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)

#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vii;
typedef vector<ll> vll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define pb push_back
#define odd(x) ((x)&1)
#define even(x) (!odd(x))
#define all(v) (v).begin(),(v).end()
#define rep(i,n) for(auto i=0;i<n;++i)
#define FASTIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define TEST_CASE int tc;cin>>tc;while(tc--)

template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

#ifdef LOCAL
#define cerr cout
#else
#endif

#define TRACE

#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif


void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}


template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
#define JD

/******************************************************************************************************************************/

#include <chrono>
#define Clock std::chrono::high_resolution_clock::now()
// #define TIME

const ll inf = 2e18;
const ll N = 2e5+3;
const ll mod = 2019;

#define vl2 vector<vll>


ll sum(vll &v){
	ll tot = v[0]+v[1]+v[2];
	return tot;
}

vll func(vll v,ll x){
	sort(all(v));
	chmax(v[0],x);
	return v;
}


int main()
{
    auto start_time = Clock;
    FASTIO

    #ifndef JD
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif

    

    ll n,m,k;
    cin>>n>>m>>k;
    vector<vll> val(n+1,vll(m+1)),dp1(n+1,vll(m+1,0));
    vector<vl2> dp(n+1,vl2(m+1,vll(3,0)));

    rep(i,k){
    	ll a,b,c; cin>>a>>b>>c;
    	val[a][b]=c;
    }

    for(ll i=1;i<=n;++i){
    	for(ll j=1;j<=m;++j){

    		vll tmp1 = func(dp[i][j],val[i][j]);
    		ll nw1 = sum(tmp1);
    		vll tmp2 = func(dp[i][j-1],val[i][j]);
    		ll nw2 = sum(tmp2);

    		ll dif = nw2 - sum(dp[i][j-1]);

    		if(nw1+dp1[i-1][j]>=dp1[i][j-1]+dif){
    			dp[i][j] = tmp1;
    			dp1[i][j] = nw1+dp1[i-1][j];
    		}
    		else{
    			dp[i][j] = tmp2;
    			dp1[i][j] = dp1[i][j-1] + dif;
    		}
    	}
    }
    ll ans = dp1[n][m];
    cout<<ans;


    auto end_time = Clock;
    #ifndef TIME
    	return 0;
    #endif
	cout << "\nTime elapsed: "
	<< (double)chrono::duration_cast<chrono::milliseconds>(end_time-start_time).count() 
	<< "ms\n";


    return 0;
}