后缀数组 POJ 2406 Power Strings

 

题目链接

题意:连续重复子串。给定一个字符串 L,已知这个字符串是由某个字符串 S 重复 R 次而得到的(L = S^R ), 求 R 的最大值。

分析:枚举长度,判断条件是能被总长度整除且LCP (suffix (0), suffix (i)) = n - i,预处理出lcp,方法是,lcp[i] = min (height[rank[i]] to height[rank[0]]); DC3算法,C++提交才能AC。

#include<cstdio>
#include<cstring>
#include <algorithm>
const int N = 2e6 + 5;
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
int wa[N],wb[N],wv[N],ws[N];
int rank[N],height[N];    
int sa[N],r[N];
char c[N];
int lcp[N]; //记录到height[rank[0]]的最小值 

int Min(int a,int b)
{
    return a<b?a:b;
}

int c0(int *y,int a,int b)
{
    return y[a]==y[b]&&y[a+1]==y[b+1]&&y[a+2]==y[b+2];
}
int c12(int k,int *y,int a,int b)
{
    if(k==2) return y[a]<y[b]||y[a]==y[b]&&c12(1,y,a+1,b+1);
    else return y[a]<y[b]||y[a]==y[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)
{
    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 calc_height(int n) {
    int i, k = 0;
    for (i=0; i<=n; ++i) rank[sa[i]] = i;
    for (i=0; i<n; ++i) {
        if (k) k--;
        int j = sa[rank[i]-1];
        while (r[i+k] == r[j+k]) k++;
        height[rank[i]] = k;
    }
}

int main() {
    while (scanf ("%s", c) != EOF) {
        if (strcmp (c, ".") == 0) {
            break;
        }
        int n = strlen (c);
        for (int i=0; i<n; ++i) {
            r[i] = c[i] + 1;
        }
        r[n] = 0;
        dc3 (r, sa, n + 1, 256);
        calc_height (n);
        
        memset (lcp, 0, sizeof (lcp));
        lcp[rank[0]] = N;
        for (int i=rank[0]-1; i>=0; --i) {
            lcp[i] = std::min (lcp[i+1], height[i+1]);
        }
        for (int i=rank[0]+1; i<=n; ++i) {
            lcp[i] = std::min (lcp[i-1], height[i]);
        }
        for (int i=1; i<=n; ++i) {
            if (n % i == 0 && lcp[rank[i]] == n - i) {
                printf ("%d\n", n / i);
                break;
            }
        }
    }
    return 0;
}

  

posted @ 2016-05-01 09:29  Running_Time  阅读(149)  评论(0编辑  收藏  举报