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 C (GCC 9.2.1) (judge verdict: Accepted)

#include <float.h>
#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

// 内部定数
#define D_SIZE_MAX		3000										// 最大サイズ
#define D_ICNT_MAX		3											// 行内最大アイテム数

// 内部変数
static FILE *szpFpI;												// 入力
static int si2Itm[D_SIZE_MAX][D_SIZE_MAX];							// アイテム
static long long sl3Val[D_SIZE_MAX][D_SIZE_MAX][D_ICNT_MAX + 1];	// 価値
static int siRow, siCol;											// 行・列

// 内部変数 - テスト用
#ifdef D_TEST
	static int siRes;
	static FILE *szpFpA;
	static int siTNo;
#endif

// １行出力
int
fOutLine(
	char *pcpLine				// <I> １行
)
{
	char lc1Buf[1024];

#ifdef D_TEST
	lc1Buf[0] = '\0';
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, pcpLine)) {
		siRes = -1;
	}
#else
	printf("%s", pcpLine);
#endif

	return 0;
}

// 最大値 - 取得
long long
fGetMaxL(
	long long plVal1			// <I> 値１
	, long long plVal2			// <I> 値２
)
{
	if (plVal1 > plVal2) {
		return plVal1;
	}
	else {
		return plVal2;
	}
}

// 実行メイン
int
fMain(
)
{
	int i, j, k;
	char lc1Buf[1024];

	// データ - 初期化
	memset(si2Itm, 0, sizeof(si2Itm));
	memset(sl3Val, 0, sizeof(sl3Val));

	// 行・列・アイテム数 - 取得
	int liICnt;
	fgets(lc1Buf, sizeof(lc1Buf), szpFpI);
	sscanf(lc1Buf, "%d%d%d", &siRow, &siCol, &liICnt);

	// アイテム - 取得
	for (i = 0; i < liICnt; i++) {
		int liRow, liCol, liVal;
		fgets(lc1Buf, sizeof(lc1Buf), szpFpI);
		sscanf(lc1Buf, "%d%d%d", &liRow, &liCol, &liVal);
		si2Itm[liRow - 1][liCol - 1] = liVal;
	}

	// 価値 - セット
	for (i = 0; i < siRow; i++) {
		for (j = 0; j < siCol; j++) {

			// アイテム - 取得
			if (si2Itm[i][j] > 0) {
				for (k = D_ICNT_MAX - 1; k >= 0; k--) {
					sl3Val[i][j][k + 1] = fGetMaxL(sl3Val[i][j][k + 1], sl3Val[i][j][k] + (long long)si2Itm[i][j]);
				}
			}

			// 右へ
			if (j < siCol - 1) {
				for (k = 0; k <= D_ICNT_MAX; k++) {
					sl3Val[i][j + 1][k] = fGetMaxL(sl3Val[i][j + 1][k], sl3Val[i][j][k]);
				}
			}

			// 下へ
			if (i < siRow - 1) {
				for (k = 0; k <= D_ICNT_MAX; k++) {
					sl3Val[i + 1][j][0] = fGetMaxL(sl3Val[i + 1][j][0], sl3Val[i][j][k]);
				}
			}
		}
	}

	// 最大値 - 取得
	long long llMax = 0;
	for (i = 0; i <= D_ICNT_MAX; i++) {
		llMax = fGetMaxL(llMax, sl3Val[siRow - 1][siCol - 1][i]);
	}

	// 出力
	sprintf(lc1Buf, "%lld\n", llMax);
	fOutLine(lc1Buf);

	return 0;
}

// １回実行
int
fOne(
)
{
	int liRet;
	char lc1Buf[1024];

	// 入力 - セット
#ifdef D_TEST
	sprintf(lc1Buf, ".\\Test\\T%d.txt", siTNo);
	szpFpI = fopen(lc1Buf, "r");
	sprintf(lc1Buf, ".\\Test\\A%d.txt", siTNo);
	szpFpA = fopen(lc1Buf, "r");
	siRes = 0;
#else
	szpFpI = stdin;
#endif

	// 実行メイン
	liRet = fMain();

	// 残データ有無
#ifdef D_TEST
	lc1Buf[0] = '\0';
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, "")) {
		siRes = -1;
	}
#endif

	// テストファイルクローズ
#ifdef D_TEST
	fclose(szpFpI);
	fclose(szpFpA);
#endif

	// テスト結果
#ifdef D_TEST
	if (siRes == 0) {
		printf("OK %d\n", siTNo);
	}
	else {
		printf("NG %d\n", siTNo);
	}
#endif

	return 0;
}

// プログラム開始
int
main()
{

#ifdef D_TEST
	int i;
	for (i = D_TEST_SNO; i <= D_TEST_ENO; i++) {
		siTNo = i;
		fOne();
	}
#else
	fOne();
#endif

	return 0;
}