最长公共子序列

来源：互联网发布：2016年网络黑客攻击编辑：程序博客网时间：2024/04/28 12:32

动态规划的一个计算最长公共子序列的方法如下，以两个序列 $X$ 、 $Y$ 为例子：

设有二维数组 $f[i][j]$ 表示 $X$ 的 $i$ 位和 $Y$ 的 $j$ 位之前的最长公共子序列的长度，则有：

$f[1][1] = same(1,1)$

$f[i][j] = max\{f[i-1][j-1] + same(i,j), f[i-1][j],f[i][j-1]\}$

其中， $same(a,b)$ 当 $X$ 的第 $a$ 位与 $Y$ 的第 $b$ 位完全相同时为“1”，否则为“0”。

此时， $f[i][j]$ 中最大的数便是 $X$ 和 $Y$ 的最长公共子序列的长度，依据该数组回溯，便可找出最长公共子序列。

该算法的空间、时间复杂度均为 $O(n^{2})$ ，经过优化后，空间复杂度可为 $O(n)$ ，时间复杂度为 $O(n\log n)$ 。

#include <stdio.h>/**  * This is an implementation, in C, of the Longest Common Subsequence algorithm.  * That is, given two strings A and B, this program will find the longest sequence  * of letters that are common and ordered in A and B.  *  * There are only two reasons you are reading this:  *   - you don't care what the algorithm is but you need a piece of code to do it  *   - you're trying to understand the algorithm, and a piece of code might help  * In either case, you should either read an entire chapter of an algorithms textbook  * on the subject of dynamic programming, or you should consult a webpage that describes  * this particular algorithm.   It is important, for example, that we use arrays of size  * |A|+1 x |B|+1.   *  * This code is provided AS-IS.  * You may use this code in any way you see fit, EXCEPT as the answer to a homework  * problem or as part of a term project in which you were expected to arrive at this  * code yourself.    *   * Copyright (C) 2005 Neil Jones.  *  *//**  * Takes two NULL-terminated strings.  Returns the longest common subsequence to both.  */char* LCS(char* a, char* b);#define NEITHER       0#define UP            1#define LEFT          2#define UP_AND_LEFT   3int main(int argc, char* argv[]) {printf("%s\n", LCS(argv[1],argv[2]));}/*  * Note that this will allocate space for the result, and you will need  * to free it yourself to avoid a memory leak.  */char* LCS(char* a, char* b) {int n = strlen(a);int m = strlen(b);int** S;int** R;int ii;int jj;int pos;char* lcs;/*  * Memory allocation.  Sigh.  What a quaint little idea. Nobody does this any more... * A much better way to do this is to allocate one big array for S (and one for R)  * and then do your own pointer indexing.  However, that makes the card just slightly  * harder to read, so I avoid it. */S = (int **)malloc( (n+1) * sizeof(int) );R = (int **)malloc( (n+1) * sizeof(int) );for(ii = 0; ii <= n; ++ii) {S[ii] = (int*) malloc( (m+1) * sizeof(int) );R[ii] = (int*) malloc( (m+1) * sizeof(int) );}/* It is important to use <=, not <.  The next two for-loops are initialization */for(ii = 0; ii <= n; ++ii) {S[ii][0] = 0;R[ii][0] = UP;}for(jj = 0; jj <= m; ++jj) {S[0][jj] = 0;R[0][jj] = LEFT;}/* This is the main dynamic programming loop that computes the score and *//* backtracking arrays. */for(ii = 1; ii <= n; ++ii) {for(jj = 1; jj <= m; ++jj) { if( a[ii-1] == b[jj-1] ) {S[ii][jj] = S[ii-1][jj-1] + 1;R[ii][jj] = UP_AND_LEFT;}else {S[ii][jj] = S[ii-1][jj-1] + 0;R[ii][jj] = NEITHER;}if( S[ii-1][jj] >= S[ii][jj] ) {S[ii][jj] = S[ii-1][jj];R[ii][jj] = UP;}if( S[ii][jj-1] >= S[ii][jj] ) {S[ii][jj] = S[ii][jj-1];R[ii][jj] = LEFT;}}}/* The length of the longest substring is S[n][m] */ii = n; jj = m;pos = S[ii][jj];lcs = (char *) malloc( (pos+1) * sizeof(char) );lcs[pos--] = (char)NULL;/* Trace the backtracking matrix. */while( ii > 0 || jj > 0 ) {if( R[ii][jj] == UP_AND_LEFT ) {ii--;jj--;lcs[pos--] = a[ii];}else if( R[ii][jj] == UP ) {ii--;}else if( R[ii][jj] == LEFT ) {jj--;}}for(ii = 0; ii <= n; ++ii ) {free(S[ii]);free(R[ii]);}free(S);free(R);return lcs;}

http://www.cs.sunysb.edu/~algorith/implement/bioalgorithms/distrib/