实用算法实现-第 8 篇后缀树和后缀数组 [3 两个字符串的最长公共子串]

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原创最新推荐文章于 2026-08-03 12:22:34 发布 1.4k 阅读 AI 写同款· GEO 优化 ›

实用算法实现-第 8 篇 后缀树和后缀数组 [2 最长公共子串] 8.3 最长公共子串求N个串的最长公共子串，可以转化为求一些后缀的最长公共前缀的最大值，这些后缀应分属于N个串。设N个串分别为S1，S2，S3，…，SN。具体方法如下： 1. 建立字符串S，使得S = S1[P1]S2[P2]S3…SN-1[PN-1]SN。其中P1，P2，…，PN-1应为不同的N - 1个不在字符集中的字符，作为分隔符。插入分隔符的目的是为了防止S的后缀的公共前缀阅读详情

8.4 两个字符串的最长公共子串

两个串的最长公共字串是相对于多个串要简单一点，不需要二分A。只需要判断相邻两个Height是不是分属两个字符串即可。

8.4.1 实例

PKU JudgeOnline, 2774, Long Long Message.

8.4.2 问题描述

给两个小写ASCII字母组成的字符串，求出它们最大公共子串的长度。

相比PKUJudgeOnline, 3450, Corporate Identity，这个题目比较简单，而且测试数据并不强，没有测出求Height数组的一个错误。

8.4.3 输入

yeshowmuchiloveyoumydearmotherreallyicannotbelieveit

yeaphowmuchiloveyoumydearmother

8.4.4 输出

8.4.5 分析

采用的算法是DC3算法，源程序可以在[i]下载。该算法实现得相当巧妙，所以只添加注释和计算height的函数，就可以使用了。

8.4.6 程序

#include <stdio.h>
#include <iostream>
#include <stdlib.h>
using namespace std;
inline bool leq(int a1, int a2,   int b1, int b2) { // lexic. orderfor 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 fromr
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;         // resetcounters
  for (int i = 0;  i< n;  i++) c[r[a[i]]]++;    // countoccurences
  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) {
  intn0=(n+2)/3, n1=(n+1)/3, n2=n/3, n02=n0+n2;
  //n0是字符串中模为的下标的个数，n1，n2依此类推
  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];
 
  // generatepositions 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;
  //将所有模不为的下标存入s12中
 
  // lsb radix sortthe 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);
  //radixPass实际是一个计数排序
  //对后缀的前三个字符进行三次计数排序完成了对SA12数组的基数排序
  //这个排序是初步的，没有将SA12数组真正地排好序，因为：
  //若SA12数组中几个后缀的前三个字符相等，则起始位置靠后的排在后面
 
  // findlexicographic 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];
    }
     //name是计算后缀数组SA12中前三个字符不完全相同的后缀个数
     //这么判断的原因是：SA12有序，故只有相邻后缀的前三个字符才可能相同
    if (SA12[i]% 3 == 1) { s12[SA12[i]/3]      = name; }// left half
    else                  { s12[SA12[i]/3 + n0] = name;} // right half
     //SA12[i]模不是就是，s12保存的是后缀数组SA12中前三个字符的排位
  }
 
  // recurse if namesare not yet unique
  if (name <n02) {
     //如果name等于n02，意味着SA12前三个字母均不相等，即SA12已有序
     //否则，根据s12的后缀数组与SA12等价，对s12的后缀数组进行排序即可
    suffixArray(s12, SA12, n02, name);
    // store uniquenames 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;
  //s12保存的是后缀数组SA12中前三个字符的排位
  //在所有后缀前三个字符都不一样的情况下，s12就是后缀的排位
 
  //至此SA12排序完毕
  //SA12[i]是第i小的后缀的序号(序号从到n02)，s12[i]是序号为i的后缀的排位
  //使用后缀序号而不是实际位置的原因是递归调用suffixArray时不能保留该信息
 
  // stably sort themod 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];
  //将SA12中所有的模为的后缀的实际位置减去按序存储在s0中
  //注意后缀序号到实际位置的转化需将前者乘
  //这意味着首先已经利用模为的后缀对SA0进行了初步排序
  //只需要采用一次计数排序即可对SA0完成基数排序的最后一步
  radixPass(s0, SA0, s, n0, K);
 
  //最后一步，对有序表SA12和SA0进行归并
  // merge sorted SA0suffixes 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 fromSA12 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;
}
void suffixArrayHeight(int *s, int *SA, int n, int K, int *rank, int *height)
{
    int i, j,h;
     for(i = 0;i < n; i++){
         rank[SA[i]] = i;
         //rank和SA互逆，即SA[rank[i]] == i&&rank[SA[i]] == i
     }
     h = 0;
     for(i = 0;i < n; i++){
         if(rank[i]== 0){
            height[rank[i]] = 0;
         }else{
              j = SA[rank[i] - 1];
              //如果用前缀的第一个字符的下标来标识前缀
              //那么，j是前缀i == SA[rank[i]]的左邻前缀
              while(s[i+ h] == s[j + h]){
                   h++;
                   //如果没有关于h[i]和h[i+1]的大小关系的定理，h需要从开始
              }
              height[rank[i]] = h;
              //求出了h[i]的值
              if(h> 0)
              {
                   h--;
                   //h[i+1]的值大于或等于h[i]-1
              }
        }
     }
}
#define maxNum (100002 * 2 + 3)
/*
int s[maxNum];
int SA[maxNum];
int rank[maxNum];
int height[maxNum];*/
bool isPermutation(int *SA, int n) {
  bool *seen = new bool[n];
  for (int i = 0;  i< n;  i++) seen[i] = 0;
  for (int i = 0;  i< n;  i++) seen[SA[i]] = 1;
  for (int i = 0;  i< n;  i++) if(!seen[i]) return 0;
  return 1;
}
bool sleq(int *s1, int *s2) {
  if (s1[0]< s2[0]) return 1;
  if (s1[0]> s2[0]) return 0;
  returnsleq(s1+1, s2+1);
}
// is SA a sorted suffix array for s?
bool isSorted(int *SA, int *s, int n) {
  for (int i = 0;  i< n-1;  i++) {
    if(!sleq(s+SA[i], s+SA[i+1])) return 0;
  }
  //每一个后缀都比其后的那个后缀小，那么后缀数组是升序的
  return1; 
}
#define Assert(c) if(!(c))\
  {cout << "\nAssertionviolation " << __FILE__ << ":"<< __LINE__ << endl;}
int main()
{
     int i, j;
     int n;
     int b = 'z' - 'a' + 2;
    
     int* s;
     int* SA;
     int* rank;
     int*height;
     charstr1[100002];
     charstr2[100002];
     intlength1;
     intlength2;
     int max;
     scanf("%s%s",str1, str2);
     length1 = strlen(str1);
     length2 = strlen(str2);
     n = length1 + length2 + 1; 
     s = new int[n+3];
     SA = new int[n+3];
     rank = new int[n];
     height = newint[n];
     s[n] = s[n+1] = s[n+2] = SA[n] = SA[n+1] =SA[n+2] = 0;
     for(i = 0;i < length1; i++){
         s[i] = str1[i] - 'a' + 1;
     }
     s[i++] = 27;
     for(j = 0;j < length2; j++){
         s[i++] = str2[j] - 'a' + 1;
     }
     suffixArray(s, SA, n, b);
     //构建后缀数组
     /*
     Assert(s[n] == 0);
     Assert(s[n+1] ==0);
     //s是字符串数组，求后缀数组时不应被改变
     Assert(SA[n] == 0);
     Assert(SA[n+1] ==0);
     //长度为n的字符串有n个后缀
     Assert(isPermutation(SA,n));
     //后缀数组是以第i(0<= i < n)个字符作为起点的后缀的数组
     Assert(isSorted(SA,s, n));
     //后缀数组必须有序
     */
     suffixArrayHeight(s, SA, n, b, rank,height);
     max = 0;
     for(i = 1;i < n; i++){
         if((SA[i]< length1 && SA[i - 1] > length1)||
         (SA[i] > length1 && SA[i -1] < length1))
        {
            if(max< height[i])
                max = height[i];
        }
     }
     cout << max << endl;
     delete []s; delete [] SA; delete[] rank; delete [] height;
}

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[i] http://www.mpi-sb.mpg.de/~sanders/programs/suffix/

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