📜  制作长度为k的最长公共子序列的最低成本

📅  最后修改于: 2021-04-23 22:44:28             🧑  作者: Mango

给定两个字符串XY和一个整数k 。现在的任务是以最小的成本转换字符串X,以使转换后的X和Y的最长公共子序列的长度为k。转换成本按旧字符值和新字符值的XOR进行计算。 “ a”的字符值是0,“ b”的字符值是1,依此类推。

例子:

输入:X =“ abble”,Y =“ pie”,k = 2输出:25 如果将“ a”更改为“ z”,则费用为0 XOR 25。

该问题可以通过略微改变最长递增子序列的动态编程问题来解决。我们维持三个状态,而不是两个状态。
请注意,如果k> min(n,m),则不可能获得至少k个长度的LCS,否则总是有可能的。
令dp [i] [j] [p]在x [0…i]和y [0….j]中存储实现长度为p的LCS的最小成本。
将基本步长设为dp [i] [j] [0] = 0,因为我们无需花费任何成本即可获得0长度的LCS,并且在这种情况下i <0或j 0。
否则,有3种情况:
1.将x [i]转换为y [j]。
2.从x跳过i字符。
3.从y跳过j字符。

如果将x [i]转换为y [j],则cost = f(x [i])XOR将添加f(y [j]),LCS将减少1。f(x)将返回字符值的x。
请注意,将字符’a’转换为任何字符’c’的最低成本始终为f(a)XOR f(c),因为f(a)XOR f(c)<=(f(a)XOR f(b )+ f(b)对所有a,b,c进行XOR f(c))。
如果您从x跳过i字符,那么我将被减1,不会增加任何费用,LCS将保持不变。
如果从x跳过j字符,则j将减少1,不会增加成本,LCS将保持不变。

所以,

dp[i][j][k] = min(cost + dp[i - 1][j - 1][k - 1], 
                  dp[i - 1][j][k], 
                  dp[i][j - 1][k])
The minimum cost to make the length of their
LCS atleast k is dp[n - 1][m - 1][k]

C++
#include 
using namespace std;
const int N = 30;
  
// Return Minimum cost to make LCS of length k
int solve(char X[], char Y[], int l, int r, 
                     int k, int dp[][N][N])
{
    // If k is 0.
    if (!k)
        return 0;
  
    // If length become less than 0, return
    // big number.
    if (l < 0 | r < 0)
        return 1e9;
  
    // If state already calculated.
    if (dp[l][r][k] != -1)
        return dp[l][r][k];
  
    // Finding the cost
    int cost = (X[l] - 'a') ^ (Y[r] - 'a');
  
    // Finding minimum cost and saving the state value
    return dp[l][r][k] = min({cost +
                      solve(X, Y, l - 1, r - 1, k - 1, dp),
                             solve(X, Y, l - 1, r, k, dp), 
                             solve(X, Y, l, r - 1, k, dp)});
}
  
// Driven Program
int main()
{
    char X[] = "abble";
    char Y[] = "pie";
    int n = strlen(X);
    int m = strlen(Y);
    int k = 2;
  
    int dp[N][N][N];
    memset(dp, -1, sizeof dp);
    int ans = solve(X, Y, n - 1, m - 1, k, dp);
  
    cout << (ans == 1e9 ? -1 : ans) << endl;
    return 0;
}


Java
class GFG 
{
  
    static int N = 30;
  
    // Return Minimum cost to make LCS of length k
    static int solve(char X[], char Y[], int l, int r,
                                    int k, int dp[][][])
    {
        // If k is 0.
        if (k == 0) 
        {
            return 0;
        }
  
        // If length become less than 0, return
        // big number.
        if (l < 0 | r < 0) 
        {
            return (int) 1e9;
        }
  
        // If state already calculated.
        if (dp[l][r][k] != -1) 
        {
            return dp[l][r][k];
        }
  
        // Finding the cost
        int cost = (X[l] - 'a') ^ (Y[r] - 'a');
  
        // Finding minimum cost and saving the state value
        return dp[l][r][k] = Math.min(Math.min(cost + 
                solve(X, Y, l - 1, r - 1, k - 1, dp),
                solve(X, Y, l - 1, r, k, dp)),
                solve(X, Y, l, r - 1, k, dp));
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        char X[] = "abble".toCharArray();
        char Y[] = "pie".toCharArray();
        int n = X.length;
        int m = Y.length;
        int k = 2;
  
        int[][][] dp = new int[N][N][N];
        for (int i = 0; i < N; i++) 
        {
            for (int j = 0; j < N; j++)
            {
                for (int l = 0; l < N; l++)
                {
                    dp[i][j][l] = -1;
                }
            }
        }
        int ans = solve(X, Y, n - 1, m - 1, k, dp);
  
        System.out.println(ans == 1e9 ? -1 : ans);
    }
}
  
// This code contributed by Rajput-Ji


Python3
# Python3 program to calculate Minimum cost 
# to make Longest Common Subsequence of length k
N = 30
  
# Return Minimum cost to make LCS of length k
def solve(X, Y, l, r, k, dp):
  
    # If k is 0
    if k == 0:
        return 0
  
    # If length become less than 0, 
    # return big number
    if l < 0 or r < 0:
        return 1000000000
  
    # If state already calculated
    if dp[l][r][k] != -1:
        return dp[l][r][k]
  
    # Finding cost
    cost = ((ord(X[l]) - ord('a')) ^ 
            (ord(Y[r]) - ord('a')))
  
    dp[l][r][k] = min([cost + solve(X, Y, l - 1, 
                                          r - 1, k - 1, dp),
                              solve(X, Y, l - 1, r, k, dp),
                              solve(X, Y, l, r - 1, k, dp)])
  
    return dp[l][r][k]
  
# Driver Code
if __name__ == "__main__":
    X = "abble"
    Y = "pie"
    n = len(X)
    m = len(Y)
    k = 2
    dp = [[[-1] * N for __ in range(N)] 
                    for ___ in range(N)]
    ans = solve(X, Y, n - 1, m - 1, k, dp)
  
    print(-1 if ans == 1000000000 else ans)
  
# This code is contributed
# by vibhu4agarwal


C#
// C# program to find subarray with
// sum closest to 0
using System;
      
class GFG 
{
  
    static int N = 30;
  
    // Return Minimum cost to make LCS of length k
    static int solve(char []X, char []Y, int l, int r,
                                    int k, int [,,]dp)
    {
        // If k is 0.
        if (k == 0) 
        {
            return 0;
        }
  
        // If length become less than 0, return
        // big number.
        if (l < 0 | r < 0) 
        {
            return (int) 1e9;
        }
  
        // If state already calculated.
        if (dp[l,r,k] != -1) 
        {
            return dp[l,r,k];
        }
  
        // Finding the cost
        int cost = (X[l] - 'a') ^ (Y[r] - 'a');
  
        // Finding minimum cost and saving the state value
        return dp[l,r,k] = Math.Min(Math.Min(cost + 
                solve(X, Y, l - 1, r - 1, k - 1, dp),
                solve(X, Y, l - 1, r, k, dp)),
                solve(X, Y, l, r - 1, k, dp));
    }
  
    // Driver code
    public static void Main(String[] args) 
    {
        char []X = "abble".ToCharArray();
        char []Y = "pie".ToCharArray();
        int n = X.Length;
        int m = Y.Length;
        int k = 2;
  
        int[,,] dp = new int[N, N, N];
        for (int i = 0; i < N; i++) 
        {
            for (int j = 0; j < N; j++)
            {
                for (int l = 0; l < N; l++)
                {
                    dp[i,j,l] = -1;
                }
            }
        }
        int ans = solve(X, Y, n - 1, m - 1, k, dp);
  
        Console.WriteLine(ans == 1e9 ? -1 : ans);
    }
}
  
// This code is contributed by Princi Singh


输出:

3