最长公共子序列，Longest-Common-Subsequence(LCS)

来源：互联网发布：算法第四版中文电子书编辑：程序博客网时间：2024/06/07 10:46

公共子序列有别于公共子串，子序列可以不连续，如 X = [abcbdab], Y = [bdcaba]，则 bcba 为其一个公共最长子序列。同样考虑DP方案，假设 X = <x1, x2, ..., xm>， Y = <y1, y2, ..., yn> ，LCS(X, Y) 表示二者的公共最长子序列，则有

如果 xm == yn，则 LCS(X, Y) = xm + LCS(Xm-1, Yn-1)；
如果 xm != yn，则 LCS(X, Y) = max{LCS(Xm-1, Y), LCS(X, Yn-1)}；

于是LCS问题具有重叠子问题性质。

#include <iostream>#include <vector>#include <string>using namespace std;#define INF (~(1<<31))int main(){string s1, s2;cin >> s1 >> s2;int len1 = (int)s1.size(), len2 = (int)s2.size();vector<vector<int>> mat(len1+1, vector<int>(len2+1, 0));for(int i = 1; i <= len1; ++i){for(int j = 1; j <= len2; ++j){if(s1[i-1] == s2[j-1]){mat[i][j] = mat[i-1][j-1] + 1;}else if(mat[i-1][j] > mat[i][j-1]){mat[i][j] = mat[i-1][j];}else{mat[i][j] = mat[i][j-1];}}}string outs;int i = len1, j = len2;while(i && j){if(s1[i-1] == s2[j-1]){outs = s1[i-1] + outs;--i; --j;}else if(mat[i-1][j] > mat[i][j-1]){--i;}else{--j;}}cout << mat[len1][len2] << ": " << outs;return 0;}

0 0