stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:


#include<bits/stdc++.h>

using namespace std;



#define MAX 100000



int dp[MAX+5][4];

int val[MAX+5][4];

int n;



void initialize()

{

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

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

            dp[i][j] = -1;

}



int vacation(int i, int j)

{

    if(i<=0) return 0;



  /*  if(i==1)

    {

        return val[i][j];

    }

*/

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



    int mx = INT_MIN;



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

    {

        if(j==k) continue;



        mx = max(mx, vacation(i-1, k)+ val[i][j]);

    }





    dp[i][j] = mx;



    return mx;

}



int main()

{

    initialize();



    cin>>n;

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

    {

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

            cin>>val[i][j];

    }



    int retVal1 = vacation(n, 1);



    initialize();



    int retVal2 = vacation(n, 2);



    initialize();



    int retVal3 = vacation(n, 3);



    cout<<max({retVal1, retVal2, retVal3})<<endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 18.946083874436106x):
#include<cstdio>

int n;

int hap[100002][3]={};

int max(int a,int b)

{

	return a>b? a:b;

}

int main()

{

	scanf("%d",&n);

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

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

		{

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

			hap[i][j]+=max(hap[i-1][(j+1)%3],hap[i-1][(j+2)%3]);

		}

	printf("%d",max(hap[n][0],max(hap[n][1],hap[n][2])));

	return 0;

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

using namespace std;



#define fr(i,n) _back

#define pb	push_back

#define eb	emplace_back

#define mk	make_pair

#define fi	first

#define se	second

#define endl	'\n'



typedef long long ll;

typedef pair<int,int> ii;

typedef vector<ii> vii;

const int INF = 0x3f3f3f3f;

const double PI = acos(-1.0);

int dp[2][3];



int main() {

    ios_base::sync_with_stdio(false);

    int n; cin >> n;

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

        for(int j = 0; j  < 3; j++) cin >> dp[1][j];

        dp[1][0] += max(dp[0][1],dp[0][2]);

        dp[1][1] += max(dp[0][0],dp[0][2]);

        dp[1][2] += max(dp[0][1],dp[0][0]);

        for(int j = 0; j < 3; j++) dp[0][j] = dp[1][j];

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.2779708257482145x):
#include<cstdio>

int n;

int hap[100002][3]={};

int max(int a,int b)

{

	return a>b? a:b;

}

int main()

{

	scanf("%d",&n);

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

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

		{

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

			hap[i][j]+=max(hap[i-1][(j+1)%3],hap[i-1][(j+2)%3]);

		}

	printf("%d",max(hap[n][0],max(hap[n][1],hap[n][2])));

	return 0;

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

using namespace std;

int arr[100010][5];

int dp[100010][5];



int vac(int a, int b){

	if(a == 0) return 0;

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

	

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

		if(i != b){

			dp[a][b] = max(dp[a][b], vac(a - 1,i) + arr[a][i]);

		}

	}

	

	return dp[a][b];

}



int main (){

	memset(dp,-1,sizeof dp);

	int n;

	cin >> n;

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

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

	}

	

	

	int ans = 0;

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

		ans = max(ans , vac(n - 1, i) + arr[n][i]);

	}

	

	cout << ans << endl;

	

}

// Faster C++ solution for the same problem (measured speedup: 15.686246474577073x):
#include<cstdio>

int n;

int hap[100002][3]={};

int max(int a,int b)

{

	return a>b? a:b;

}

int main()

{

	scanf("%d",&n);

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

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

		{

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

			hap[i][j]+=max(hap[i-1][(j+1)%3],hap[i-1][(j+2)%3]);

		}

	printf("%d",max(hap[n][0],max(hap[n][1],hap[n][2])));

	return 0;

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

using namespace std;

int n,a[100010][5],dp[100010][5];



int main()

{

	scanf("%d",&n);

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

	{

		scanf("%d %d %d",&a[i][0],&a[i][1],&a[i][2]);

	}

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

	{

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

	}

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

	{

	//	printf("%d. ",i);

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

		{

			dp[i][j] = -1;

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

			{

			//	printf("%d %d : %d %d\n",i,j,dp[i-1][j]+a[i][k],a[i][k]);

				if(j!=k)

				{

					dp[i][j] = max(dp[i][j],dp[i-1][k]+a[i][j]);

				}

			}

			//printf("%d ",dp[i][j]);

		}

		//printf("\n");

	}

	printf("%d\n",max(dp[n][0],max(dp[n][1],dp[n][2])));

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.149832204828738x):
#include<cstdio>

int n;

int hap[100002][3]={};

int max(int a,int b)

{

	return a>b? a:b;

}

int main()

{

	scanf("%d",&n);

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

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

		{

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

			hap[i][j]+=max(hap[i-1][(j+1)%3],hap[i-1][(j+2)%3]);

		}

	printf("%d",max(hap[n][0],max(hap[n][1],hap[n][2])));

	return 0;

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

#include<iostream>

#include<algorithm>

#include<queue>

#define inf 0x3f3f3f3f

using namespace std;

typedef long long LL;

const int maxn = 1e5+5;



int a[maxn],b[maxn],c[maxn],dp[maxn][3];



int main(){

	int n;

	scanf("%d",&n);

	int m;

	for(int i = 1; i <= n; i++) scanf("%d %d %d", &a[i],&b[i],&c[i]);

	dp[1][0] = a[1]; dp[1][1] = b[1]; dp[1][2] = c[1];

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

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

			if(j == 0) dp[i][j] = max(dp[i-1][1] + a[i], dp[i-1][2] + a[i]);

			if(j == 1) dp[i][j] = max(dp[i-1][0] + b[i], dp[i-1][2] + b[i]);

			if(j == 2) dp[i][j] = max(dp[i-1][0] + c[i], dp[i-1][1] + c[i]);

		}

	}

	m = max(dp[n][0], dp[n][1]);

	m = max(m, dp[n][2]);

	printf("%d",m);

}

// Faster C++ solution for the same problem (measured speedup: 5.146617993214956x):
#include<cstdio>

int n;

int hap[100002][3]={};

int max(int a,int b)

{

	return a>b? a:b;

}

int main()

{

	scanf("%d",&n);

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

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

		{

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

			hap[i][j]+=max(hap[i-1][(j+1)%3],hap[i-1][(j+2)%3]);

		}

	printf("%d",max(hap[n][0],max(hap[n][1],hap[n][2])));

	return 0;

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



using namespace std;



int main() {

    vector<vector<int>> anpan;

    int n, k = -1;

    cin >> n;

    vector<int> zerad = {0};

        anpan.push_back(zerad);

        anpan.push_back(zerad);

        anpan.push_back(zerad);



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

        int a, b, c;

        cin >> a >> b >> c;

        anpan[0].push_back(max(anpan[1][i], anpan[2][i]) + a);

        anpan[1].push_back(max(anpan[0][i], anpan[2][i]) + b);

        anpan[2].push_back(max(anpan[0][i], anpan[1][i]) + c);

    }

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

}

// Faster C++ solution for the same problem (measured speedup: 5.288995099805646x):
#include<cstdio>

int n;

int hap[100002][3]={};

int max(int a,int b)

{

	return a>b? a:b;

}

int main()

{

	scanf("%d",&n);

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

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

		{

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

			hap[i][j]+=max(hap[i-1][(j+1)%3],hap[i-1][(j+2)%3]);

		}

	printf("%d",max(hap[n][0],max(hap[n][1],hap[n][2])));

	return 0;

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

using namespace std;



int felicidade(int n){

    int a[n+1],b[n+1],c[n+1];

    

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

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

    }



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

        a[i]=max(b[i-1],c[i-1])+a[i];

        b[i]=max(a[i-1],c[i-1])+b[i];

        c[i]=max(a[i-1],b[i-1])+c[i];

    }



    if(a[n]>=b[n] and a[n]>=c[n]){

        return a[n];

    } else if(b[n]>=a[n] and b[n]>=c[n]){

        return b[n];

    }else if(c[n]>=a[n] and c[n]>=b[n]){

        return c[n];

    }

    

    return 0;

}



int main() {

    int n;



    cin >> n;



    cout << felicidade(n) << endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.225034226221752x):
#include<cstdio>

int n;

int hap[100002][3]={};

int max(int a,int b)

{

	return a>b? a:b;

}

int main()

{

	scanf("%d",&n);

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

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

		{

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

			hap[i][j]+=max(hap[i-1][(j+1)%3],hap[i-1][(j+2)%3]);

		}

	printf("%d",max(hap[n][0],max(hap[n][1],hap[n][2])));

	return 0;

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

#include <algorithm>

using namespace std;



int main(){

	int n;

	int result;

	int x = 0, y = 0, z = 0;

	

	scanf("%d", &n);

	

	int m[n+1][3];

	

	m[0][0] = m[0][1] = m[0][2] = 0;

	

	for(int i = 0; i < n+1; i++){

		if(i > 0){

			scanf("%d", &x);

			scanf("%d", &y);

			scanf("%d", &z);

			

			//Caso não possa escolher a atividade a

			m[i][0] = max(m[i - 1][1], m[i - 1][2]) + x;

			//Caso não possa escolher a atividade b

			m[i][1] = max(m[i - 1][0], m[i - 1][2]) + y;

			//Caso não possa escolher a atividade c

			m[i][2] = max(m[i - 1][1], m[i - 1][0]) + z;

		}

	}

	

	result = max(max(m[n][0], m[n][1]), m[n][2]);

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

}

// Faster C++ solution for the same problem (measured speedup: 5.150688506816224x):
#include<cstdio>

int n;

int hap[100002][3]={};

int max(int a,int b)

{

	return a>b? a:b;

}

int main()

{

	scanf("%d",&n);

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

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

		{

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

			hap[i][j]+=max(hap[i-1][(j+1)%3],hap[i-1][(j+2)%3]);

		}

	printf("%d",max(hap[n][0],max(hap[n][1],hap[n][2])));

	return 0;

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

#define ll long long

#define ull unsigned ll

#define ld long double

#define pll pair<ll, ll>

#define pii pair<int, int>

#define f first

#define se second

#define pb push_back

#define ft fflush(stdout)

 

 

using namespace std;

 

 

const int N = 2e5 + 123;

const int PLS = 3e5 + 123;

const int mod = 1e9 + 7;

const ll inf = 1e16;





ll n, a[N], c[N], b[N], dp[N][3];





int main() {

    ios_base::sync_with_stdio(NULL);

    cin.tie(NULL); cout.tie(NULL);

    cin >> n;

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

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

    dp[1][0] = a[1];

    dp[1][1] = b[1];

    dp[1][2] = c[1];

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

        dp[i][0] = a[i] + max(dp[i - 1][1], dp[i - 1][2]);

        dp[i][1] = b[i] + max(dp[i - 1][0], dp[i - 1][2]);

        dp[i][2] = c[i] + max(dp[i - 1][1], dp[i - 1][0]);

    }

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

}

// Faster C++ solution for the same problem (measured speedup: 5.276114426768966x):
#include<cstdio>

int n;

int hap[100002][3]={};

int max(int a,int b)

{

	return a>b? a:b;

}

int main()

{

	scanf("%d",&n);

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

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

		{

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

			hap[i][j]+=max(hap[i-1][(j+1)%3],hap[i-1][(j+2)%3]);

		}

	printf("%d",max(hap[n][0],max(hap[n][1],hap[n][2])));

	return 0;

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

using namespace std;



int ar[100010][5],n;

int dp[100010][5];



int call(int i,int f)

{



    if(i==n)

        return 0;



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

        return dp[i][f];



    int ans=0;



    if(f==0)

        ans+=max(ar[i][1]+call(i+1,1),ar[i][2]+call(i+1,2));



    else if(f==1)

        ans+=max(ar[i][0]+call(i+1,0),ar[i][2]+call(i+1,2));

    else

        ans+=max(ar[i][1]+call(i+1,1),ar[i][0]+call(i+1,0));



        dp[i][f]=ans;

        return ans;



}



int main()

{



    cin>>n;



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

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



    memset(dp,-1,sizeof dp);



    int ans=max(max(call(0,0),call(0,1)),call(0,2));



    cout<<ans<<endl;



}


// Faster C++ solution for the same problem (measured speedup: 3.0421724397158374x):
#include<cstdio>

#include<iostream>

using namespace std;

const int MAXN = 100010;

int n, a[MAXN][3], f[MAXN][3];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[i][0],&a[i][1],&a[i][2]);

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

	{

		f[i][0] = max(f[i-1][1], f[i-1][2])+a[i][0];

		f[i][1] = max(f[i-1][0], f[i-1][2])+a[i][1];

		f[i][2] = max(f[i-1][1], f[i-1][0])+a[i][2];

	}

	int ans = max(f[n][0], max(f[n][1], f[n][2]));

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

	return 0;

}
// Slower C++ solution:


#include<bits/stdc++.h>

using namespace std;



#define MAX 100000



int dp[MAX+5][4];

int val[MAX+5][4];

int n;



void initialize()

{

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

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

            dp[i][j] = -1;

}



int vacation(int i, int j)

{

    if(i<=0) return 0;



  /*  if(i==1)

    {

        return val[i][j];

    }

*/

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



    int mx = INT_MIN;



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

    {

        if(j==k) continue;



        mx = max(mx, vacation(i-1, k)+ val[i][j]);

    }





    dp[i][j] = mx;



    return mx;

}



int main()

{

    initialize();



    cin>>n;

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

    {

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

            cin>>val[i][j];

    }



    int retVal1 = vacation(n, 1);



    initialize();



    int retVal2 = vacation(n, 2);



    initialize();



    int retVal3 = vacation(n, 3);



    cout<<max({retVal1, retVal2, retVal3})<<endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 3.67783865595594x):
#include<cstdio>

#include<iostream>

using namespace std;

const int MAXN = 100010;

int n, a[MAXN][3], f[MAXN][3];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[i][0],&a[i][1],&a[i][2]);

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

	{

		f[i][0] = max(f[i-1][1], f[i-1][2])+a[i][0];

		f[i][1] = max(f[i-1][0], f[i-1][2])+a[i][1];

		f[i][2] = max(f[i-1][1], f[i-1][0])+a[i][2];

	}

	int ans = max(f[n][0], max(f[n][1], f[n][2]));

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

	return 0;

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

using namespace std;

int arr[100010][5];

int dp[100010][5];



int vac(int a, int b){

	if(a == 0) return 0;

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

	

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

		if(i != b){

			dp[a][b] = max(dp[a][b], vac(a - 1,i) + arr[a][i]);

		}

	}

	

	return dp[a][b];

}



int main (){

	memset(dp,-1,sizeof dp);

	int n;

	cin >> n;

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

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

	}

	

	

	int ans = 0;

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

		ans = max(ans , vac(n - 1, i) + arr[n][i]);

	}

	

	cout << ans << endl;

	

}

// Faster C++ solution for the same problem (measured speedup: 3.045034743506815x):
#include<cstdio>

#include<iostream>

using namespace std;

const int MAXN = 100010;

int n, a[MAXN][3], f[MAXN][3];

int main()

{

	scanf("%d",&n);

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

		scanf("%d%d%d",&a[i][0],&a[i][1],&a[i][2]);

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

	{

		f[i][0] = max(f[i-1][1], f[i-1][2])+a[i][0];

		f[i][1] = max(f[i-1][0], f[i-1][2])+a[i][1];

		f[i][2] = max(f[i-1][1], f[i-1][0])+a[i][2];

	}

	int ans = max(f[n][0], max(f[n][1], f[n][2]));

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

	return 0;

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

using namespace std;



int ar[100010][5],n;

int dp[100010][5];



int call(int i,int f)

{



    if(i==n)

        return 0;



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

        return dp[i][f];



    int ans=0;



    if(f==0)

        ans+=max(ar[i][1]+call(i+1,1),ar[i][2]+call(i+1,2));



    else if(f==1)

        ans+=max(ar[i][0]+call(i+1,0),ar[i][2]+call(i+1,2));

    else

        ans+=max(ar[i][1]+call(i+1,1),ar[i][0]+call(i+1,0));



        dp[i][f]=ans;

        return ans;



}



int main()

{



    cin>>n;



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

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



    memset(dp,-1,sizeof dp);



    int ans=max(max(call(0,0),call(0,1)),call(0,2));



    cout<<ans<<endl;



}


// Faster C++ solution for the same problem (measured speedup: 1.3675722875957919x):
#include <bits/stdc++.h> 

using namespace std;

#define INF INT_MAX

#define N 100005

#define K 3

int n, k;

int arr[N][K];

int dp[N][K];

int solve(int idx, int last_k){

    int res;

    if(idx == n)

        return 0;

    if(~last_k && ~dp[idx][last_k])

        return dp[idx][last_k];

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

        if(last_k != i){

            dp[idx][last_k] = max(dp[idx][last_k], arr[idx][i]+solve(idx+1, i));

        }

    }

    return dp[idx][last_k];

}

int main()

{

    cin >> n;

    memset(dp, -1, sizeof(dp[0][0]) * N * K);

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

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

            cin >> arr[i][j];

    cout << solve(0, -1);

    return 0;

}
// Slower C++ solution:


#include<bits/stdc++.h>

using namespace std;



#define MAX 100000



int dp[MAX+5][4];

int val[MAX+5][4];

int n;



void initialize()

{

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

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

            dp[i][j] = -1;

}



int vacation(int i, int j)

{

    if(i<=0) return 0;



  /*  if(i==1)

    {

        return val[i][j];

    }

*/

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



    int mx = INT_MIN;



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

    {

        if(j==k) continue;



        mx = max(mx, vacation(i-1, k)+ val[i][j]);

    }





    dp[i][j] = mx;



    return mx;

}



int main()

{

    initialize();



    cin>>n;

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

    {

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

            cin>>val[i][j];

    }



    int retVal1 = vacation(n, 1);



    initialize();



    int retVal2 = vacation(n, 2);



    initialize();



    int retVal3 = vacation(n, 3);



    cout<<max({retVal1, retVal2, retVal3})<<endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.65332844334219x):
#include <bits/stdc++.h> 

using namespace std;

#define INF INT_MAX

#define N 100005

#define K 3

int n, k;

int arr[N][K];

int dp[N][K];

int solve(int idx, int last_k){

    int res;

    if(idx == n)

        return 0;

    if(~last_k && ~dp[idx][last_k])

        return dp[idx][last_k];

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

        if(last_k != i){

            dp[idx][last_k] = max(dp[idx][last_k], arr[idx][i]+solve(idx+1, i));

        }

    }

    return dp[idx][last_k];

}

int main()

{

    cin >> n;

    memset(dp, -1, sizeof(dp[0][0]) * N * K);

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

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

            cin >> arr[i][j];

    cout << solve(0, -1);

    return 0;

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

using namespace std;

int arr[100010][5];

int dp[100010][5];



int vac(int a, int b){

	if(a == 0) return 0;

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

	

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

		if(i != b){

			dp[a][b] = max(dp[a][b], vac(a - 1,i) + arr[a][i]);

		}

	}

	

	return dp[a][b];

}



int main (){

	memset(dp,-1,sizeof dp);

	int n;

	cin >> n;

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

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

	}

	

	

	int ans = 0;

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

		ans = max(ans , vac(n - 1, i) + arr[n][i]);

	}

	

	cout << ans << endl;

	

}

// Faster C++ solution for the same problem (measured speedup: 1.3688590020805194x):
#include <bits/stdc++.h> 

using namespace std;

#define INF INT_MAX

#define N 100005

#define K 3

int n, k;

int arr[N][K];

int dp[N][K];

int solve(int idx, int last_k){

    int res;

    if(idx == n)

        return 0;

    if(~last_k && ~dp[idx][last_k])

        return dp[idx][last_k];

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

        if(last_k != i){

            dp[idx][last_k] = max(dp[idx][last_k], arr[idx][i]+solve(idx+1, i));

        }

    }

    return dp[idx][last_k];

}

int main()

{

    cin >> n;

    memset(dp, -1, sizeof(dp[0][0]) * N * K);

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

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

            cin >> arr[i][j];

    cout << solve(0, -1);

    return 0;

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

using namespace std;



int ar[100010][5],n;

int dp[100010][5];



int call(int i,int f)

{



    if(i==n)

        return 0;



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

        return dp[i][f];



    int ans=0;



    if(f==0)

        ans+=max(ar[i][1]+call(i+1,1),ar[i][2]+call(i+1,2));



    else if(f==1)

        ans+=max(ar[i][0]+call(i+1,0),ar[i][2]+call(i+1,2));

    else

        ans+=max(ar[i][1]+call(i+1,1),ar[i][0]+call(i+1,0));



        dp[i][f]=ans;

        return ans;



}



int main()

{



    cin>>n;



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

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



    memset(dp,-1,sizeof dp);



    int ans=max(max(call(0,0),call(0,1)),call(0,2));



    cout<<ans<<endl;



}


// Faster C++ solution for the same problem (measured speedup: 3.0059175618806626x):
#include <bits/stdc++.h>

using namespace std;

int f[100010][4];

const int e=110000;

int a[e][4];

int main(){

	int n;

	cin>>n;

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

	{

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

	}

	f[1][1]=a[1][1];

	f[1][2]=a[1][2];

	f[1][3]=a[1][3];

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

	{

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

		{

			int t1,t2;

			if(j==1){t1=2;t2=3;}

			else if(j==2){t1=1;t2=3;}

			else{t1=1;t2=2;}

			f[i][j]=max(f[i-1][t1],f[i-1][t2])+a[i][j];

		}

	}

	int ans=-1;

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

		ans=max(ans,f[n][i]);

	cout<<ans;

	return 0;

}
// Slower C++ solution:


#include<bits/stdc++.h>

using namespace std;



#define MAX 100000



int dp[MAX+5][4];

int val[MAX+5][4];

int n;



void initialize()

{

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

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

            dp[i][j] = -1;

}



int vacation(int i, int j)

{

    if(i<=0) return 0;



  /*  if(i==1)

    {

        return val[i][j];

    }

*/

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



    int mx = INT_MIN;



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

    {

        if(j==k) continue;



        mx = max(mx, vacation(i-1, k)+ val[i][j]);

    }





    dp[i][j] = mx;



    return mx;

}



int main()

{

    initialize();



    cin>>n;

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

    {

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

            cin>>val[i][j];

    }



    int retVal1 = vacation(n, 1);



    initialize();



    int retVal2 = vacation(n, 2);



    initialize();



    int retVal3 = vacation(n, 3);



    cout<<max({retVal1, retVal2, retVal3})<<endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 3.634008270331376x):
#include <bits/stdc++.h>

using namespace std;

int f[100010][4];

const int e=110000;

int a[e][4];

int main(){

	int n;

	cin>>n;

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

	{

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

	}

	f[1][1]=a[1][1];

	f[1][2]=a[1][2];

	f[1][3]=a[1][3];

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

	{

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

		{

			int t1,t2;

			if(j==1){t1=2;t2=3;}

			else if(j==2){t1=1;t2=3;}

			else{t1=1;t2=2;}

			f[i][j]=max(f[i-1][t1],f[i-1][t2])+a[i][j];

		}

	}

	int ans=-1;

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

		ans=max(ans,f[n][i]);

	cout<<ans;

	return 0;

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

using namespace std;

int arr[100010][5];

int dp[100010][5];



int vac(int a, int b){

	if(a == 0) return 0;

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

	

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

		if(i != b){

			dp[a][b] = max(dp[a][b], vac(a - 1,i) + arr[a][i]);

		}

	}

	

	return dp[a][b];

}



int main (){

	memset(dp,-1,sizeof dp);

	int n;

	cin >> n;

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

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

	}

	

	

	int ans = 0;

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

		ans = max(ans , vac(n - 1, i) + arr[n][i]);

	}

	

	cout << ans << endl;

	

}

// Faster C++ solution for the same problem (measured speedup: 3.008745754365879x):
#include <bits/stdc++.h>

using namespace std;

int f[100010][4];

const int e=110000;

int a[e][4];

int main(){

	int n;

	cin>>n;

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

	{

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

	}

	f[1][1]=a[1][1];

	f[1][2]=a[1][2];

	f[1][3]=a[1][3];

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

	{

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

		{

			int t1,t2;

			if(j==1){t1=2;t2=3;}

			else if(j==2){t1=1;t2=3;}

			else{t1=1;t2=2;}

			f[i][j]=max(f[i-1][t1],f[i-1][t2])+a[i][j];

		}

	}

	int ans=-1;

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

		ans=max(ans,f[n][i]);

	cout<<ans;

	return 0;

}