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: Wrong Answer)

// #Sazaの1日1AC
#include<bits/stdc++.h>
using namespace std;
const long long mod=1000000000+7;
const long long INF=9e15;
const double PI=3.14159265358979323846;

//マクロ
//型エイリアス
using ll = long long;
using P=pair<long long,long long>;
using vl=vector<long long>;
using vvl=vector<vector<long long>>;
using vP=vector<pair<long long, long long>>;
//ショートカット
#define rep(i,n) for(long long i = 0; i < n; i++)
#define rep2(i,k,n) for(long long i = k; i < n; i++) //半開区間
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define mkp make_pair
//入力
#define vin(v,N) for(long long i=0;i<N;i++) cin>>v.at(i)
#define lin(n) long long n;cin>>n
#define chin(x) char x;cin>>x;
#define sin(s) string s;cin>>s;
#define vlin(v,N) vector<long long>v(N);for(long long i=0;i<N;i++)cin>>v.at(i)

#define max(a,b) max((ll)a,(ll)b)
#define min(a,b) min((ll)a,(ll)b)

//関数
//最大公約数
long long gcd(long long m, long long n){
  long long a=max(m,n);
  long long b=min(m,n);
  long long r=a%b;
  while(r!=0){
    a=b; b=r; r=a%b;
  }
  return b;
}
// 最小公倍数
long long lcd(long long m, long long n){
  return m*n/gcd(m,n);
}
//xのn乗
long long power(long long x, long long N){
  long long ret=1;
  for(long long i=0;i<N;i++)   
    ret*=x;
  return ret;
}
//繰り返し二乗法
long long repeat_squaring(long long x, long long n){
  if(n==0) return 1;
  else if(n%2==0){
    long long t=repeat_squaring(x,n/2);
    return t*t%mod;
  }
  else
    return x*repeat_squaring(x,n-1);
}
//素因数分解(mapで返す)
map<long long,long long> factor(long long p){
  ll p1=p;
  map<long long,long long>ret;
  for(long long i=2;i*i<=p1;i++){
     while(p%i==0){
      ret[i]++;
      p/=i;
    }
  }
  if(p!=1)
  ret[p]++;
  return ret;
}
//素数判定
bool is_prime(long long N){
  for(long long i=2;i*i<=N;i++){
    if(N%i==0)
      return false;
  }
  return true;
}
//最大値更新
void chmax(long long &a,long long b){
  a=max(a,b);
}
//最小値更新
void chmin(long long &a,long long b){
  a=min(a,b);
}

//構造体

//Union-Find木
struct UnionFind{
  //メンバ変数 
  vector<long long>par; //par[i]:=頂点iの親
  vector<long long>s; //s[i]:=頂点iが属する集合の個数(iが根のとき)
  
  //コンストラクタ
  UnionFind(long long N):par(N),s(N){
    for(long long i=0;i<N;i++){
      par[i]=i; //最初はすべてが根として初期化
      s[i]=1;
    }
  }
  //メンバ関数
  //root(i):=頂点iの根
  long long root(long long i){
    if(par[i]==i) return i;
    par[i]=root(par[i]); //経路圧縮
    return root(par[i]); //再帰
  }
  //size[i]:=iが属する集合の個数
  long long size(long long i){
    return s[root(i)];
  }
  //same(x,y) xとyが同じ根を持つか(同じ集合に含まれるか)
  bool same(long long x, long long y){
    return root(x)==root(y);
  }
  //unite(x,y) xの根をyの根に繋げる(集合を合併)
  void unite(long long x, long long y){
    if(!same(x,y)){
      long long rx=root(x);
      long long ry=root(y);
      par[rx]=par[ry];
      s[ry]=s[rx]+s[ry];
    }
  }
};

/////////////////⊂('ω'⊂ )))Σ≡GO!/////////////////

//メイン関数
int main(){
  //doubleの桁数
  cout<<fixed<<setprecision(15);

  lin(H); lin(W); lin(K);
  vector<vector<ll>> item(H, vector<ll>(W, 0));
  rep(i, K){
    lin(h);lin(w);lin(c);
    h--;w--;
    item[h][w] = c;
  }
  vector<ll> v = {0, 0, 0};

  vector<vector<ll>> dp(H, vector<ll>(W, 0));

  dp[0][0] = item[0][0];

  rep(i, H) {
    rep(j, W) {
      ll m = INF;
      ll s = 0;
      rep(k, 3){
        m = min(v[k], m);
        s += v[k];
      }
      if(m < item[i][j]){
        rep(k, 3){
          if(v[k] == m){
            v[k] = item[i][j];
            break;
          }
        }
      }
      ll S = 0;
      rep(k, 3) S += v[k];


      if(j > 0) dp[i][j] = dp[i][j - 1] + S - s;
      if(i > 0 && dp[i][j] < dp[i-1][j] + item[i][j]){
        v = {0, 0, 0};
        v[0] = item[i][j];
        dp[i][j] = dp[i-1][j] + item[i][j];
      }
    }
  }

  cout << dp[H-1][W-1] << endl;
}