Question

任何人都可以给我提示如何开发后缀数组部分吗？我知道这个概念; LCP阵列设计，但我没有得到如何在C中实现它？有人可以帮忙吗？我知道后缀数组的用途，算法，因为我已经阅读了很多内容。我想要我必须对字符串的后缀进行排序的部分的实现提示。

例如，如果字符串为'banana'，则为：

数据结构应如下所示：（$ - ＆gt; mnemonic）

banana
anana
nana
ana
na
a
$

然后，在保留它之后，我需要对它进行排序，这意味着最低的子串应该位于最顶点。那怎么办呢？字符串可以很长。怎么做这件事？你能给点提示或链接吗？我已经尝试过，现在在你的帮助下思考。

Answer 1

您可能需要查看此article

在本文的最后，您将找到后缀树的这种实现：

注意：以下代码是C ++，但如果用新的[]和delete []运算符替换C类堆分配，则可以非常轻松地重用它。

inline bool leq(int a1, int a2,   int b1, int b2) { // lexic. order for pairs
  return(a1 < b1 || a1 == b1 && a2 <= b2); 
}                                                   // and triples
inline bool leq(int a1, int a2, int a3,   int b1, int b2, int b3) {
  return(a1 < b1 || a1 == b1 && leq(a2,a3, b2,b3)); 
}
// stably sort a[0..n-1] to b[0..n-1] with keys in 0..K from r
static void radixPass(int* a, int* b, int* r, int n, int K) 
{ // count occurrences
  int* c = new int[K + 1];                          // counter array
  for (int i = 0;  i <= K;  i++) c[i] = 0;         // reset counters
  for (int i = 0;  i < n;  i++) c[r[a[i]]]++;    // count occurences
  for (int i = 0, sum = 0;  i <= K;  i++) { // exclusive prefix sums
     int t = c[i];  c[i] = sum;  sum += t;
  }
  for (int i = 0;  i < n;  i++) b[c[r[a[i]]]++] = a[i];      // sort
  delete [] c;
}

// find the suffix array SA of s[0..n-1] in {1..K}^n
// require s[n]=s[n+1]=s[n+2]=0, n>=2
void suffixArray(int* s, int* SA, int n, int K) {
  int n0=(n+2)/3, n1=(n+1)/3, n2=n/3, n02=n0+n2; 
  int* s12  = new int[n02 + 3];  s12[n02]= s12[n02+1]= s12[n02+2]=0; 
  int* SA12 = new int[n02 + 3]; SA12[n02]=SA12[n02+1]=SA12[n02+2]=0;
  int* s0   = new int[n0];
  int* SA0  = new int[n0];

  // generate positions of mod 1 and mod  2 suffixes
  // the "+(n0-n1)" adds a dummy mod 1 suffix if n%3 == 1
  for (int i=0, j=0;  i < n+(n0-n1);  i++) if (i%3 != 0) s12[j++] = i;

  // lsb radix sort the mod 1 and mod 2 triples
  radixPass(s12 , SA12, s+2, n02, K);
  radixPass(SA12, s12 , s+1, n02, K);  
  radixPass(s12 , SA12, s  , n02, K);

  // find lexicographic names of triples
  int name = 0, c0 = -1, c1 = -1, c2 = -1;
  for (int i = 0;  i < n02;  i++) {
    if (s[SA12[i]] != c0 || s[SA12[i]+1] != c1 || s[SA12[i]+2] != c2) { 
      name++;  c0 = s[SA12[i]];  c1 = s[SA12[i]+1];  c2 = s[SA12[i]+2];
    }
    if (SA12[i] % 3 == 1) { s12[SA12[i]/3]      = name; } // left half
    else                  { s12[SA12[i]/3 + n0] = name; } // right half
  }

  // recurse if names are not yet unique
  if (name < n02) {
    suffixArray(s12, SA12, n02, name);
    // store unique names in s12 using the suffix array 
    for (int i = 0;  i < n02;  i++) s12[SA12[i]] = i + 1;
  } else // generate the suffix array of s12 directly
    for (int i = 0;  i < n02;  i++) SA12[s12[i] - 1] = i; 

  // stably sort the mod 0 suffixes from SA12 by their first character
  for (int i=0, j=0;  i < n02;  i++) if (SA12[i] < n0) s0[j++] = 3*SA12[i];
  radixPass(s0, SA0, s, n0, K);

  // merge sorted SA0 suffixes and sorted SA12 suffixes
  for (int p=0,  t=n0-n1,  k=0;  k < n;  k++) {
#define GetI() (SA12[t] < n0 ? SA12[t] * 3 + 1 : (SA12[t] - n0) * 3 + 2)
    int i = GetI(); // pos of current offset 12 suffix
    int j = SA0[p]; // pos of current offset 0  suffix
    if (SA12[t] < n0 ? 
        leq(s[i],       s12[SA12[t] + n0], s[j],       s12[j/3]) :
        leq(s[i],s[i+1],s12[SA12[t]-n0+1], s[j],s[j+1],s12[j/3+n0]))
    { // suffix from SA12 is smaller
      SA[k] = i;  t++;
      if (t == n02) { // done --- only SA0 suffixes left
        for (k++;  p < n0;  p++, k++) SA[k] = SA0[p];
      }
    } else { 
      SA[k] = j;  p++; 
      if (p == n0)  { // done --- only SA12 suffixes left
        for (k++;  t < n02;  t++, k++) SA[k] = GetI(); 
      }
    }  
  } 
  delete [] s12; delete [] SA12; delete [] SA0; delete [] s0; 
}

Answer 2

这可能会对你有帮助。

http://code.google.com/p/code-share/source/browse/trunk/cpp/algo/suffix_array/SuffixArray.h

#ifndef _SUFFIX_ARRAY_H
#define _SUFFIX_ARRAY_H

#include<algorithm>
#include<cstring>
#include <stdexcept>

using namespace std;
template<class T>
struct comp_func
{
    bool operator()(const T l,const T r)
    {
        return strcmp(l,r) < 0;

    }
};
template<class T =char>
class SuffixArray
{
    int len_;
    T **data_;

    public:
    T *operator[](int i)
    {
        if(i<0 || i>len_)
            throw std::out_of_range("Out of range error\n");
        return data_[i];
    }
    SuffixArray(T *str):len_(strlen(str)),data_(new T*[len_])
    {
        //len_ = strlen(str);
        //data_= new T*[len];
        for(int i =0;i<len_;++i)
        {
            data_[i] = &str[i];
            cout << data_[i] << endl;
        }
        std::sort(&data_[0],&data_[len_],comp_func<T *>());
    }
    void Print()
    {
        for(int i =0;i<len_;++i)
        {
            cout << data_[i] << endl;
        }

    }
};


#endif

后缀数组实现

2 个答案: