stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <bits/stdc++.h>

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n, k;

int a[MXN], b[MXN], c[MXN];

int dp[MXN][4];



int Dp(int d, int lst) {

    if(d == n) return 0;

    if(dp[d][lst] != -1) return dp[d][lst];

    int ans = 0;

    if(lst != 0) ans = Dp(d + 1, 0) + a[d];

    if(lst != 1) ans = max(ans, Dp(d + 1, 1) + b[d]);

    if(lst != 2) ans = max(ans, Dp(d + 1, 2) + c[d]);

    return dp[d][lst] = ans;

}



int main() {

    ios_base::sync_with_stdio(false); cin.tie(NULL);

    cin >> n;

    for(int i = 0; i < n; i++) {

        cin >> a[i] >> b[i] >> c[i];

    }

    memset(dp, -1, sizeof dp);

    cout << Dp(0, 3) << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.609315785463649x):
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

typedef long long ll;

const int maxn=100025;

const int inf=0x3f3f3f3f;

ll dp[maxn][5];

struct node

{

    ll x,y,z;

}arr[maxn];

int main()

{

    int n;

    cin>>n;

    for(int i=1;i<=n;++i) 

	cin>>arr[i].x>>arr[i].y>>arr[i].z;

    dp[1][0]=arr[1].x;

	dp[1][1]=arr[1].y;

	dp[1][2]=arr[1].z;

    for(int i=2;i<=n;++i)

    {

    	dp[i][0]=max(dp[i-1][1],dp[i-1][2])+arr[i].x;

	    dp[i][1]=max(dp[i-1][0],dp[i-1][2])+arr[i].y;

	    dp[i][2]=max(dp[i-1][0],dp[i-1][1])+arr[i].z;

	}

    cout<<max(dp[n][0],max(dp[n][1],dp[n][2]))<<endl;

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;

#define tam 100001

int A[tam][3],n;

int dp[tam][3];

int solver(int y, int x)

{

    if(dp[y][x]!=-1)

        return dp[y][x];

    if(y == 0 )

    {

    	dp[y][x] = A[y][x];

        return A[y][x];

    }

    return dp[y][x] = max(solver((y-1),(x+1)%3)+ A[y][x],

	solver((y-1),(x+2)%3) + A[y][x]);

}

int main()

{

    int bas;

    cin>>n;

    for(int i = 0;i < n ;i++)

        for(int j = 0 ; j < 3; j++)

            cin >>A[i][j];

    memset(dp,-1,sizeof dp);

     for(int i = 0; i < 3; i ++)

    	bas = solver(n-1,i);

    cout <<max(max(dp[n-1][0],dp[n-1][1]),max(dp[n-1][0],dp[n-1][2]))<<"\n";

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.2005827664274378x):
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

typedef long long ll;

const int maxn=100025;

const int inf=0x3f3f3f3f;

ll dp[maxn][5];

struct node

{

    ll x,y,z;

}arr[maxn];

int main()

{

    int n;

    cin>>n;

    for(int i=1;i<=n;++i) 

	cin>>arr[i].x>>arr[i].y>>arr[i].z;

    dp[1][0]=arr[1].x;

	dp[1][1]=arr[1].y;

	dp[1][2]=arr[1].z;

    for(int i=2;i<=n;++i)

    {

    	dp[i][0]=max(dp[i-1][1],dp[i-1][2])+arr[i].x;

	    dp[i][1]=max(dp[i-1][0],dp[i-1][2])+arr[i].y;

	    dp[i][2]=max(dp[i-1][0],dp[i-1][1])+arr[i].z;

	}

    cout<<max(dp[n][0],max(dp[n][1],dp[n][2]))<<endl;

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

 

#define pb push_back

#define fst first

#define snd second

#define INF 0x7fffffff

#define INFLL 0x7FFFFFFFFFFFFFFF

#define lenA(ar) sizeof(ar)/sizeof(*ar)

#define lenrows(mat) sizeof(mat)/sizeof(*mat)

#define lencols(mat) sizeof(*mat)/sizeof(**mat)

#define fill(a,c) memset(&a, c, sizeof(a))

#define fore(i,x,y) for(ll i=x;i<y;i++)

#define foreI(i,x,y) for(ll i=x-1;i>=y;i--)

#define max2(a,b) max(a, b)

#define max3(a,b,c) max2(max2(a,b),c)

#define min2(a,b) min(a, b)

#define min3(a,b,c) min2(min2(a,b),c)

#define MOD 1000000009

#define all(x) (x).begin(), (x).end()

#define rall(x) (x).rbegin(), (x).rend()

#define FIN ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)

 

typedef long long ll;

typedef unsigned long long ull;

 

using namespace std;

ll arr[100010][3];

ll dp[100010][3];

int n;

ll solve(int i, int j){

    if(i==n-1) return arr[i][j];

    ll &r = dp[i][j];

    if(r>-1) return r;

    ll a = 0; ll b = 0;ll c = 0; ll act=arr[i][j];

    if(j!=0) a=solve(i+1, 0)+act;

    if(j!=1) b=solve(i+1, 1)+act;

    if(j!=2) c=solve(i+1, 2)+act;

   return r=max3(a,b,c);

}

 

int main() {

     

    cin>>n;

    arr[n][3];

    memset(dp,-1,sizeof(dp));

    fore(i,0,n){

        fore(j,0,3){

            cin>>arr[i][j];

        }

    }

 

    cout<<max3(solve(0,0),solve(0,1),solve(0,2))<<"\n";

 

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.3967082381232085x):
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

typedef long long ll;

const int maxn=100025;

const int inf=0x3f3f3f3f;

ll dp[maxn][5];

struct node

{

    ll x,y,z;

}arr[maxn];

int main()

{

    int n;

    cin>>n;

    for(int i=1;i<=n;++i) 

	cin>>arr[i].x>>arr[i].y>>arr[i].z;

    dp[1][0]=arr[1].x;

	dp[1][1]=arr[1].y;

	dp[1][2]=arr[1].z;

    for(int i=2;i<=n;++i)

    {

    	dp[i][0]=max(dp[i-1][1],dp[i-1][2])+arr[i].x;

	    dp[i][1]=max(dp[i-1][0],dp[i-1][2])+arr[i].y;

	    dp[i][2]=max(dp[i-1][0],dp[i-1][1])+arr[i].z;

	}

    cout<<max(dp[n][0],max(dp[n][1],dp[n][2]))<<endl;

    return 0;

}
// Slower C++ solution:
#include <cstdio>

#include <algorithm>

#include <cstring>

const int maxn = 3e5 + 5;

using namespace std;



int N;

int a[maxn][4];

int dp[maxn][4];

int main()

{

	scanf("%d", &N);

	for(int i = 1;i <= N;i++)

	{

		for(int j = 1;j <= 3;j++)		

		{

			scanf("%d", &a[i][j]);

		}

	}

	memset(dp, 0, sizeof dp);

	for(int i = 1;i <= 3;i++)

		dp[1][i] = a[1][i];

	for(int i = 2;i <= N;i++)

	{

		dp[i][1] = max(dp[i - 1][2] + a[i][1], dp[i - 1][3] + a[i][1]);

		dp[i][2] = max(dp[i - 1][1] + a[i][2], dp[i - 1][3] + a[i][2]);

		dp[i][3] = max(dp[i - 1][1] + a[i][3], dp[i - 1][2] + a[i][3]);

	}

	int ans = 0;

	for(int i = 1;i <= 3;i++)

		ans = max(dp[N][i], ans);

	printf("%d\n", ans);

	return 0;	

}

// Faster C++ solution for the same problem (measured speedup: 5.115712642256576x):
#include <bits/stdc++.h>

using namespace std;

#define ll long long int

ll N, x, y;

ll memo[(ll)1e5][3];

ll activities[(ll)1e5][3];



ll hp(int i, int ac){

	if (i >= N) return 0;

	if (memo[i][ac]) return memo[i][ac];

	ll maxv = 0;

	for (int au = 0; au < 3; au++)

		if (ac != au)

			maxv = max(maxv, hp(i + 1, au));

	return memo[i][ac] = activities[i][ac] + maxv;

}



int main(){

	cin >> N;

	for (x = 0; x < N; x++)

		for (y = 0; y < 3; y++)

			cin >> activities[x][y];

	y = 0;

	for (x = 0; x < 3; x++)

		y = max(y, hp(0 , x));

	cout << y;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

#define db(x) cerr<<__LINE__<<": "<<#x<<" = "<<(x)<<'\n'

const int N = 1e5;

int A[N+7][5], mem[N+7][5];



int dp(int i, int j) {

	if(i==0) return 0;

	if(mem[i][j]!=-1) return mem[i][j];

	if(j==1) return mem[i][j] = A[i][j] + max(dp(i-1, 2), dp(i-1, 3));

	if(j==2) return mem[i][j] = A[i][j] + max(dp(i-1, 3), dp(i-1, 1));

	else return mem[i][j] = A[i][j] + max(dp(i-1, 1), dp(i-1, 2));

}

int main(){

	ios_base::sync_with_stdio(0);

	int a,b,n,T,cas=0;

	cin>>n;

	for(int i=1;i<=n;++i) {

		cin>>A[i][1]>>A[i][2]>>A[i][3];

	}

	memset(mem, -1, sizeof(mem));

	int res = max(dp(n,1), max(dp(n,2), dp(n,3)));

	cout<<res<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.018358994910628x):
#include <bits/stdc++.h>

using namespace std;

#define ll long long int

ll N, x, y;

ll memo[(ll)1e5][3];

ll activities[(ll)1e5][3];



ll hp(int i, int ac){

	if (i >= N) return 0;

	if (memo[i][ac]) return memo[i][ac];

	ll maxv = 0;

	for (int au = 0; au < 3; au++)

		if (ac != au)

			maxv = max(maxv, hp(i + 1, au));

	return memo[i][ac] = activities[i][ac] + maxv;

}



int main(){

	cin >> N;

	for (x = 0; x < N; x++)

		for (y = 0; y < 3; y++)

			cin >> activities[x][y];

	y = 0;

	for (x = 0; x < 3; x++)

		y = max(y, hp(0 , x));

	cout << y;

}
// Slower C++ solution:
/*|In The Name Of Allah|*/



#include <bits/stdc++.h>



using namespace std;



const int N = 1e5+5;



int n;

int a[N][3];

int mem[N][3];



int go(int idx , int lst){

    if(idx == n + 1)

        return 0;

    if(~mem[idx][lst])

        return mem[idx][lst];

    int best = 0;

    for(int i = 0; i < 3; i++){

        if(idx == 1 || lst != i){

            best = max(best , go(idx + 1 , i) + a[idx][i]);

        }

    }

    return mem[idx][lst] = best;

}



int main(){

    scanf("%d" , &n);

    for(int i = 1; i <= n; i++){

        for(int j = 0; j < 3; j++){

            scanf("%d" , &a[i][j]);

        }

    }

    memset(mem , -1 , sizeof mem);

    printf("%d\n" , go(1 , 0));

}


// Faster C++ solution for the same problem (measured speedup: 2.1907813911746645x):
#include <bits/stdc++.h>

using namespace std;

#define ll long long int

ll N, x, y;

ll memo[(ll)1e5][3];

ll activities[(ll)1e5][3];



ll hp(int i, int ac){

	if (i >= N) return 0;

	if (memo[i][ac]) return memo[i][ac];

	ll maxv = 0;

	for (int au = 0; au < 3; au++)

		if (ac != au)

			maxv = max(maxv, hp(i + 1, au));

	return memo[i][ac] = activities[i][ac] + maxv;

}



int main(){

	cin >> N;

	for (x = 0; x < N; x++)

		for (y = 0; y < 3; y++)

			cin >> activities[x][y];

	y = 0;

	for (x = 0; x < 3; x++)

		y = max(y, hp(0 , x));

	cout << y;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;

//const int mod = 1e9+9;

/*ll gcd(ll x, ll y){return (y ? gcd(y, x%y) : x); }

*ll lcm(ll x, ll y){return x*(y/gcd(x,y));}

*void exEuclid(int a, int b){

	if(b==0){x = 1; y=0; d=a; return ;}

	exEuclid(b, a%b);

	int x1 = y;

	int y1 = x-(a/b)*y;

	x = x1; 100000

	y = y1;

} 

* */

//===========================================================//

int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i){

	if(i == n) return 0;

	if(dp[a][i] != -1) return dp[a][i];

	int& ans = dp[a][i];

	

	if(a==0){

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1){

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2){

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

}

int main(){

	cin>>n;

	for(int i=0; i<n; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

	memset(dp, -1, sizeof dp);

	cout<<max(go(0,0),max(go(1,0),go(2,0)))<<"\n";

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.2094123756775965x):
#include <bits/stdc++.h>

using namespace std;

#define ll long long int

ll N, x, y;

ll memo[(ll)1e5][3];

ll activities[(ll)1e5][3];



ll hp(int i, int ac){

	if (i >= N) return 0;

	if (memo[i][ac]) return memo[i][ac];

	ll maxv = 0;

	for (int au = 0; au < 3; au++)

		if (ac != au)

			maxv = max(maxv, hp(i + 1, au));

	return memo[i][ac] = activities[i][ac] + maxv;

}



int main(){

	cin >> N;

	for (x = 0; x < N; x++)

		for (y = 0; y < 3; y++)

			cin >> activities[x][y];

	y = 0;

	for (x = 0; x < 3; x++)

		y = max(y, hp(0 , x));

	cout << y;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



int m;

const int mx = 1e5+5;

int tasks[mx][3];

int diario[3][mx];

int cosa(int a, int i){

	if(i == m){

     return 0;

     }



	if(diario[a][i] != -1) {

        return diario[a][i];

    }

	int& ans = diario[a][i];



        if(a==0){

		ans=max(ans, cosa(1, i+1)+tasks[i][0]);

		ans=max(ans, cosa(2, i+1)+tasks[i][0]);

	}

        if(a==1){

		ans=max(ans, cosa(0, i+1)+tasks[i][1]);

		ans=max(ans, cosa(2, i+1)+tasks[i][1]);

        }

	if(a==2){

		ans=max(ans, cosa(0, i+1)+tasks[i][2]);

		ans=max(ans, cosa(1, i+1)+tasks[i][2]);

	}



	return ans;

}

int main(){

	cin>>m;

	for(int i=0; i<m; i++) {

        cin>>tasks[i][0]>>tasks[i][1]>>tasks[i][2];

	}

	memset(diario, -1, sizeof diario);

	cout<<max(cosa(0,0),max(cosa(1,0),cosa(2,0)))<<"\n";

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.2094387061645784x):
#include <bits/stdc++.h>

using namespace std;

#define ll long long int

ll N, x, y;

ll memo[(ll)1e5][3];

ll activities[(ll)1e5][3];



ll hp(int i, int ac){

	if (i >= N) return 0;

	if (memo[i][ac]) return memo[i][ac];

	ll maxv = 0;

	for (int au = 0; au < 3; au++)

		if (ac != au)

			maxv = max(maxv, hp(i + 1, au));

	return memo[i][ac] = activities[i][ac] + maxv;

}



int main(){

	cin >> N;

	for (x = 0; x < N; x++)

		for (y = 0; y < 3; y++)

			cin >> activities[x][y];

	y = 0;

	for (x = 0; x < 3; x++)

		y = max(y, hp(0 , x));

	cout << y;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;



int nn;

const int nxx = 1e5+5;

int acc[nxx][3];

int dpp[3][nxx];

int go(int aa, int ii)

{

	if(ii==nn) return 0;

	if(dpp[aa][ii] != -1) return dpp[aa][ii];

	int& ans = dpp[aa][ii];

	if(aa==0)

	{

		ans=max(ans, go(1, ii+1)+acc[ii][0]);

		ans=max(ans, go(2, ii+1)+acc[ii][0]);

	}

	if(aa==1)

	{

		ans=max(ans, go(0, ii+1)+acc[ii][1]);

		ans=max(ans, go(2, ii+1)+acc[ii][1]);

	}

	if(aa==2)

	{

		ans=max(ans, go(0, ii+1)+acc[ii][2]);

		ans=max(ans, go(1, ii+1)+acc[ii][2]);

	}



	return ans;



}

int main()

{

    cin>>nn;

    for(int ii=0; ii<nn; ii++) cin>>acc[ii][0]>>acc[ii][1]>>acc[ii][2];

    memset(dpp, -1, sizeof dpp);

    cout<<max(go(0,0), max(go(1,0),go(2,0)))<<"\n";

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.2094364873188392x):
#include <bits/stdc++.h>

using namespace std;

#define ll long long int

ll N, x, y;

ll memo[(ll)1e5][3];

ll activities[(ll)1e5][3];



ll hp(int i, int ac){

	if (i >= N) return 0;

	if (memo[i][ac]) return memo[i][ac];

	ll maxv = 0;

	for (int au = 0; au < 3; au++)

		if (ac != au)

			maxv = max(maxv, hp(i + 1, au));

	return memo[i][ac] = activities[i][ac] + maxv;

}



int main(){

	cin >> N;

	for (x = 0; x < N; x++)

		for (y = 0; y < 3; y++)

			cin >> activities[x][y];

	y = 0;

	for (x = 0; x < 3; x++)

		y = max(y, hp(0 , x));

	cout << y;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;



int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i)

{

	if(i==n) return 0;

	if(dp[a][i] != -1) return dp[a][i];

	int& ans = dp[a][i];

	if(a==0)

	{

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1)

	{

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2)

	{

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

	

}

int main()

{

    cin>>n;

    for(int i=0; i<n; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

    memset(dp, -1, sizeof dp);

    cout<<max(go(0,0), max(go(1,0),go(2,0)))<<"\n";

    return 0;

}

	

// Faster C++ solution for the same problem (measured speedup: 2.2094123756775965x):
#include <bits/stdc++.h>

using namespace std;

#define ll long long int

ll N, x, y;

ll memo[(ll)1e5][3];

ll activities[(ll)1e5][3];



ll hp(int i, int ac){

	if (i >= N) return 0;

	if (memo[i][ac]) return memo[i][ac];

	ll maxv = 0;

	for (int au = 0; au < 3; au++)

		if (ac != au)

			maxv = max(maxv, hp(i + 1, au));

	return memo[i][ac] = activities[i][ac] + maxv;

}



int main(){

	cin >> N;

	for (x = 0; x < N; x++)

		for (y = 0; y < 3; y++)

			cin >> activities[x][y];

	y = 0;

	for (x = 0; x < 3; x++)

		y = max(y, hp(0 , x));

	cout << y;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

int v[100009][3];

int dp[100009][3];

int n;

int vacation(int i, int j)

{

    if(i==n-1)

        return dp[i][j]=v[i][j];

    if(i>=n || j>=3)

        return 0;

    if(dp[i][j]!=-1)

        return dp[i][j];

    int temp = INT_MIN;

        if(j==0)

        {

             temp = max(vacation(i+1,j+1)+v[i][j],temp);

        temp = max(vacation(i+1,j+2)+v[i][j],temp);

 return dp[i][j]= temp;

        }

      else  if(j==1)

        {

             temp = max(vacation(i+1,0)+v[i][j],temp);

        temp = max(vacation(i+1,2)+v[i][j],temp);

 return dp[i][j]= temp;

        }

        else

        {

             temp = max(vacation(i+1,1)+v[i][j],temp);

        temp = max(vacation(i+1,0)+v[i][j],temp);

         return dp[i][j]= temp;

        }

}



int main()

{

    cin>>n;

    memset(dp,-1,sizeof(dp));

    for(int i=0; i<n; i++)

        cin>>v[i][0]>>v[i][1]>>v[i][2];

    cout<<max(vacation(0,0),max(vacation(0,1),vacation(0,2)))<<endl;

}


// Faster C++ solution for the same problem (measured speedup: 2.209534431535604x):
#include <bits/stdc++.h>

using namespace std;

#define ll long long int

ll N, x, y;

ll memo[(ll)1e5][3];

ll activities[(ll)1e5][3];



ll hp(int i, int ac){

	if (i >= N) return 0;

	if (memo[i][ac]) return memo[i][ac];

	ll maxv = 0;

	for (int au = 0; au < 3; au++)

		if (ac != au)

			maxv = max(maxv, hp(i + 1, au));

	return memo[i][ac] = activities[i][ac] + maxv;

}



int main(){

	cin >> N;

	for (x = 0; x < N; x++)

		for (y = 0; y < 3; y++)

			cin >> activities[x][y];

	y = 0;

	for (x = 0; x < 3; x++)

		y = max(y, hp(0 , x));

	cout << y;

}
// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;

typedef long long ll;

typedef double dd;

#define all(v) v.begin(),v.end()

#define endl "\n"

#define clr(n, r) memset(n,r,sizeof(n));



void fast() {

    cin.tie(0);

    cin.sync_with_stdio(0);

}

const int MAX=1e5+10;

int n,a[MAX],b[MAX],c[MAX],dp[MAX][3];

int ans(int i,int prev){

    if(i==n)return 0;

    int path1=0,path2=0,path3=0;

    int &ret=dp[i][prev];

    if(prev!=-1&&ret!=-1)return ret;

    if(prev!=0){

        path1=a[i]+ans(i+1,0);

    }

    if(prev!=1){

        path2=b[i]+ans(i+1,1);

    }

    if(prev!=2){

        path3=c[i]+ans(i+1,2);

    }

    return ret=max(path1,max(path2,path3));

}



int main() {

    fast();

    cin>>n;

    for (int i = 0; i < n; ++i) {

    cin>>a[i]>>b[i]>>c[i];

    }

    clr(dp,-1);

    cout<<ans(0,-1);

}


// Faster C++ solution for the same problem (measured speedup: 2.2198542333298446x):
#include <bits/stdc++.h>

using namespace std;

#define ll long long int

ll N, x, y;

ll memo[(ll)1e5][3];

ll activities[(ll)1e5][3];



ll hp(int i, int ac){

	if (i >= N) return 0;

	if (memo[i][ac]) return memo[i][ac];

	ll maxv = 0;

	for (int au = 0; au < 3; au++)

		if (ac != au)

			maxv = max(maxv, hp(i + 1, au));

	return memo[i][ac] = activities[i][ac] + maxv;

}



int main(){

	cin >> N;

	for (x = 0; x < N; x++)

		for (y = 0; y < 3; y++)

			cin >> activities[x][y];

	y = 0;

	for (x = 0; x < 3; x++)

		y = max(y, hp(0 , x));

	cout << y;

}
// Slower C++ solution:
/*

** author:mahmoud magdy

*/

#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define endl "\n"



inline int D()

{

    int m ;

    cin>>m;

    return m ;

}

inline ll lD()

{

    ll m ;

    cin>>m;

    return m ;

}

inline double dD()

{

    double m;

    cin>>m;

    return m;

}

void fast()

{

    ios_base :: sync_with_stdio( 0 ) ;

    cin.tie( 0 ) ;

    cout.tie( 0 ) ;

}

int n;

ll a[100005],b[100005],c[100005];

ll dp[100005][5];

ll solve(int i,int prv){

 if(i==n)return 0;

 if(dp[i][prv]!=-1)return dp[i][prv];

 ll x=0;

 if(prv!=1){

 x=a[i]+solve(i+1,1);

 }

 ll y=0;

 if(prv!=2)

 y=b[i]+solve(i+1,2);

 ll z=0;

 if(prv!=3)

 z=c[i]+solve(i+1,3);

 return dp[i][prv]=max({x,y,z});

}

int main()

{

    memset(dp,-1,sizeof dp);

    fast();

    n=D();

    for(int i=0;i<n;++i){

        a[i]=D();b[i]=D();c[i]=D();

    }

    cout<<solve(0,10)<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.106203327059987x):
#include <bits/stdc++.h>

using namespace std;

#define ll long long int

ll N, x, y;

ll memo[(ll)1e5][3];

ll activities[(ll)1e5][3];



ll hp(int i, int ac){

	if (i >= N) return 0;

	if (memo[i][ac]) return memo[i][ac];

	ll maxv = 0;

	for (int au = 0; au < 3; au++)

		if (ac != au)

			maxv = max(maxv, hp(i + 1, au));

	return memo[i][ac] = activities[i][ac] + maxv;

}



int main(){

	cin >> N;

	for (x = 0; x < N; x++)

		for (y = 0; y < 3; y++)

			cin >> activities[x][y];

	y = 0;

	for (x = 0; x < 3; x++)

		y = max(y, hp(0 , x));

	cout << y;

}
// Slower C++ solution:
#include <iostream>

#include <queue>

#include <string>

#include <map>

#include <stdio.h>

#include <iomanip>

#include <math.h>

#include <vector>

#include <stack>

#include <algorithm>

#include <string.h>

#include <set>

#include <cstdlib>

#include <iostream>

#include <bits/stdc++.h>

#define pii pair<int,int> 

#define ll long long int 

#define sl(n) scanf("%lld",&n)

#define s(n) scanf("%d",&n)

#define f(i,n) for(int i=0;i<n;++i)

#define fo(i,n) for(int i=1;i<=n;++i)

#define fnm(n,m) for(ll i=n;i<m+1;++i)

#define pb push_back

#define pf printf

#define F first

#define S second

#define srt(v) sort(v.begin(),v.end());

#define srtc(v,cmp) sort(v.begin(),v.end(),cmp);

#define srta(v,n) sort(v,v+n);

const ll LL_INF=0x3f3f3f3f3f3f;

const ll MOD=1073741824;





using namespace std;





int T;

int d,arr[100005][5];

int dp[100005][5];



int gen(int i,int prev){ // swim:0, bugs:1, homework:2

    if(i == d)return 0;

    int &ret = dp[i][prev];

    if(~ret)return ret;

    ret = gen(i+1,(prev+1)%3)+arr[i][(prev+1)%3];

    ret = max( ret , gen(i+1,(prev+2)%3)+arr[i][(prev+2)%3]);

    if(!i)ret = max( ret , gen(i+1,(prev)%3)+arr[i][(prev)%3]);

    return ret;

}



int main(){

    // pf("\t\t\t***************************************************************************\n\n");

    // pf("\t\t\t***************************************************************************\n\n");

    // freopen("input.txt","r",stdin);

    // freopen("output.txt","w",stdout);

    memset(dp,-1,sizeof dp);

    s(d);

    f(i,d){

        s(arr[i][0]);

        s(arr[i][1]);

        s(arr[i][2]);

    }

    pf("%d\n",gen(0,0));



    return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 2.9909986485018893x):
#include <bits/stdc++.h>

using namespace std;

#define ll long long int

ll N, x, y;

ll memo[(ll)1e5][3];

ll activities[(ll)1e5][3];



ll hp(int i, int ac){

	if (i >= N) return 0;

	if (memo[i][ac]) return memo[i][ac];

	ll maxv = 0;

	for (int au = 0; au < 3; au++)

		if (ac != au)

			maxv = max(maxv, hp(i + 1, au));

	return memo[i][ac] = activities[i][ac] + maxv;

}



int main(){

	cin >> N;

	for (x = 0; x < N; x++)

		for (y = 0; y < 3; y++)

			cin >> activities[x][y];

	y = 0;

	for (x = 0; x < 3; x++)

		y = max(y, hp(0 , x));

	cout << y;

}
// Slower C++ solution:


#include <bits/stdc++.h>

using namespace  std;



int N;

const int SIZE = 100005;

int mem [SIZE][4];

int arr [SIZE][3];





int dp (int i, int activ) {

  

  if (i == N) return 0;

  



 if (mem[i][activ] != -1) return mem[i][activ];  

  

  int ret = -1; 



  if (activ != 0) {

    ret = max (ret, arr[i][0] + dp(i+1, 0));

  }



  if (activ != 1) {

    ret = max (ret, arr[i][1] + dp(i+1, 1));

    

  }



  if (activ != 2) {

    ret = max (ret, arr[i][2] + dp(i+1, 2));

    

  }



  return  mem[i][activ] = ret;

}



int main (){ 



  cin  >> N ;

  for (int i =0; i< N; i++) {

    cin >> arr[i][0] >> arr [i][1] >> arr [i][2];

  }



  memset(mem, -1, sizeof mem);

  





  cout << dp(0,3);



}

// Faster C++ solution for the same problem (measured speedup: 2.6046800424858945x):
#include <bits/stdc++.h>

using namespace std;

#define ll long long int

ll N, x, y;

ll memo[(ll)1e5][3];

ll activities[(ll)1e5][3];



ll hp(int i, int ac){

	if (i >= N) return 0;

	if (memo[i][ac]) return memo[i][ac];

	ll maxv = 0;

	for (int au = 0; au < 3; au++)

		if (ac != au)

			maxv = max(maxv, hp(i + 1, au));

	return memo[i][ac] = activities[i][ac] + maxv;

}



int main(){

	cin >> N;

	for (x = 0; x < N; x++)

		for (y = 0; y < 3; y++)

			cin >> activities[x][y];

	y = 0;

	for (x = 0; x < 3; x++)

		y = max(y, hp(0 , x));

	cout << y;

}