程序员编程艺术:最长公共子序列(LCS)问题

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鸣谢:感谢@July,您的博客给予我很大的帮助,增强了我学习的动力。希望分享的力量永传!!!

问题来源:http://blog.csdn.net/v_july_v/article/details/6695482

问题来源作者:@July

本文作者:@baoquanz

本文仅给予一种动态规划下LCS的C++实现。

#include <iostream>#include <string>#include <ctime>using namespace std;// 最长公共子序列函数,输出最长公共子序列的长度int LCS_len(const string x, const string y){int m, n;m = x.length();n = y.length();int** c;c = new int*[m+1]; //动态地定义二维数组:c[m+1][n+1]for (int i=0; i<m+1; ++i){c[i] = new int[n+1];}for (int i=0; i<m+1; ++i)  //初始化第0列{c[i][0] = 0;}for (int j=0; j<n+1; ++j)  //初始化第0行{c[0][j] = 0;}for (int i=1; i<m+1; ++i)  //计算二维数组c[m+1][n+1]的值,其值记录了各子序列的长度{for (int j=1; j<n+1; ++j){if (x[i] == y[j]){c[i][j] = c[i-1][j-1] + 1;}else if (c[i-1][j] >= c[i][j-1]){c[i][j] = c[i-1][j];}else{c[i][j] = c[i][j-1];}}}int ret = c[m][n];  //保存公共子序列的最大长度for (int i=0; i<m+1; ++i){delete[] c[i];}delete[] c;return ret;}int main(){string x, y;cin>>x>>y;cout<<"x: "<<x<<endl;cout<<"y: "<<y<<endl;clock_t begin = clock();cout << "MAX_LEN: " << LCS_len(x, y) << endl;clock_t end = clock();cout << "TIME: " << (end - begin)/CLK_TCK << "s" << endl;cout << "TIME: " << (end - begin) << "ms" << endl;return 0;}


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