poj 2406 Power Strings (后缀数组 || KMP)
Power Strings
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 28859 | Accepted: 12045 |
Description
Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).
Input
Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
Output
For each s you should print the largest n such that s = a^n for some string a.
Sample Input
abcd aaaa ababab .
Sample Output
1 4 3
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source
1 //42704K 2829MS C++ 2709B 2013-12-18 13:31:43 2 /* 3 4 题意： 5 给出一个由某个子串重复R次组成的字符串，求R的最大值 6 7 后缀数组： 8 KMP应该会简单些，因为此处要练习后缀数组，故用后缀数组。首先考虑用DA的话会TLE， 9 因为其时间复杂度为O(nlgn)，数据太大(n=2000000)。 10 此处用dc3，dc3的算法没研究，只是套用其模板，dc3算法会比DA快些，在此处勉勉强强 11 的过了。先用dc3求出后缀数组，然后再求height数组，再穷举字符串长度len。求k能整除len 12 且suffix(1) 与 suffix(k+1) 最长前缀等于len-k。 13 在求最长公共前缀时，由于suffix(1)是固定的，利用height数组的特性，求出height数组 14 中每一个数到height[rank[0]]间的最小值即可。 15 16 */ 17 #include<stdio.h> 18 #include<string.h> 19 #define N 2000005 20 #define F(x) ((x)/3+((x)%3==1?0:tb)) 21 #define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2) 22 int wa[N],wb[N],wv[N],ws[N]; 23 int rank[N],height[N]; 24 int sa[N],r[N]; 25 char c[N]; 26 int lcp[N]; //记录到height[rank[0]]的最小值 27 int Max(int a,int b) 28 { 29 return a>b?a:b; 30 } 31 int Min(int a,int b) 32 { 33 return a<b?a:b; 34 } 35 int cmp(int *y,int a,int b,int l) 36 { 37 return y[a]==y[b]&&y[a+l]==y[b+l]; 38 } 39 40 int c0(int *y,int a,int b) 41 { 42 return y[a]==y[b]&&y[a+1]==y[b+1]&&y[a+2]==y[b+2]; 43 } 44 int c12(int k,int *y,int a,int b) 45 { 46 if(k==2) return y[a]<y[b]||y[a]==y[b]&&c12(1,y,a+1,b+1); 47 else return y[a]<y[b]||y[a]==y[b]&&wv[a+1]<wv[b+1]; 48 } 49 void sort(int *r,int *a,int *b,int n,int m) 50 { 51 int i; 52 for(i=0;i<n;i++) wv[i]=r[a[i]]; 53 for(i=0;i<m;i++) ws[i]=0; 54 for(i=0;i<n;i++) ws[wv[i]]++; 55 for(i=1;i<m;i++) ws[i]+=ws[i-1]; 56 for(i=n-1;i>=0;i--) b[--ws[wv[i]]]=a[i]; 57 return; 58 } 59 void dc3(int *r,int *sa,int n,int m) 60 { 61 int i,j,*rn=r+n,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p; 62 r[n]=r[n+1]=0; 63 for(i=0;i<n;i++) if(i%3!=0) wa[tbc++]=i; 64 sort(r+2,wa,wb,tbc,m); 65 sort(r+1,wb,wa,tbc,m); 66 sort(r,wa,wb,tbc,m); 67 for(p=1,rn[F(wb[0])]=0,i=1;i<tbc;i++) 68 rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++; 69 if(p<tbc) dc3(rn,san,tbc,p); 70 else for(i=0;i<tbc;i++) san[rn[i]]=i; 71 for(i=0;i<tbc;i++) if(san[i]<tb) wb[ta++]=san[i]*3; 72 if(n%3==1) wb[ta++]=n-1; 73 sort(r,wb,wa,ta,m); 74 for(i=0;i<tbc;i++) wv[wb[i]=G(san[i])]=i; 75 for(i=0,j=0,p=0;i<ta && j<tbc;p++) 76 sa[p]=c12(wb[j]%3,r,wa[i],wb[j])?wa[i++]:wb[j++]; 77 for(;i<ta;p++) sa[p]=wa[i++]; 78 for(;j<tbc;p++) sa[p]=wb[j++]; 79 return; 80 } 81 void get_height(int n) 82 { 83 int i,j,k=0; 84 for(i=0;i<=n;i++) rank[sa[i]]=i; 85 for(i=0;i<n;height[rank[i++]]=k) 86 for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++); 87 return; 88 } 89 int main(void) 90 { 91 while(scanf("%s",c)!=EOF) 92 { 93 if(strcmp(c,".")==0) break; 94 int n=strlen(c); 95 for(int i=0;i<n;i++) 96 r[i]=c[i]+1; 97 r[n]=0; 98 dc3(r,sa,n+1,256); 99 get_height(n); 100 //for(int i=0;i<n;i++) printf("%d %d %d\n",i,rank[i],height[i]); 101 memset(lcp,0,sizeof(lcp)); 102 lcp[rank[0]]=N; 103 for(int i=rank[0]-1;i>=0;i--) lcp[i]=Min(lcp[i+1],height[i+1]); 104 for(int i=rank[0]+1;i<=n;i++) lcp[i]=Min(lcp[i-1],height[i]); 105 //for(int i=0;i<=n;i++) printf("%d %d %d\n",rank[i],height[i],lcp[i]); 106 for(int k=1;k<=n;k++) //遍历所有值 107 if(n%k==0 && lcp[rank[k]]==n-k){ 108 printf("%d\n",n/k); 109 break; 110 } 111 } 112 return 0; 113 } 114 /* 115 abcd 116 aaaa 117 ababab 118 */
