Problem:
Score : 300 points

Problem Statement

Given are positive integers N, M, Q, and Q quadruples of integers ( a_i , b_i , c_i , d_i ).

Consider a sequence A satisfying the following conditions:

A is a sequence of N positive integers.

1 \leq A_1 \leq A_2 \le \cdots \leq A_N \leq M.

Let us define a score of this sequence as follows:

The score is the sum of d_i over all indices i such that A_{b_i} - A_{a_i} = c_i. (If there is no such i, the score is 0.)

Find the maximum possible score of A.

Constraints

All values in input are integers.

2 ≤ N ≤ 10

1 \leq M \leq 10

1 \leq Q \leq 50

1 \leq a_i < b_i \leq N ( i = 1, 2, ..., Q )

0 \leq c_i \leq M - 1 ( i = 1, 2, ..., Q )

(a_i, b_i, c_i) \neq (a_j, b_j, c_j) (where i \neq j)

1 \leq d_i \leq 10^5 ( i = 1, 2, ..., Q )

Input

Input is given from Standard Input in the following format:

N M Q
a_1 b_1 c_1 d_1
:
a_Q b_Q c_Q d_Q

Output

Print the maximum possible score of A.

Sample Input 1

3 4 3
1 3 3 100
1 2 2 10
2 3 2 10

Sample Output 1

110

When A = \{1, 3, 4\}, its score is 110. Under these conditions, no sequence has a score greater than 110, so the answer is 110.

Sample Input 2

4 6 10
2 4 1 86568
1 4 0 90629
2 3 0 90310
3 4 1 29211
3 4 3 78537
3 4 2 8580
1 2 1 96263
1 4 2 2156
1 2 0 94325
1 4 3 94328

Sample Output 2

357500

Sample Input 3

10 10 1
1 10 9 1

Sample Output 3

1

Solution in Python (3.8.2) (judge verdict: Runtime Error)

import sys
from math import ceil, floor, sqrt, sin, cos, pi
from itertools import accumulate, permutations, combinations
from fractions import gcd # 最大公約数
from collections import deque, Counter
from operator import itemgetter
from heapq import heappop,heappush
sys.setrecursionlimit(10**7)
def lcm(x, y): return ((x * y) // gcd(x, y)) # 最小公倍数
# list(map(int, input().split()))
 
n, m, q = map(int, input().split())
vec = []
a = []
b = []
c = []
d = []
for _ in range(q):
    i, j, k, l = map(int, input().split())
    a.append(i - 1)
    b.append(j - 1)
    c.append(k)
    d.append(l)
it = a + b
it = list(set(it))
for i in range(1, m+1):
    for j in range(len(it)):
        vec.append(i)
na = []
nb = []
for i in range(q):
    for j in range(len(it)):
        if a[i] == it[j]:
            na.append(j)
        if b[i] == it[j]:
            nb.append(j)
ans = 0
for x in combinations(vec, len(it)):
    cnt = 0
    for i in range(q):
        if x[nb[i]] - x[na[i]] == c[i]:
            cnt += d[i]
    #print(x, cnt)
    if ans < cnt:
        ans = cnt
print(ans)
import sys
from math import ceil, floor, sqrt, sin, cos, pi
from itertools import accumulate, permutations, combinations, combinations_with_replacement
from fractions import gcd # 最大公約数
from collections import deque, Counter
from operator import itemgetter
from heapq import heappop,heappush
sys.setrecursionlimit(10**7)
def lcm(x, y): return ((x * y) // gcd(x, y)) # 最小公倍数
# list(map(int, input().split()))

n, m, q = map(int, input().split(' '))
a = []
b = []
c = []
d = []
for _ in range(q):
    i, j, k, l = map(int, input().split())
    a.append(i - 1)
    b.append(j - 1)
    c.append(k)
    d.append(l)
it = a + b
it = list(set(it))
na = []
nb = []
for i in range(q):
    for j in range(len(it)):
        if a[i] == it[j]:
            na.append(j)
        if b[i] == it[j]:
            nb.append(j)
ans = 0
for x in combinations_with_replacement(range(1, m+1), len(it)):
    cnt = 0
    for i in range(q):
        if x[nb[i]] - x[na[i]] == c[i]:
            cnt += d[i]
    #print(x, cnt)
    if ans < cnt:
        ans = cnt
print(ans)