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

#define MAX 100005

#define pb push_back

#define mp make_pair

#define pii pair<long long,long long>

#define read(a) scanf("%d",&a)

#define p1(a) cout<<"Check "<<a<<endl;

#define p2(a,b) cout<<"Check "<<a<<' '<<b<<endl;

#define p3(a,b,c) cout<<"Check "<<a<<' '<<b<<' '<<c<<endl;

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

#define ll long long

#define clr(a) memset(a,0,sizeof(a))

#define MOD 10000000000

#define INF 10000000000000;



using namespace std;



int main()

{

    int n,m;

    cin>>n;

    int ar[n+5][4];

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

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

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

        }

    }

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

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

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

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

    }

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



    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.198828984284808x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

#define ll long long

#define sz 100105

using namespace std;

ll dp[sz][10];

void solve(ll n , ll i){

   if(i<n){

        ll mx;

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

            mx = 0;

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

                if(k!=j)mx = max(mx,dp[i-1][k]);

            dp[i][j] += mx;

        }

        solve(n,i+1);

   }

}



int main(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    ll i , j , n , k , res = 0;

    cin>>n;

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

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

            cin>>dp[i][j];

    }

    solve(n,1);

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

        res = max(res,dp[n-1][j]);

    cout<<res<<endl;

    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 5.256701246501406x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

#define ll long long

using namespace std;

struct actv{

ll f,s,t;

};

int main(){

    ll n;

    cin>>n;

    vector < actv > v,dp;

    v.resize(n);

    dp.resize(n);

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

        cin>>v[i].f>>v[i].s>>v[i].t;

        dp[i].f=dp[i].s=dp[i].t=0;

    }

    dp[0].f=v[0].f;

    dp[0].t=v[0].t;

    dp[0].s=v[0].s;

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

        dp[i].f=v[i].f+max(dp[i-1].s,dp[i-1].t);

        dp[i].s=v[i].s+max(dp[i-1].f,dp[i-1].t);

        dp[i].t=v[i].t+max(dp[i-1].s,dp[i-1].f);

    }

    cout<<max(max(dp[n-1].f,dp[n-1].s),dp[n-1].t);

return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.238597603841017x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

using namespace std;

typedef long long ll;



ll arr[100005][3];

ll dp[100005][3];





int main(){

    ll n;

    cin>>n;

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

        cin>>arr[i][0];

        cin>>arr[i][1];

        cin>>arr[i][2];

    }

    dp[0][0]=arr[0][0];

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

    dp[0][2]=arr[0][2];

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

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

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

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

    }

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

}


// Faster C++ solution for the same problem (measured speedup: 5.1857701563495695x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

 }
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 5.403585493557641x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

 }
// Slower C++ solution:
#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;

}

// Faster C++ solution for the same problem (measured speedup: 5.257496823136622x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

 }
// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 5.2560976507067165x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

 }
// 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: 11.522650949471833x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

 }
// 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: 26.50489285798141x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	return 0;

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

using namespace std;



#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)

template <typename Arg1>

void __f(const char* name, Arg1&& arg1) {

  cerr << name << " : " << arg1 << "\n";

}

template <typename Arg1, typename... Args>

void __f(const char* na, Arg1&& arg1, Args&&... args) {

  const char* c = strchr(na + 1, ',');

  cerr.write(na, c - na) << " : " << arg1 << " , ";

  __f(c + 1, args...);

}

#define endl "\n"

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

#define UNIQUE(v) sort(all(v)),v.erase(unique(all(v)),v.end())

typedef long long LL;

typedef pair <int,int> pii;

const double pi = acos(-1.0);

const double eps = 1e-8;

const int mod = 1e9 + 7;

const int inf = 1e9;

const LL INF = 1e18;

const int N = 1e5 + 5;



LL a[ N ][ 5 ];

LL dp[ N ][5 ];



int main() {

  ios_base::sync_with_stdio(false);

  cin.tie(0); cout.tie(0);

  #ifdef LOCAL

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

  #endif // LOCAL

  int n;

  cin >> n;

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

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

      cin >> a[i][j];

    }

  }

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

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

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

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

      }

    }

  }

  cout << max(dp[n][1], max(dp[n][2], dp[n][3])) << endl;

  return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.272946913360445x):
#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int dp[maxn][3];

int abc[3];

int main()

{

	int n;

	while(~scanf("%d",&n)){

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

		dp[0][0] = abc[0];

		dp[0][1] = abc[1];

		dp[0][2] = abc[2];

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

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

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

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

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

		} 

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

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

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

	}

	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: 5.113709308156017x):
#include<bits/stdc++.h>

#include<iostream>

#include<vector>

#include<algorithm>

#include<map>



using namespace std;



const int N=100500;

int OO=INT_MAX;

int n,a[N],b[N],c[N];

int mem[N][4];

int many=0;



int solve(int idx,int last)

{

    if(idx==n)

        return 0;

    if(mem[idx][last]!=-1)

        return mem[idx][last];

    int op1=-OO,op2=-OO;

    if(last==1)

    {

        op1=solve(idx+1,2)+b[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==2)

    {

        op1=solve(idx+1,1)+a[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==3)

    {

        op1=solve(idx+1,1)+a[idx];

        op2=solve(idx+1,2)+b[idx];

    }

    return mem[idx][last] = max(op1,op2);

}



int main() {



    cin>>n;

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

    {

        mem[i][1]=-1;

        mem[i][2]=-1;

        mem[i][3]=-1;

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

    }

    int fcall,scall,thcall;



    fcall=solve(1,1)+a[0];

    scall=solve(1,2)+b[0];

    thcall=solve(1,3)+c[0];



    cout<< max(fcall,max(scall,thcall));



	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: 3.017176993120245x):
#include<bits/stdc++.h>

#include<iostream>

#include<vector>

#include<algorithm>

#include<map>



using namespace std;



const int N=100500;

int OO=INT_MAX;

int n,a[N],b[N],c[N];

int mem[N][4];

int many=0;



int solve(int idx,int last)

{

    if(idx==n)

        return 0;

    if(mem[idx][last]!=-1)

        return mem[idx][last];

    int op1=-OO,op2=-OO;

    if(last==1)

    {

        op1=solve(idx+1,2)+b[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==2)

    {

        op1=solve(idx+1,1)+a[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==3)

    {

        op1=solve(idx+1,1)+a[idx];

        op2=solve(idx+1,2)+b[idx];

    }

    return mem[idx][last] = max(op1,op2);

}



int main() {



    cin>>n;

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

    {

        mem[i][1]=-1;

        mem[i][2]=-1;

        mem[i][3]=-1;

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

    }

    int fcall,scall,thcall;



    fcall=solve(1,1)+a[0];

    scall=solve(1,2)+b[0];

    thcall=solve(1,3)+c[0];



    cout<< max(fcall,max(scall,thcall));



	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: 2.189923472175939x):
#include<bits/stdc++.h>

#include<iostream>

#include<vector>

#include<algorithm>

#include<map>



using namespace std;



const int N=100500;

int OO=INT_MAX;

int n,a[N],b[N],c[N];

int mem[N][4];

int many=0;



int solve(int idx,int last)

{

    if(idx==n)

        return 0;

    if(mem[idx][last]!=-1)

        return mem[idx][last];

    int op1=-OO,op2=-OO;

    if(last==1)

    {

        op1=solve(idx+1,2)+b[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==2)

    {

        op1=solve(idx+1,1)+a[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==3)

    {

        op1=solve(idx+1,1)+a[idx];

        op2=solve(idx+1,2)+b[idx];

    }

    return mem[idx][last] = max(op1,op2);

}



int main() {



    cin>>n;

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

    {

        mem[i][1]=-1;

        mem[i][2]=-1;

        mem[i][3]=-1;

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

    }

    int fcall,scall,thcall;



    fcall=solve(1,1)+a[0];

    scall=solve(1,2)+b[0];

    thcall=solve(1,3)+c[0];



    cout<< max(fcall,max(scall,thcall));



	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: 2.208547160708751x):
#include<bits/stdc++.h>

#include<iostream>

#include<vector>

#include<algorithm>

#include<map>



using namespace std;



const int N=100500;

int OO=INT_MAX;

int n,a[N],b[N],c[N];

int mem[N][4];

int many=0;



int solve(int idx,int last)

{

    if(idx==n)

        return 0;

    if(mem[idx][last]!=-1)

        return mem[idx][last];

    int op1=-OO,op2=-OO;

    if(last==1)

    {

        op1=solve(idx+1,2)+b[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==2)

    {

        op1=solve(idx+1,1)+a[idx];

        op2=solve(idx+1,3)+c[idx];

    }

    if(last==3)

    {

        op1=solve(idx+1,1)+a[idx];

        op2=solve(idx+1,2)+b[idx];

    }

    return mem[idx][last] = max(op1,op2);

}



int main() {



    cin>>n;

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

    {

        mem[i][1]=-1;

        mem[i][2]=-1;

        mem[i][3]=-1;

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

    }

    int fcall,scall,thcall;



    fcall=solve(1,1)+a[0];

    scall=solve(1,2)+b[0];

    thcall=solve(1,3)+c[0];



    cout<< max(fcall,max(scall,thcall));



	return 0;

}