<tbc) dc3(rn,san,tbc,p); else for(i=0; i<tbc; i++) san[rn[i]]=i; for(i=0; i<tbc; i++) if(san[i]<tb) wb[ta++]=san[i]*3; if(n%3==1) wb[ta++]=n-1; sort(r,wb,wa,ta,m); for(i=0; i<tbc; i++) wv[wb[i]=G(san[i])]=i; for(i=0,j=0,p=0; i<ta && j<tbc; p++) sa[p]=c12(wb[j]%3,r,wa[i],wb[j])?wa[i++]:wb[j++]; for(; i<ta; p++) sa[p]=wa[i++]; for(; j<tbc; p++) sa[p]=wb[j++]; return; } void getheight(int *r,int n)//n不保存最后的0 { int i,j,k=0; for(i=1; i<=n; i++) rank[sa[i]]=i; for(i=0; i<n; i++) { if(k) k--; else k=0; j=sa[rank[i]-1]; while(r[i+k]==r[j+k]) k++; height[rank[i]]=k; } } int rm[N]; void RMQ(int n) { int k = rank[0]; rm[k] = N; int i; DOWN(i,k-1,0) { if(height[i+1]<rm[i+1]) rm[i]=height[i+1]; else rm[i]=rm[i+1]; } UP(i,k+1,n) { if(height[i]<rm[i-1]) rm[i]=height[i]; else rm[i]=rm[i-1]; } } int solve(int n) { int i; UP(i,1,n/2) { if(n%i) continue; if(rm[rank[i]]==n-i) return n/i; } return 1; } int main() { int n,len,i,j,k; W(~scanf("%s",str)) { if(str[0]=='.') break; len = strlen(str); UP(i,0,len-1) s[i]=str[i]; s[len] = 0; dc3(s,sa,len+1,300); getheight(s,len); RMQ(len); printf("%d\n",solve(len)); } }
POJ2406:Power Strings(后缀数组DC3)
原文地址:http://blog.csdn.net/libin56842/article/details/46310425