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 C++ (GCC 9.2.1) (judge verdict: Accepted)


#include <bits/stdc++.h>
using namespace std;

vector<vector<int>> data2(50, vector<int>(4));
long max_wk=0;

void dfs(vector<int> &vec,int N, int M, int A_n, int Q)
{
    //終了条件
    //数列のサイズがNになったら
    if(vec.size()==N){
    //条件チェック
    //Q個の条件を調べる。
        long point=0;
        for(int i=0;i<Q;i++){
            if(vec.at(data2.at(i).at(1)-1)-vec.at(data2.at(i).at(0)-1)==data2.at(i).at(2))
                point+=data2.at(i).at(3);
        }
        max_wk=max(max_wk,point);
        return;
    }

    //再帰条件
    //Ai以上の値をAi+1に追加する。
    for(int i=A_n;i<=M;i++){
        vector<int> wk=vec;
        wk.push_back(i);
        dfs(wk,N,M,i,Q);
    }
    
    return;
}

int main()
{
    int n,m,q;
    cin>>n>>m>>q;

    for(int i=0;i<q;i++){
        cin>>data2.at(i).at(0);
        cin>>data2.at(i).at(1);
        cin>>data2.at(i).at(2);
        cin>>data2.at(i).at(3);
    }

    vector<int> wk;

    dfs(wk,n,m,1,q);

    cout<<max_wk<<endl;

    return 0;
}