stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 2.122403140449843x):
#include <bits/stdc++.h>



using namespace std;



#define ll long long

#define PI acos(-1.0)



int main()

{

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

	

	scanf("%lld", &t);

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

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

	{

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

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

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

	}

	printf("%lld\n",max(a[t],max(b[t],c[t])));

	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: 2.1224010090102374x):
#include <bits/stdc++.h>



using namespace std;



#define ll long long

#define PI acos(-1.0)



int main()

{

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

	

	scanf("%lld", &t);

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

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

	{

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

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

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

	}

	printf("%lld\n",max(a[t],max(b[t],c[t])));

	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: 2.122377847189567x):
#include <bits/stdc++.h>



using namespace std;



#define ll long long

#define PI acos(-1.0)



int main()

{

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

	

	scanf("%lld", &t);

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

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

	{

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

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

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

	}

	printf("%lld\n",max(a[t],max(b[t],c[t])));

	return 0;

}
// 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: 2.1224950949482952x):
#include <bits/stdc++.h>



using namespace std;



#define ll long long

#define PI acos(-1.0)



int main()

{

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

	

	scanf("%lld", &t);

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

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

	{

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

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

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

	}

	printf("%lld\n",max(a[t],max(b[t],c[t])));

	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: 5.050740681961272x):
#include <bits/stdc++.h>

using namespace std;



int mat[100002][3], dp[100002][3];





int main() {

  int n;

  cin >> n;

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

    for (int j = 0; j < 3; ++j) cin >> mat[i][j];



  for (int i = 0; i < 3; ++i) dp[0][i] = mat[0][i];



  vector<vector<int>> now {{1, 2}, {0, 2}, {0, 1}};



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

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

      for (auto el: now[col]) {

        dp[pos][col] = max(dp[pos][col], mat[pos][col] + dp[pos - 1][el]);

      }

    }

  }



  int res = 0;

  for (int i = 0; i < 3; ++i) res = max(res, dp[n - 1][i]);

  cout << res << endl;  

  return 0;

}
// 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: 2.9800244138877576x):
#include <bits/stdc++.h>

using namespace std;



int mat[100002][3], dp[100002][3];





int main() {

  int n;

  cin >> n;

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

    for (int j = 0; j < 3; ++j) cin >> mat[i][j];



  for (int i = 0; i < 3; ++i) dp[0][i] = mat[0][i];



  vector<vector<int>> now {{1, 2}, {0, 2}, {0, 1}};



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

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

      for (auto el: now[col]) {

        dp[pos][col] = max(dp[pos][col], mat[pos][col] + dp[pos - 1][el]);

      }

    }

  }



  int res = 0;

  for (int i = 0; i < 3; ++i) res = max(res, dp[n - 1][i]);

  cout << res << 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: 2.1629574355467924x):
#include <bits/stdc++.h>

using namespace std;



int mat[100002][3], dp[100002][3];





int main() {

  int n;

  cin >> n;

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

    for (int j = 0; j < 3; ++j) cin >> mat[i][j];



  for (int i = 0; i < 3; ++i) dp[0][i] = mat[0][i];



  vector<vector<int>> now {{1, 2}, {0, 2}, {0, 1}};



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

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

      for (auto el: now[col]) {

        dp[pos][col] = max(dp[pos][col], mat[pos][col] + dp[pos - 1][el]);

      }

    }

  }



  int res = 0;

  for (int i = 0; i < 3; ++i) res = max(res, dp[n - 1][i]);

  cout << res << 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: 2.1813517977704766x):
#include <bits/stdc++.h>

using namespace std;



int mat[100002][3], dp[100002][3];





int main() {

  int n;

  cin >> n;

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

    for (int j = 0; j < 3; ++j) cin >> mat[i][j];



  for (int i = 0; i < 3; ++i) dp[0][i] = mat[0][i];



  vector<vector<int>> now {{1, 2}, {0, 2}, {0, 1}};



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

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

      for (auto el: now[col]) {

        dp[pos][col] = max(dp[pos][col], mat[pos][col] + dp[pos - 1][el]);

      }

    }

  }



  int res = 0;

  for (int i = 0; i < 3; ++i) res = max(res, dp[n - 1][i]);

  cout << res << 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: 2.181377793847871x):
#include <bits/stdc++.h>

using namespace std;



int mat[100002][3], dp[100002][3];





int main() {

  int n;

  cin >> n;

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

    for (int j = 0; j < 3; ++j) cin >> mat[i][j];



  for (int i = 0; i < 3; ++i) dp[0][i] = mat[0][i];



  vector<vector<int>> now {{1, 2}, {0, 2}, {0, 1}};



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

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

      for (auto el: now[col]) {

        dp[pos][col] = max(dp[pos][col], mat[pos][col] + dp[pos - 1][el]);

      }

    }

  }



  int res = 0;

  for (int i = 0; i < 3; ++i) res = max(res, dp[n - 1][i]);

  cout << res << 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: 2.1813756031825178x):
#include <bits/stdc++.h>

using namespace std;



int mat[100002][3], dp[100002][3];





int main() {

  int n;

  cin >> n;

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

    for (int j = 0; j < 3; ++j) cin >> mat[i][j];



  for (int i = 0; i < 3; ++i) dp[0][i] = mat[0][i];



  vector<vector<int>> now {{1, 2}, {0, 2}, {0, 1}};



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

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

      for (auto el: now[col]) {

        dp[pos][col] = max(dp[pos][col], mat[pos][col] + dp[pos - 1][el]);

      }

    }

  }



  int res = 0;

  for (int i = 0; i < 3; ++i) res = max(res, dp[n - 1][i]);

  cout << res << 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: 2.1813517977704766x):
#include <bits/stdc++.h>

using namespace std;



int mat[100002][3], dp[100002][3];





int main() {

  int n;

  cin >> n;

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

    for (int j = 0; j < 3; ++j) cin >> mat[i][j];



  for (int i = 0; i < 3; ++i) dp[0][i] = mat[0][i];



  vector<vector<int>> now {{1, 2}, {0, 2}, {0, 1}};



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

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

      for (auto el: now[col]) {

        dp[pos][col] = max(dp[pos][col], mat[pos][col] + dp[pos - 1][el]);

      }

    }

  }



  int res = 0;

  for (int i = 0; i < 3; ++i) res = max(res, dp[n - 1][i]);

  cout << res << endl;  

  return 0;

}
// 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: 2.1814723034615935x):
#include <bits/stdc++.h>

using namespace std;



int mat[100002][3], dp[100002][3];





int main() {

  int n;

  cin >> n;

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

    for (int j = 0; j < 3; ++j) cin >> mat[i][j];



  for (int i = 0; i < 3; ++i) dp[0][i] = mat[0][i];



  vector<vector<int>> now {{1, 2}, {0, 2}, {0, 1}};



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

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

      for (auto el: now[col]) {

        dp[pos][col] = max(dp[pos][col], mat[pos][col] + dp[pos - 1][el]);

      }

    }

  }



  int res = 0;

  for (int i = 0; i < 3; ++i) res = max(res, dp[n - 1][i]);

  cout << res << 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: 5.052085188395974x):
#include<bits/stdc++.h>

#define ll long long

using namespace std;

#define mx 100009

ll ar[mx][4];

ll br[mx][4];

ll n;



ll dp(){

    br[n][1]=ar[n][1];

    br[n][2]=ar[n][2];

    br[n][3]=ar[n][3];

    for(ll i=n-1;i>=1;i--){

        br[i][1]=ar[i][1]+max(br[i+1][2],br[i+1][3]);

        br[i][2]=ar[i][2]+max(br[i+1][1],br[i+1][3]);

        br[i][3]=ar[i][3]+max(br[i+1][1],br[i+1][2]);

    }

    return max(br[1][1],max(br[1][2],br[1][3]));

}



int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

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

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



    //ll n;

    cin>>n;

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

        for(ll j=1;j<=3;j++){

            cin>>ar[i][j];

        }

    }

    cout<<dp()<<endl;

    return 0;

}
// 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: 2.9808176959531685x):
#include<bits/stdc++.h>

#define ll long long

using namespace std;

#define mx 100009

ll ar[mx][4];

ll br[mx][4];

ll n;



ll dp(){

    br[n][1]=ar[n][1];

    br[n][2]=ar[n][2];

    br[n][3]=ar[n][3];

    for(ll i=n-1;i>=1;i--){

        br[i][1]=ar[i][1]+max(br[i+1][2],br[i+1][3]);

        br[i][2]=ar[i][2]+max(br[i+1][1],br[i+1][3]);

        br[i][3]=ar[i][3]+max(br[i+1][1],br[i+1][2]);

    }

    return max(br[1][1],max(br[1][2],br[1][3]));

}



int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

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

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



    //ll n;

    cin>>n;

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

        for(ll j=1;j<=3;j++){

            cin>>ar[i][j];

        }

    }

    cout<<dp()<<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: 2.1635332145014448x):
#include<bits/stdc++.h>

#define ll long long

using namespace std;

#define mx 100009

ll ar[mx][4];

ll br[mx][4];

ll n;



ll dp(){

    br[n][1]=ar[n][1];

    br[n][2]=ar[n][2];

    br[n][3]=ar[n][3];

    for(ll i=n-1;i>=1;i--){

        br[i][1]=ar[i][1]+max(br[i+1][2],br[i+1][3]);

        br[i][2]=ar[i][2]+max(br[i+1][1],br[i+1][3]);

        br[i][3]=ar[i][3]+max(br[i+1][1],br[i+1][2]);

    }

    return max(br[1][1],max(br[1][2],br[1][3]));

}



int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

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

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



    //ll n;

    cin>>n;

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

        for(ll j=1;j<=3;j++){

            cin>>ar[i][j];

        }

    }

    cout<<dp()<<endl;

    return 0;

}