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]

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

27

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; }

本文章欢迎转载，请保留原始博客链接http://blog.csdn.net/fsdev/article

---

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