Hot Start Up (easy version)
This is an easy version of the problem. The constraints of $$$t$$$, $$$n$$$, $$$k$$$ are the only difference between versions.You have a device with two CPUs. You also have $$$k$$$ programs, numbered $$$1$$$ through $$$k$$$, that you can run on the CPUs. The $$$i$$$-th program ($$$1 \le i \le k$$$) takes $$$cold_i$$$ seconds to run on some CPU. However, if the last program we ran on this CPU was also program $$$i$$$, it only takes $$$hot_i$$$ seconds ($$$hot_i \le cold_i$$$). Note that this only applies if we run program $$$i$$$ multiple times consecutively  — if we run program $$$i$$$, then some different program, then program $$$i$$$ again, it will take $$$cold_i$$$ seconds the second time.You are given a sequence $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$, consisting of integers from $$$1$$$ to $$$k$$$. You need to use your device to run programs $$$a_1, a_2, \ldots, a_n$$$ in sequence. For all $$$2 \le i \le n$$$, you cannot start running program $$$a_i$$$ until program $$$a_{i - 1}$$$ has completed.Find the minimum amount of time needed to run all programs $$$a_1, a_2, \ldots, a_n$$$ in sequence.
In the first test case, we can do the following:   Run program $$$a_1 = 1$$$ on CPU $$$1$$$. It takes $$$cold_1 = 3$$$ seconds to run.  Run program $$$a_2 = 2$$$ on CPU $$$2$$$. It takes $$$cold_2 = 2$$$ seconds to run.  Run program $$$a_3 = 2$$$ on CPU $$$2$$$. The last program run on this CPU was also program $$$2$$$, so it takes $$$hot_2 = 1$$$ second to run. In total, we need $$$3 + 2 + 1 = 6$$$ seconds to run them all. We can show this is optimal.In the second test case, we can use do the following:   Run program $$$a_1 = 1$$$ on CPU $$$1$$$. It takes $$$cold_1 = 5$$$ seconds to run.  Run program $$$a_2 = 2$$$ on CPU $$$2$$$. It takes $$$cold_2 = 3$$$ seconds to run.  Run program $$$a_3 = 1$$$ on CPU $$$1$$$. The last program run on this CPU was also program $$$1$$$, so it takes $$$hot_1 = 2$$$ seconds to run.  Run program $$$a_4 = 2$$$ on CPU $$$2$$$. The last program run on this CPU was also program $$$2$$$, so it takes $$$hot_2 = 1$$$ second to run. In total, we need $$$5 + 3 + 2 + 1 = 11$$$ seconds. We can show this is optimal.
Input consists of multiple test cases. The first line contains a single integer $$$t$$$, the number of test cases ($$$1 \le t \le 5000$$$).The first line of each test case contains $$$n$$$ and $$$k$$$ ($$$1 \le n, k \le 5000$$$).The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le k$$$).The third line of each test case contains $$$k$$$ integers $$$cold_1, cold_2, \ldots, cold_k$$$ ($$$1 \le cold_i \le 10^9$$$).The fourth line of each test case contains $$$k$$$ integers $$$hot_1, hot_2, \ldots, hot_k$$$ ($$$1 \le hot_i \le cold_i$$$).It is guaranteed the sum of $$$n$$$ and the sum of $$$k$$$ over all test cases do not exceed $$$5000$$$.
For each test case, print the minimum time needed to run all programs in the given order.
import sys

def read_integers():
    return map(int, sys.stdin.readline().split())
for _ in range(int(sys.stdin.readline())):
    (n, k) = read_integers()
    (a, c, h) = ([0], [0], [0])
    a += list(read_integers())
    c += list(read_integers())
    h += list(read_integers())
    dp = [0] + [float('inf')] * k
    s = c[a[1]]
    for i in range(2, n + 1):
        if a[i] == a[i - 1]:
            s += h[a[i]]
            continue
        s += c[a[i]]
        k = dp[a[i]]
        dp[a[i]] = float('inf')
        dp[a[i - 1]] = min(min(dp), k - c[a[i]] + h[a[i]])
    print(min(dp) + s)
import sys

def read_integers():
    return map(int, sys.stdin.readline().split())
for _ in range(int(sys.stdin.readline())):
    (n, k) = read_integers()
    (a, c, h) = ([0], [0], [0])
    a += list(read_integers())
    c += list(read_integers())
    h += list(read_integers())
    dp = [0] + [float('inf')] * k
    s = c[a[1]]
    for i in range(2, n + 1):
        if a[i] == a[i - 1]:
            s += h[a[i]]
            continue
        s += c[a[i]]
        k = dp[a[i]]
        dp[a[i]] = float('inf')
        dp[a[i - 1]] = min(min(dp), k - c[a[i]] + h[a[i]])
    print(min(dp) + s)
import sys

def read_integers():
    return map(int, sys.stdin.readline().split())
for _ in range(int(sys.stdin.readline())):
    (n, k) = read_integers()
    (a, c, h) = ([0], [0], [0])
    a += list(read_integers())
    c += list(read_integers())
    h += list(read_integers())
    dp = [0] + [float('inf')] * k
    s = c[a[1]]
    for i in range(2, n + 1):
        if a[i] == a[i - 1]:
            s += h[a[i]]
            continue
        s += c[a[i]]
        k = dp[a[i]]
        dp[a[i]] = float('inf')
        dp[a[i - 1]] = min(min(dp), k - c[a[i]] + h[a[i]])
    print(min(dp) + s)