Problem:
Score : 500 points

Problem Statement

There are K items placed on a grid of squares with R rows and C columns. Let (i, j) denote the square at the i-th row (1 \leq i \leq R) and the j-th column (1 \leq j \leq C). The i-th item is at (r_i, c_i) and has the value v_i.

Takahashi will begin at (1, 1), the start, and get to (R, C), the goal. When he is at (i, j), he can move to (i + 1, j) or (i, j + 1) (but cannot move to a non-existent square).

He can pick up items on the squares he visits, including the start and the goal, but at most three for each row. It is allowed to ignore the item on a square he visits.

Find the maximum possible sum of the values of items he picks up.

Constraints

1 \leq R, C \leq 3000

1 \leq K \leq \min(2 \times 10^5, R \times C)

1 \leq r_i \leq R

1 \leq c_i \leq C

(r_i, c_i) \neq (r_j, c_j) (i \neq j)

1 \leq v_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

R C K
r_1 c_1 v_1
r_2 c_2 v_2
:
r_K c_K v_K

Output

Print the maximum possible sum of the values of items Takahashi picks up.

Sample Input 1

2 2 3
1 1 3
2 1 4
1 2 5

Sample Output 1

8

He has two ways to get to the goal:

Visit (1, 1), (1, 2), and (2, 2), in this order. In this case, the total value of the items he can pick up is 3 + 5 = 8.

Visit (1, 1), (2, 1), and (2, 2), in this order. In this case, the total value of the items he can pick up is 3 + 4 = 7.

Thus, the maximum possible sum of the values of items he picks up is 8.

Sample Input 2

2 5 5
1 1 3
2 4 20
1 2 1
1 3 4
1 4 2

Sample Output 2

29

We have four items in the 1-st row. The optimal choices are as follows:

Visit (1, 1) (1, 2), (1, 3), (1, 4), (2, 4), and (2, 5), in this order, and pick up all items except the one on (1, 2). Then, the total value of the items he picks up will be 3 + 4 + 2 + 20 = 29.

Sample Input 3

4 5 10
2 5 12
1 5 12
2 3 15
1 2 20
1 1 28
2 4 26
3 2 27
4 5 21
3 5 10
1 3 10

Sample Output 3

142

Solution in PyPy3 (7.3.0) (judge verdict: Wrong Answer)

## coding: UTF-8
R,C,K = map(int,input().split())
rcv = []
for i in range(K):
    rcv.append(list(map(int,input().split())))

#print(rcv)

available = [False] * (R*C+1) #アイテムの有無を示す
item = [0] * (R*C+1)
for i in range(K):
    index = (rcv[i][0] - 1) * C + (rcv[i][1] - 1) +1
    available[index] = True
    item[index] = rcv[i][2]
#print(available, item)


def solver(R,C,K,rcv):
    #その場所にいる段階での最大スコアを出力
    #その行で獲得した個数も格納
    #DPの都合上0番目の要素を追加
    #グリッドは1次元で処理(1行目は1~C, 2行目はC+1~2C, ...)
    dp = []
    for i in range(R*C+1):
        dp.append([0] * 4)
    #print(dp)
    for i in range(1, R*C+1):
        #print(item[i])
        if(i == 1):
            #左上なら
            #dp[i][0] = dp[i-1][0] #左から(何も持ってこない)
            if(available[i]):
                #itemがある場合は取得するときと取得しない二通りに分けられる
                dp[i][1] = dp[i-1][0] + item[i]
        elif(i <= C):
            #左上以外の上(左からくる場合に限られる)
            dp[i][0] = dp[i-1][0] #itemを取得しない場合
            if(available[i]):
                dp[i][1] = max(dp[i-1][1], dp[i-1][0] + item[i])
                #if(dp[i-1][1] > 0): #左ですでに一個以上所持している場合にのみ、このマスで二個目ということが可能
                dp[i][2] = max(dp[i-1][2], dp[i-1][1] + item[i])
                #else:
                    #dp[i][2] = 0
                #if(dp[i-1][2] > 0): #左ですでに二個以上所持している場合にのみ、このマスで三個目ということが可能
                dp[i][3] = max(dp[i-1][3], dp[i-1][2] + item[i])
                #else:
                    #dp[i][3] = 0
            else:
                dp[i][1] = dp[i-1][1]
                dp[i][2] = dp[i-1][2]
                dp[i][3] = dp[i-1][3]
        elif(i % C == 1):
            #左上以外の左(上からくる場合に限られる)
            dp[i][0] = max(dp[i-C]) #下に降りてくる場合は、行で取得したアイテム数はリセットされる
            if(available[i]):
                #dp[i][1] = max(dp[i-C]) + item[i]
                dp[i][1] = dp[i][0] + item[i]
        else:
            #左でも上でもない場合
            dp[i][0] = max(dp[i-1][0], max(dp[i-C])) #左からくる場合は0限定だが、上からくる場合は、上の行で何個拾っててもいい
            if(available[i]):
                #dp[i][1] = max(dp[i-1][1], dp[i-1][0] + item[i], max(dp[i-C]) + item[i] ) #左から一個持った状態で来て拾わないor左から何も持たずに来て拾うor上から来て(所持数関係無し)拾う
                dp[i][1] = max(dp[i-1][1], dp[i][0] + item[i])
                #if(dp[i-1][1] > 0): #左ですでに一個以上所持している場合にのみ、このマスで二個目ということが可能
                dp[i][2] = max(dp[i-1][2], dp[i-1][1] + item[i]) #上から来て二個目は不可能
                #else:
                    #dp[i][2] = 0
                #if(dp[i-1][2] > 0):
                dp[i][3] = max(dp[i-1][3], dp[i-1][2] + item[i])
                #else:
                    #dp[i][3] = 0
            else:
                dp[i][1] = dp[i-1][1]
                dp[i][2] = dp[i-1][2]
                dp[i][3] = dp[i-1][3]

         

        #print(i, dp)
    print(max(dp[R*C]))


#solver(R,C,K,rcv)
print(-1)