给定两个字符串X ， Y和一个整数k 。现在的任务是以最小成本转换字符串X，使得转换后 X 和 Y 的最长公共子序列的长度为 k。转换成本计算为旧字符值和新字符值的异或。 ‘a’ 的字符值为 0，’b’ 为 1，依此类推。
例子：
Input : X = "abble",
Y = "pie",
k = 2
Output : 25
If you changed 'a' to 'z', it will cost 0 XOR 25.这个问题可以通过对最长递增子序列的动态规划问题稍加改动来解决。我们维护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 长度的 LCS，否则总是可能的。
让 dp[i][j][p] 存储在 x[0…i] 和 y[0….j] 中实现长度为 p 的 LCS 的最小成本。
基本步长为 dp[i][j][0] = 0，因为我们可以在没有任何成本的情况下实现 0 长度的 LCS，并且在这种情况下，对于 i < 0 或 j 0)。
否则有3种情况：
1. 将 x[i] 转换为 y[j]。
2. 从 x 中跳过第i个字符。
3. 从 y 中跳过第j个字符。
如果我们将 x[i] 转换为 y[j]，那么 cost = f(x[i]) XOR f(y[j]) 将被添加并且 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’ 的最低成本始终是 f(a) XOR f(c) 因为 f(a) XOR f(c) <= (f(a) XOR f(b) ) + f(b) XOR f(c)) 对所有 a、b、c。
如果您从 x 中跳过第i个字符，则 i 将减 1，不会增加任何成本并且 LCS 将保持不变。
如果您从 x 中跳过第j个字符，则 j 将减 1，不会增加任何成本并且 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：
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