Power Strings POJ - 2406 后缀数组

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.

 

 

 

题意:

输入一个字符串，让你找出来这个串的最小循环节

 

 

题解：

对于答案就用for循环从1到n枚举（为什么不能用二分？因为这个判断没有单调性。即i不是循环节的话，你不能说i-1或者i+1也肯定不是循环节）

 

下一步就是判断枚举的这个i是不是循环节，怎么判断？

满足这三个条件：n%i==0 && Rank[0]-Rank[i]==1 && height[Rank[0]]==n-i

 

1、n%i==0这就肯定不需要说了

2、你要注意Rank[0]表示的是啥，他代表后缀[0...n]的排名是多少，如果循环节是i，那么Rank[0]-Rank[i]==Rank[i]-Rank[2*i]......==1

3、如果循环节是i，那么height[Rank[0]]==n-i,   height[Rank[0]]代表后缀[0..n]与后缀[i...n]的最长公共前缀，既然循环节是i，那么height[Rank[0]]的值也就等于n-i

 

还要注意这道题用后缀数组来做的话要用dc3模板，如果用Da模板会TLE

 

dc3模板AC代码：

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 3000010;
#define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb))
#define G(x) ((x) < tb ? (x) * 3 + 1 : ((x) - tb) * 3 + 2)
int wa[maxn], wb[maxn], Ws[maxn], wv[maxn], sa[maxn];
int Rank[maxn], height[maxn],r[maxn];
char s[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);
    return r[a] < r[b] || r[a] == r[b] && wv[a + 1] < wv[b + 1];
}
void Rsort(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, *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;
    Rsort(r + 2, wa, wb, tbc, m);
    Rsort(r + 1, wb, wa, tbc, m);
    Rsort(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;
    Rsort(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 get_height(int n)
{
    int i, j, k = 0;
    for (i = 1; i <= n; i++) Rank[sa[i]] = i;
    for (i = 0; i < n; height[Rank[i++]] = k)
        for (k ? k-- : 0, j = sa[Rank[i] - 1]; r[i + k] == r[j + k]; k++);
}
int main()
{
    while(~scanf("%s",s))
    {
        int n=strlen(s);
        if(n==1 && s[0]=='.') break;
        for(int i=0;i<n;++i)
            r[i]=s[i];
        r[n]='0';
        dc3(r,sa,n+1,200);
        get_height(n);
        int  flag=0;
        for(int i=1;i<n;++i)  //枚举循环节长度
        {
            if(n%i==0 && Rank[0]-Rank[i]==1 && height[Rank[0]]==n-i)
            {
                flag=1;
                printf("%d\n",n/i);
                break;
            }
        }
        if(!flag)
            printf("1\n");
    }
    return 0;
}

 

Da模板TLE代码：

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

const int N = 1000010;
int x[N], y[N], c[N];
int rank[N], height[N];
int sa[N],n,k;
char s[N];
bool pan(int *x,int i,int j,int k,int n)
{
    int ti=i+k<n?x[i+k]:-1;
    int tj=j+k<n?x[j+k]:-1;
    return x[i]==x[j]&&ti==tj;
}
void build_SA(int n,int r)
{
    int *x=rank,*y=height;
    for(int i=0; i<r; i++)c[i]=0;
    for(int i=0; i<n; i++)c[s[i]]++;
    for(int i=1; i<r; i++)c[i]+=c[i-1];
    for(int i=n-1; i>=0; i--)sa[--c[s[i]]]=i;
    r=1;
    x[sa[0]]=0;
    for(int i=1; i<n; i++)
        x[sa[i]]=s[sa[i]]==s[sa[i-1]]?r-1:r++;
    for(int k=1; r<n; k<<=1)
    {
        int yn=0;
        for(int i=n-k; i<n; i++)y[yn++]=i;
        for(int i=0; i<n; i++)
            if(sa[i]>=k)y[yn++]=sa[i]-k;
        for(int i=0; i<r; i++)c[i]=0;
        for(int i=0; i<n; i++)++c[x[y[i]]];
        for(int i=1; i<r; i++)c[i]+=c[i-1];
        for(int i=n-1; i>=0; i--)sa[--c[x[y[i]]]]=y[i];
        swap(x,y);
        r=1;
        x[sa[0]]=0;
        for(int i=1; i<n; i++)
            x[sa[i]]=pan(y,sa[i],sa[i-1],k,n)?r-1:r++;
    }
    for(int i=0; i<n; i++)rank[i]=x[i];
}
void get_height(int n)
{
    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(s[i+k]==s[j+k])k++;
        height[rank[i]]=k;
    }
}
int check(int len)
{
    int i=2,cnt=0;
    while(1)
    {
        while(i<=n && height[i]>=len)
            cnt++,i++;
        if(cnt+1>=k)return 1;
        if(i>=n)return 0;
        while(i <=n &&height[i]<len)
            i++;
        cnt=0;
    }
}

int main()
{
    while(~scanf("%s",s))
    {
        n=strlen(s);
        if(n==1 && s[0]=='.') break;
        s[n]='0';
        build_SA(n+1,200);
        get_height(n);
        //printf("%d %d %d\n",rank[0],rank[3],height[rank[0]]);
        int  flag=0;
        for(int i=1;i<n;++i)  //枚举循环节长度
        {
            if(n%i==0 && rank[0]-rank[i]==1 && height[rank[0]]==n-i)
            {
                flag=1;
                printf("%d\n",n/i);
                break;
            }
        }
        if(!flag)
            printf("1\n");
    }
    return 0;
}