数据结构：LCS 最长公共子序列动态规划DP + DFS深度优先遍历

素颜马尾好姑娘i 2022-05-31 12:23 142阅读 0赞

# 问题是什么？ #

就是求解两个字符串的最长的公共子序列，这是一个经典的DP动态规划问题

# 代码如下 #

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    #include <iomanip>
    
    using namespace std;
    
    
    string a, b;
    int lena, lenb;
    int LCSbyRecursion(int i, int j)
    {
        if (i == lena || j == lenb)
            return 0;
        if (a[i] == b[j])
            return 1 + LCSbyRecursion(i + 1, j + 1);
        else
            return LCSbyRecursion(i + 1, j)>LCSbyRecursion(i, j + 1) ? LCSbyRecursion(i + 1, j) : LCSbyRecursion(i, j + 1);
    }
    
    const int SIZE = 500;
    int flag[SIZE][SIZE] = { 0 };
    int cal[SIZE][SIZE] = { 0 };
    void LCSbyDP()//动态规划
    {
        for (int i = 1; i <= lena; i++)
        {
            for (int j = 1; j <= lenb; j++)
            {
                if (a[i - 1] == b[j - 1])
                {
                    cal[i][j] = cal[i - 1][j - 1] + 1;
                    flag[i][j] = 1;//斜向上
                }
                else if (cal[i - 1][j] > cal[i][j - 1])
                {
                    cal[i][j] = cal[i - 1][j];
                    flag[i][j] = 2;//向下
                }
                else
                {
                    cal[i][j] = cal[i][j - 1];
                    flag[i][j] = 3;//向右
                }
            }
        }
    
        char res[SIZE];
        int i = lena, j = lenb, k = 0;
        while (i>0 && j>0)
        {
            if (flag[i][j] == 1)
            {
                res[k++] = a[i - 1];
                i--;
                j--;
            }
            else if (flag[i][j] == 2)
                i--;
            else
                j--;
        }
        cout << "最大公共子序列是：";
        for (int i = k - 1; i >= 0; i--)
        {
            cout << res[i];
        }
        cout << endl << "长度为：" << cal[lena][lenb] << endl;
    }
    int main()
    {
        a = "ABCBDAB";
        b = "BDCABA";
        lena = a.length();
        lenb = b.length();
        cout << "递归解法：" << LCSbyRecursion(0, 0) << endl;;
        LCSbyDP();
        system("pause");
        return 0;
    }