POJ2406——Power Strings(字尾陣列)
阿新 • • 發佈:2018-12-24
Power Strings
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.
Time Limit: 3000MS | Memory Limit: 65536K |
Total Submissions: 51218 | Accepted: 21386 |
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
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
題意:給出一個字串由一個子串重複n次形成,求n的最大值。
思路:列舉子串長度k,首先需要被總長度n整除,然後判斷sa[0]和sa[k]的公共字首是否n-k即可
#include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<vector> #include<algorithm> #define F(x) ((x)/3+((x)%3==1?0:tb)) #define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2) using namespace std; const int MAXN = 3e6+10; const int N = MAXN; int wa[MAXN],wb[MAXN],wv[MAXN],ws[MAXN]; char s[MAXN]; int a[MAXN]; int sa[MAXN],height[MAXN],RANK[MAXN]; int len[MAXN]; 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++) ws[i]=0; for(i=0;i<n;i++) ws[wv[i]]++; for(i=1;i<m;i++) ws[i]+=ws[i-1]; for(i=n-1;i>=0;i--) b[--ws[wv[i]]]=a[i]; return; } void dc3(int *r,int *sa,int n,int m) //涵義與DA 相同 { int i,j,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p; int *rn=r+n; 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 calheight(int *r,int *sa,int n) { // 此處N為實際長度 int i,j,k=0; // height[]的合法範圍為 1-N, 其中0是結尾加入的字元 for(i=1;i<=n;i++) RANK[sa[i]]=i; // 根據SA求RANK for(i=0;i<n; height[RANK[i++]] = k ) // 定義:h[i] = height[ RANK[i] ] for(k?k--:0,j=sa[RANK[i]-1]; r[i+k]==r[j+k]; k++); //根據 h[i] >= h[i-1]-1 來優化計算height過程 } int main(){ while(scanf("%s",s)!=EOF){ if(s[0]=='.') break; int n=(int)strlen(s); for(int i=0;i<n;i++){ a[i]=s[i]; } a[n]=0; dc3(a,sa,n+1,200); calheight(a, sa, n); int MIN=1e9+7; int flag=1; for(int i=1;i<=n;i++){ if(sa[i]==0){ flag=i; break; } } for(int i=flag+1;i<=n;i++){ MIN=min(height[i],MIN); len[sa[i]]=MIN; } MIN=1e9+7; for(int i=flag-1;i>=1;i--){ MIN=min(height[i+1],MIN); len[sa[i]]=MIN; } int k; //for(int i=0;i<=n;i++) // cout<<i<<" "<<len[i]<<endl; for(k=1;k<=n;k++){ if(n%k==0){ if(len[k]==n-k){ break; } } } printf("%d\n",n/k); } }