stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 2.9768879457945867x):
    #include <algorithm>

    #include <fstream>

    #include <iomanip>

    #include <iostream>

    #include <map>

    #include <math.h>

    #include <set>

    #include <stdio.h>

    #include <string>

    #include <utility>

    #include <vector>

     

    using namespace std;

     

    int main() {

      int n;

      cin >> n;

      vector<long long> act[3];

      act[0].resize(n);

      act[1].resize(n);

      act[2].resize(n);

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

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

     

      long long dp[n + 1][3];

      dp[0][0] = dp[0][1] = dp[0][2] = 0;

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

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

          dp[i][j] = 0;

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

            if (j == k)

              continue;

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

          }

        }

      }

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

      return 0;

    }
// 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: 2.1606809283638766x):
    #include <algorithm>

    #include <fstream>

    #include <iomanip>

    #include <iostream>

    #include <map>

    #include <math.h>

    #include <set>

    #include <stdio.h>

    #include <string>

    #include <utility>

    #include <vector>

     

    using namespace std;

     

    int main() {

      int n;

      cin >> n;

      vector<long long> act[3];

      act[0].resize(n);

      act[1].resize(n);

      act[2].resize(n);

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

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

     

      long long dp[n + 1][3];

      dp[0][0] = dp[0][1] = dp[0][2] = 0;

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

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

          dp[i][j] = 0;

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

            if (j == k)

              continue;

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

          }

        }

      }

      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;

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: 2.1790559305682473x):
    #include <algorithm>

    #include <fstream>

    #include <iomanip>

    #include <iostream>

    #include <map>

    #include <math.h>

    #include <set>

    #include <stdio.h>

    #include <string>

    #include <utility>

    #include <vector>

     

    using namespace std;

     

    int main() {

      int n;

      cin >> n;

      vector<long long> act[3];

      act[0].resize(n);

      act[1].resize(n);

      act[2].resize(n);

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

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

     

      long long dp[n + 1][3];

      dp[0][0] = dp[0][1] = dp[0][2] = 0;

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

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

          dp[i][j] = 0;

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

            if (j == k)

              continue;

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

          }

        }

      }

      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;



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: 2.179081899284836x):
    #include <algorithm>

    #include <fstream>

    #include <iomanip>

    #include <iostream>

    #include <map>

    #include <math.h>

    #include <set>

    #include <stdio.h>

    #include <string>

    #include <utility>

    #include <vector>

     

    using namespace std;

     

    int main() {

      int n;

      cin >> n;

      vector<long long> act[3];

      act[0].resize(n);

      act[1].resize(n);

      act[2].resize(n);

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

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

     

      long long dp[n + 1][3];

      dp[0][0] = dp[0][1] = dp[0][2] = 0;

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

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

          dp[i][j] = 0;

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

            if (j == k)

              continue;

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

          }

        }

      }

      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;

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: 2.179079710925153x):
    #include <algorithm>

    #include <fstream>

    #include <iomanip>

    #include <iostream>

    #include <map>

    #include <math.h>

    #include <set>

    #include <stdio.h>

    #include <string>

    #include <utility>

    #include <vector>

     

    using namespace std;

     

    int main() {

      int n;

      cin >> n;

      vector<long long> act[3];

      act[0].resize(n);

      act[1].resize(n);

      act[2].resize(n);

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

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

     

      long long dp[n + 1][3];

      dp[0][0] = dp[0][1] = dp[0][2] = 0;

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

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

          dp[i][j] = 0;

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

            if (j == k)

              continue;

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

          }

        }

      }

      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;

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: 2.1790559305682473x):
    #include <algorithm>

    #include <fstream>

    #include <iomanip>

    #include <iostream>

    #include <map>

    #include <math.h>

    #include <set>

    #include <stdio.h>

    #include <string>

    #include <utility>

    #include <vector>

     

    using namespace std;

     

    int main() {

      int n;

      cin >> n;

      vector<long long> act[3];

      act[0].resize(n);

      act[1].resize(n);

      act[2].resize(n);

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

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

     

      long long dp[n + 1][3];

      dp[0][0] = dp[0][1] = dp[0][2] = 0;

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

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

          dp[i][j] = 0;

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

            if (j == k)

              continue;

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

          }

        }

      }

      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: 2.315289849862719x):
#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;

}
// 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: 1.3660610813893945x):
#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;

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

using namespace std;



const int MaxN = 100005;



int n;

int f[MaxN][3], val[MaxN][3];



void upd(int &a, int b) {

	if (b > a) a = b;

}



int main() {

	scanf("%d", &n);

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

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

	}

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

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

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

				if (j == k) continue;

				upd(f[i][j], f[i - 1][k] + val[i][j]);

			}

		}

	}

	int Ans = 0;

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

			upd(Ans, f[n][j]);

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.18174310641152x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

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



    return 0;

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

using namespace std;

 

#define mp make_pair

#define pb push_back

#define st first

#define nd second

#define lb lower_bound

#define ub upper_bound

 

typedef long long ll;

typedef long double ld;

typedef pair<int,int> pi;

typedef pair<ll,ll> pll;

typedef vector<int> vi;

typedef vector<ll> vll;

typedef vector<pi> vpi;

typedef vector<pll> vpll;

 

const ll oo=1e18;

const int MOD=1000000007;

const int INF=1000000000;

const ld EPS=0.000000001;

 

int dp[100005][3],n,a,b,c;

 

signed main() {

    ios_base::sync_with_stdio(NULL);

    cin.tie(NULL);

    cout.tie(NULL);

    cin>>n;

    cin>>a>>b>>c;

    dp[1][0]=a;

    dp[1][1]=b;

    dp[1][2]=c;

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

        cin>>a>>b>>c;

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

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

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

    }

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.2709007028187225x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

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



    return 0;

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

using namespace std;

int x[100000][3];

int a[100001][3], mx=0;

int calc(int pos, int u)

{

	if(pos==-1) return 0;

	if(a[pos][u]==-1)

	{

		//if(pos==0) a[pos][u]=mx(x[pos][(u+1)%3],x[pos][(u+2)%3]);

		a[pos][u]=max(calc(pos-1, (u+1)%3)+x[pos][(u+1)%3],calc(pos-1, (u+2)%3)+x[pos][(u+2)%3]);

	}

	return a[pos][u];

}



int main() {

	int n;

	cin>>n;

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

	{

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

		{

			cin>>x[i][u];

		}

	}

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

	{

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

		{

			a[i][u]=-1;

		}

	}

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

	{

		if(mx<calc(n, i) ) mx=calc(n-1, i);

	}

	cout<<mx;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.217186995758108x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

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



    return 0;

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

#define LL long long

using namespace std;

const int N =1e6;



LL int dp[N][4], a[N], b[N], c[N];



int main(){

    int n;

    cin >> n;

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

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

    }

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

    dp[0][1] = b[0];

    dp[0][2] = c[0];

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

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

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

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

        //cout << dp[i][0] << " " << dp[i][1] << " " << dp[i][2] << endl;

    }

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

    cout << ans << endl;

}


// Faster C++ solution for the same problem (measured speedup: 5.2267379187831615x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

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



    return 0;

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

using namespace std;



int main(){

    int n, i, j, k;

    scanf("%d",&n);

    int a[n+5], b[n+5], c[n+5];

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

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

    }

    int dp[n+5][3];

    memset(dp, 0, sizeof(dp));

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

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

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

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

    }

    int ans  = 0;

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

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

    cout << ans << "\n";



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.208645638800865x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

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



    return 0;

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

using namespace std;



int main(){



    int n, flag;

    long long ans = 0;



    cin >> n;

    int a[n + 5], b[n + 5], c[n + 5], dp[n + 5][3];

    for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];

    memset(dp, 0, sizeof(dp));

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

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

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

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

    }

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



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.219497345002614x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

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



    return 0;

}