POJ 2406 Power Strings 【字尾陣列||KMP】
Power Strings
Time Limit: 3000MS |
Memory Limit: 65536K |
Total Submissions: 51298 |
Accepted: 21420 |
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-negativeinteger is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).
Input
Eacht est 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. Aline 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.
【題意】
給出一個字串,要把它寫成(x)n的形式,問n的最大值。
【思路】
方法一:字尾陣列 (2485MS)
用字尾陣列求出字串的各種資訊:Rank陣列,height陣列等等。
然後我們從1開始列舉迴圈節的長度i。如果字串存在長度大於等於2的迴圈節,需要滿足以下條件
-
首先迴圈節的長度必須是字串長度的因子。
-
要保證Rank[0]==Rank[i]+1,因為兩者比較字典序的時候前面的都相同,只是原串更長一些。
-
而且從i開始的字尾與原串的公共字首為前者的長度即len-i。
否則結果就為1。
PS: 此題資料範圍較大,用DA演算法會超時,用DC3演算法也只是剛剛過,所以建議用下面的方法。
方法二: KMP (141MS)
設字串的長度為len,我們知道nex[len]表示既是原串字首又是原串字尾的字串最大長度(不包括本身),所以如果字串的迴圈節長度大於等於2,那麼len-nex[len]一定是字串的最小週期,且一定能整除len。否則輸出1。
程式碼一:字尾陣列#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define rush() int T,scanf("%d",&T),while(T--)
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
typedef long long ll;
const int maxn = 1000005;
const ll mod = 1e9+7;
const int INF = 0x3f3f3f3f;
int wa[maxn*3],wb[maxn*3],wv[maxn*3],ws[maxn*3];
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));
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)
{
for(int i=0; i<n; i++) wv[i]=r[a[i]];
for(int i=0; i<m; i++) ws[i]=0;
for(int i=0; i<n; i++) ws[wv[i]]++;
for(int i=1; i<m; i++) ws[i]+=ws[i-1];
for(int i=n-1; i>=0; i--) b[--ws[wv[i]]]=a[i];
}
void dc3(int *r,int *sa,int n,int m)
{
int i,j,*rn=r+n;
int *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++];
}
void da(int *str,int *sa,int *Rank,int *height,int n,int m)
{
for(int i=n;i<n*3;i++) str[i]=0;
dc3(str,sa,n+1,m);
int i,j,k=0;
for(i=0;i<=n;i++) Rank[sa[i]]=i;
for(i=0;i<n;i++)
{
if(k) k--;
j=sa[Rank[i]-1];
while(str[i+k]==str[j+k]) k++;
height[Rank[i]]=k;
}
}
int Rank[maxn*3],height[maxn*3],sa[maxn*3];
char s[maxn*3];
int r[maxn*3];
int main()
{
while(~scanf("%s",s)&&s[0]!='.')
{
int len=strlen(s);
for(int i=0;i<len;i++)
{
r[i]=s[i]-'a'+1;
}
r[len]=0;
da(r,sa,Rank,height,len,105);
int flag=0;
for(int i=1;i<=len;i++)
{
if(len%i==0&&Rank[0]==Rank[i]+1&&height[Rank[0]]==len-i)
{
printf("%d\n",len/i);
flag=1;
break;
}
}
if(flag==0) puts("1");
}
return 0;
}
程式碼二:KMP
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define rush() int T,scanf("%d",&T),while(T--)
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
typedef long long ll;
const int maxn = 1000005;
const ll mod = 1e9+7;
const int INF = 0x3f3f3f3f;
char s[maxn];
int nex[maxn];
void kmp_pre(char *b)
{
int i=0,j=-1;
int len=strlen(b);
nex[0]=-1;
while(i<len)
{
while(j!=-1&&b[i]!=b[j]) j=nex[j];
if(b[++i]==b[++j]) nex[i]=nex[j];
else nex[i]=j;
}
}
int main()
{
while(~scanf("%s",s)&&s[0]!='.')
{
int len=strlen(s);
kmp_pre(s);
int ans=1;
if(len%(len-nex[len])==0)
{
ans=len/(len-nex[len]);
}
printf("%d\n",ans);
}
return 0;
}