给定两个字符串X ， Y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跳过第i个字符。
3.从y跳过第j个字符。
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跳过第i个字符，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跳过第j个字符，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-JiPython3
# 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 vibhu4agarwalC#
// 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