stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 2.2062733274494306x):
#include <iostream>



using namespace std;



int n;

int a[100001], b[100001], c[100001];

bool found[100001][3];

long long int dp[100001][3];



long long int m(long long int a, long long int b){

    return ((a > b) ? a : b);

}



long long int f(int now, int k){

    if (now == n){

        if (k == 0){

            return max(b[now], c[now]);

        } else if (k == 1){

            return max(a[now], c[now]);

        } else if (k == 2) {

            return max(b[now], c[now]);

        } else {

            return max(a[now], max(b[now], c[now]));

        }

    } else {

        if (found[now][k]){

            return dp[now][k];

        } else {

            if (k == 0){

                dp[now][k] = m(b[now] + f(now+1, 1), c[now] + f(now+1, 2));

            } else if (k == 1){

                dp[now][k] = m(a[now] + f(now+1, 0), c[now] + f(now+1, 2));

            } else if (k == 2){

                dp[now][k] = m(b[now] + f(now+1, 1), a[now] + f(now+1, 0));

            } else {

                dp[now][k] = m(a[now] + f(now+1, 0), m(b[now] + f(now+1, 1), c[now] + f(now+1, 2)));

            }

            found[now][k] = true;

            return dp[now][k];

        }

    }

}



int main(){

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

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

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

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

    }

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

}
// 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: 2.2063952098951636x):
#include <iostream>



using namespace std;



int n;

int a[100001], b[100001], c[100001];

bool found[100001][3];

long long int dp[100001][3];



long long int m(long long int a, long long int b){

    return ((a > b) ? a : b);

}



long long int f(int now, int k){

    if (now == n){

        if (k == 0){

            return max(b[now], c[now]);

        } else if (k == 1){

            return max(a[now], c[now]);

        } else if (k == 2) {

            return max(b[now], c[now]);

        } else {

            return max(a[now], max(b[now], c[now]));

        }

    } else {

        if (found[now][k]){

            return dp[now][k];

        } else {

            if (k == 0){

                dp[now][k] = m(b[now] + f(now+1, 1), c[now] + f(now+1, 2));

            } else if (k == 1){

                dp[now][k] = m(a[now] + f(now+1, 0), c[now] + f(now+1, 2));

            } else if (k == 2){

                dp[now][k] = m(b[now] + f(now+1, 1), a[now] + f(now+1, 0));

            } else {

                dp[now][k] = m(a[now] + f(now+1, 0), m(b[now] + f(now+1, 1), c[now] + f(now+1, 2)));

            }

            found[now][k] = true;

            return dp[now][k];

        }

    }

}



int main(){

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

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

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

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

    }

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

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



using namespace std;

typedef long long ll;

typedef double dd;

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

#define endl "\n"

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



void fast() {

    cin.tie(0);

    cin.sync_with_stdio(0);

}

const int MAX=1e5+10;

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

int ans(int i,int prev){

    if(i==n)return 0;

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

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

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

    if(prev!=0){

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

    }

    if(prev!=1){

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

    }

    if(prev!=2){

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

    }

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

}



int main() {

    fast();

    cin>>n;

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

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

    }

    clr(dp,-1);

    cout<<ans(0,-1);

}


// Faster C++ solution for the same problem (measured speedup: 2.216700349711453x):
#include <iostream>



using namespace std;



int n;

int a[100001], b[100001], c[100001];

bool found[100001][3];

long long int dp[100001][3];



long long int m(long long int a, long long int b){

    return ((a > b) ? a : b);

}



long long int f(int now, int k){

    if (now == n){

        if (k == 0){

            return max(b[now], c[now]);

        } else if (k == 1){

            return max(a[now], c[now]);

        } else if (k == 2) {

            return max(b[now], c[now]);

        } else {

            return max(a[now], max(b[now], c[now]));

        }

    } else {

        if (found[now][k]){

            return dp[now][k];

        } else {

            if (k == 0){

                dp[now][k] = m(b[now] + f(now+1, 1), c[now] + f(now+1, 2));

            } else if (k == 1){

                dp[now][k] = m(a[now] + f(now+1, 0), c[now] + f(now+1, 2));

            } else if (k == 2){

                dp[now][k] = m(b[now] + f(now+1, 1), a[now] + f(now+1, 0));

            } else {

                dp[now][k] = m(a[now] + f(now+1, 0), m(b[now] + f(now+1, 1), c[now] + f(now+1, 2)));

            }

            found[now][k] = true;

            return dp[now][k];

        }

    }

}



int main(){

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

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

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

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

    }

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

}
// Slower C++ solution:
/*

** author:mahmoud magdy

*/

#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define endl "\n"



inline int D()

{

    int m ;

    cin>>m;

    return m ;

}

inline ll lD()

{

    ll m ;

    cin>>m;

    return m ;

}

inline double dD()

{

    double m;

    cin>>m;

    return m;

}

void fast()

{

    ios_base :: sync_with_stdio( 0 ) ;

    cin.tie( 0 ) ;

    cout.tie( 0 ) ;

}

int n;

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

ll dp[100005][5];

ll solve(int i,int prv){

 if(i==n)return 0;

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

 ll x=0;

 if(prv!=1){

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

 }

 ll y=0;

 if(prv!=2)

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

 ll z=0;

 if(prv!=3)

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

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

}

int main()

{

    memset(dp,-1,sizeof dp);

    fast();

    n=D();

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

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

    }

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

}

// Faster C++ solution for the same problem (measured speedup: 5.098948629529135x):
#include <iostream>



using namespace std;



int n;

int a[100001], b[100001], c[100001];

bool found[100001][3];

long long int dp[100001][3];



long long int m(long long int a, long long int b){

    return ((a > b) ? a : b);

}



long long int f(int now, int k){

    if (now == n){

        if (k == 0){

            return max(b[now], c[now]);

        } else if (k == 1){

            return max(a[now], c[now]);

        } else if (k == 2) {

            return max(b[now], c[now]);

        } else {

            return max(a[now], max(b[now], c[now]));

        }

    } else {

        if (found[now][k]){

            return dp[now][k];

        } else {

            if (k == 0){

                dp[now][k] = m(b[now] + f(now+1, 1), c[now] + f(now+1, 2));

            } else if (k == 1){

                dp[now][k] = m(a[now] + f(now+1, 0), c[now] + f(now+1, 2));

            } else if (k == 2){

                dp[now][k] = m(b[now] + f(now+1, 1), a[now] + f(now+1, 0));

            } else {

                dp[now][k] = m(a[now] + f(now+1, 0), m(b[now] + f(now+1, 1), c[now] + f(now+1, 2)));

            }

            found[now][k] = true;

            return dp[now][k];

        }

    }

}



int main(){

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

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

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

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

    }

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

}
// Slower C++ solution:


#include <bits/stdc++.h>

using namespace  std;



int N;

const int SIZE = 100005;

int mem [SIZE][4];

int arr [SIZE][3];





int dp (int i, int activ) {

  

  if (i == N) return 0;

  



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

  

  int ret = -1; 



  if (activ != 0) {

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

  }



  if (activ != 1) {

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

    

  }



  if (activ != 2) {

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

    

  }



  return  mem[i][activ] = ret;

}



int main (){ 



  cin  >> N ;

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

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

  }



  memset(mem, -1, sizeof mem);

  





  cout << dp(0,3);



}

// Faster C++ solution for the same problem (measured speedup: 2.6009794131410455x):
#include <iostream>



using namespace std;



int n;

int a[100001], b[100001], c[100001];

bool found[100001][3];

long long int dp[100001][3];



long long int m(long long int a, long long int b){

    return ((a > b) ? a : b);

}



long long int f(int now, int k){

    if (now == n){

        if (k == 0){

            return max(b[now], c[now]);

        } else if (k == 1){

            return max(a[now], c[now]);

        } else if (k == 2) {

            return max(b[now], c[now]);

        } else {

            return max(a[now], max(b[now], c[now]));

        }

    } else {

        if (found[now][k]){

            return dp[now][k];

        } else {

            if (k == 0){

                dp[now][k] = m(b[now] + f(now+1, 1), c[now] + f(now+1, 2));

            } else if (k == 1){

                dp[now][k] = m(a[now] + f(now+1, 0), c[now] + f(now+1, 2));

            } else if (k == 2){

                dp[now][k] = m(b[now] + f(now+1, 1), a[now] + f(now+1, 0));

            } else {

                dp[now][k] = m(a[now] + f(now+1, 0), m(b[now] + f(now+1, 1), c[now] + f(now+1, 2)));

            }

            found[now][k] = true;

            return dp[now][k];

        }

    }

}



int main(){

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

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

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

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

    }

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

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



using namespace std;



int a[100007], b[100007], c[100007];

int memo[100007][4];

int n; 



int solve(int p, int last) {

    if(p == n) return 0;

    int & st = memo[p][last];

    if(st == -1) {

        st = max({

            last != 1? solve(p+1, 1) + a[p] : 0,

            last != 2? solve(p+1, 2) + b[p] : 0,

            last != 3? solve(p+1, 3) + c[p] : 0

        });

    }

    return st;

}

int main() {

    memset(memo, -1, sizeof memo);

    cin >> n;

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

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

    }



    cout << solve(0, 0);

}

// Faster C++ solution for the same problem (measured speedup: 2.60736072845947x):
#include <iostream>



using namespace std;



int n;

int a[100001], b[100001], c[100001];

bool found[100001][3];

long long int dp[100001][3];



long long int m(long long int a, long long int b){

    return ((a > b) ? a : b);

}



long long int f(int now, int k){

    if (now == n){

        if (k == 0){

            return max(b[now], c[now]);

        } else if (k == 1){

            return max(a[now], c[now]);

        } else if (k == 2) {

            return max(b[now], c[now]);

        } else {

            return max(a[now], max(b[now], c[now]));

        }

    } else {

        if (found[now][k]){

            return dp[now][k];

        } else {

            if (k == 0){

                dp[now][k] = m(b[now] + f(now+1, 1), c[now] + f(now+1, 2));

            } else if (k == 1){

                dp[now][k] = m(a[now] + f(now+1, 0), c[now] + f(now+1, 2));

            } else if (k == 2){

                dp[now][k] = m(b[now] + f(now+1, 1), a[now] + f(now+1, 0));

            } else {

                dp[now][k] = m(a[now] + f(now+1, 0), m(b[now] + f(now+1, 1), c[now] + f(now+1, 2)));

            }

            found[now][k] = true;

            return dp[now][k];

        }

    }

}



int main(){

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

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

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

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

    }

    cout << m(dp[n][0], m(dp[n][1], dp[n][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: 5.175531921617787x):
#include <cstdio>

#include <algorithm>

#include <vector>

using namespace std;

 

int n,dp[100005][5],a[100005][5],k; 

int main(){

	scanf("%d",&n);

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

		scanf("%d %d %d",&a[i][1],&a[i][2],&a[i][3]);

	}

	dp[1][1]=a[1][1]; dp[1][2]=a[1][2]; dp[1][3]=a[1][3];

	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]);

	}

	printf("%d\n",max(max(dp[n][1],dp[n][2]),dp[n][3]));

	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: 5.264582647750705x):
#include <cstdio>

#include <algorithm>

#include <vector>

using namespace std;

 

int n,dp[100005][5],a[100005][5],k; 

int main(){

	scanf("%d",&n);

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

		scanf("%d %d %d",&a[i][1],&a[i][2],&a[i][3]);

	}

	dp[1][1]=a[1][1]; dp[1][2]=a[1][2]; dp[1][3]=a[1][3];

	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]);

	}

	printf("%d\n",max(max(dp[n][1],dp[n][2]),dp[n][3]));

	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: 5.2109333255416095x):
#include <cstdio>

#include <algorithm>

#include <vector>

using namespace std;

 

int n,dp[100005][5],a[100005][5],k; 

int main(){

	scanf("%d",&n);

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

		scanf("%d %d %d",&a[i][1],&a[i][2],&a[i][3]);

	}

	dp[1][1]=a[1][1]; dp[1][2]=a[1][2]; dp[1][3]=a[1][3];

	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]);

	}

	printf("%d\n",max(max(dp[n][1],dp[n][2]),dp[n][3]));

	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: 5.220472800189805x):
#include <cstdio>

#include <algorithm>

#include <vector>

using namespace std;

 

int n,dp[100005][5],a[100005][5],k; 

int main(){

	scanf("%d",&n);

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

		scanf("%d %d %d",&a[i][1],&a[i][2],&a[i][3]);

	}

	dp[1][1]=a[1][1]; dp[1][2]=a[1][2]; dp[1][3]=a[1][3];

	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]);

	}

	printf("%d\n",max(max(dp[n][1],dp[n][2]),dp[n][3]));

	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: 5.20240220682763x):
#include <cstdio>

#include <algorithm>

#include <vector>

using namespace std;

 

int n,dp[100005][5],a[100005][5],k; 

int main(){

	scanf("%d",&n);

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

		scanf("%d %d %d",&a[i][1],&a[i][2],&a[i][3]);

	}

	dp[1][1]=a[1][1]; dp[1][2]=a[1][2]; dp[1][3]=a[1][3];

	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]);

	}

	printf("%d\n",max(max(dp[n][1],dp[n][2]),dp[n][3]));

	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: 5.21324090544657x):
#include <cstdio>

#include <algorithm>

#include <vector>

using namespace std;

 

int n,dp[100005][5],a[100005][5],k; 

int main(){

	scanf("%d",&n);

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

		scanf("%d %d %d",&a[i][1],&a[i][2],&a[i][3]);

	}

	dp[1][1]=a[1][1]; dp[1][2]=a[1][2]; dp[1][3]=a[1][3];

	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]);

	}

	printf("%d\n",max(max(dp[n][1],dp[n][2]),dp[n][3]));

	return 0;

}