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: 2.2188726303346895x):
#include<bits/stdc++.h>

#define ll long long

#define pb push_back

#define pf push_front

#define pob pop_back

#define pof pop_front

#define scn(n) scanf("%d",&n)

#define scnll(n) scanf("%lld",&n)

#define nl printf("\n")

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

using namespace std;

int p[100002][4];

int a[100002][4];

int max_sum(int c,int i,int n){

    if(c>n) return 0;

    if(p[c][i]!=-1) return p[c][i];

    int j;

    int mx=0;

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

        if(j==i) continue;

        mx=max(mx,a[c][j]+max_sum(c+1,j,n));

    }

    return p[c][i]=mx;

}

main()

{

    int n,ans;

    scanf("%d",&n);

    int i;

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

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

        p[i][0]=-1;

        p[i][1]=-1;

        p[i][2]=-1;

        p[i][3]=-1;



    }

    ans=max_sum(1,3,n);

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

}
// 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 m;

const int mx = 1e5+5;

int ac[mx][3];

int dp[3][mx];

int go(int a, int i){

	if(i == m) 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>>m;

	for(int i=0; i<m; 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: 2.2188950358997146x):
#include<bits/stdc++.h>

#define ll long long

#define pb push_back

#define pf push_front

#define pob pop_back

#define pof pop_front

#define scn(n) scanf("%d",&n)

#define scnll(n) scanf("%lld",&n)

#define nl printf("\n")

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

using namespace std;

int p[100002][4];

int a[100002][4];

int max_sum(int c,int i,int n){

    if(c>n) return 0;

    if(p[c][i]!=-1) return p[c][i];

    int j;

    int mx=0;

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

        if(j==i) continue;

        mx=max(mx,a[c][j]+max_sum(c+1,j,n));

    }

    return p[c][i]=mx;

}

main()

{

    int n,ans;

    scanf("%d",&n);

    int i;

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

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

        p[i][0]=-1;

        p[i][1]=-1;

        p[i][2]=-1;

        p[i][3]=-1;



    }

    ans=max_sum(1,3,n);

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

}
// 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";

}

	

// Faster C++ solution for the same problem (measured speedup: 2.2188726303346895x):
#include<bits/stdc++.h>

#define ll long long

#define pb push_back

#define pf push_front

#define pob pop_back

#define pof pop_front

#define scn(n) scanf("%d",&n)

#define scnll(n) scanf("%lld",&n)

#define nl printf("\n")

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

using namespace std;

int p[100002][4];

int a[100002][4];

int max_sum(int c,int i,int n){

    if(c>n) return 0;

    if(p[c][i]!=-1) return p[c][i];

    int j;

    int mx=0;

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

        if(j==i) continue;

        mx=max(mx,a[c][j]+max_sum(c+1,j,n));

    }

    return p[c][i]=mx;

}

main()

{

    int n,ans;

    scanf("%d",&n);

    int i;

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

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

        p[i][0]=-1;

        p[i][1]=-1;

        p[i][2]=-1;

        p[i][3]=-1;



    }

    ans=max_sum(1,3,n);

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

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

using namespace std;

const int mx=100010;

int arr[mx][3],n;

int dp[mx][3];

int f(int pos,int ty){

	if(pos==n)return 0;

	int &ans=dp[pos][ty];

	if(ans^-1)return ans;

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

		if(ty^i || !pos){

			ans=max(ans,f(pos+1,i)+arr[pos][i]);

		}

	}

	return ans;

}



int main(int argc, char const *argv[])

{

	memset(dp,-1,sizeof dp);

	cin>>n;

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

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

			cin>>arr[i][j];

		}

	}

	cout<<f(0,0)<<endl;



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.2202998237563936x):
#include<bits/stdc++.h>

#define ll long long

#define pb push_back

#define pf push_front

#define pob pop_back

#define pof pop_front

#define scn(n) scanf("%d",&n)

#define scnll(n) scanf("%lld",&n)

#define nl printf("\n")

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

using namespace std;

int p[100002][4];

int a[100002][4];

int max_sum(int c,int i,int n){

    if(c>n) return 0;

    if(p[c][i]!=-1) return p[c][i];

    int j;

    int mx=0;

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

        if(j==i) continue;

        mx=max(mx,a[c][j]+max_sum(c+1,j,n));

    }

    return p[c][i]=mx;

}

main()

{

    int n,ans;

    scanf("%d",&n);

    int i;

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

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

        p[i][0]=-1;

        p[i][1]=-1;

        p[i][2]=-1;

        p[i][3]=-1;



    }

    ans=max_sum(1,3,n);

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

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

#include<cstring>

using namespace std;



#define Max 100001

int dp[Max][3];



int main(){

	int n,a,b,c;

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

	cin>>n;

	for(int i=1;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][1],dp[i-1][0])+c;

	}

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

	cout<<ans<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.2196625527655356x):
#include<bits/stdc++.h>

#define ll long long

#define pb push_back

#define pf push_front

#define pob pop_back

#define pof pop_front

#define scn(n) scanf("%d",&n)

#define scnll(n) scanf("%lld",&n)

#define nl printf("\n")

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

using namespace std;

int p[100002][4];

int a[100002][4];

int max_sum(int c,int i,int n){

    if(c>n) return 0;

    if(p[c][i]!=-1) return p[c][i];

    int j;

    int mx=0;

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

        if(j==i) continue;

        mx=max(mx,a[c][j]+max_sum(c+1,j,n));

    }

    return p[c][i]=mx;

}

main()

{

    int n,ans;

    scanf("%d",&n);

    int i;

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

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

        p[i][0]=-1;

        p[i][1]=-1;

        p[i][2]=-1;

        p[i][3]=-1;



    }

    ans=max_sum(1,3,n);

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

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

using namespace std ;

#define ll long long

inline int D(){

    int t ;

    scanf( "%d" , &t ) ;

    return t ;

}

inline ll LLD(){

    ll t ;

    scanf( "%lld" , &t ) ;

    return t ;

}

//################################################################

const int N=100009;

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

int mem[N][4];

int solve(int ind ,int lastInd){

    if(ind==n)return 0;

    int  ret=INT_MIN;

    if(mem[ind][lastInd]!=-1)return mem[ind][lastInd];

    if(lastInd!=0){



        ret=max(ret,a[ind]+solve(ind+1,0));

    }

    if(lastInd!=1){



        ret=max(ret,b[ind]+solve(ind+1,1));

    }

    if(lastInd!=2){



        ret=max(ret,c[ind]+solve(ind+1,2));

    }

    return mem[ind][lastInd]=ret;



}

int main() {

    n=D();

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

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

    }

    memset(mem,-1, sizeof(mem));

    printf("%d",solve(0,3));





}

// Faster C++ solution for the same problem (measured speedup: 2.600837557983002x):
#include<bits/stdc++.h>

#define ll long long

#define pb push_back

#define pf push_front

#define pob pop_back

#define pof pop_front

#define scn(n) scanf("%d",&n)

#define scnll(n) scanf("%lld",&n)

#define nl printf("\n")

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

using namespace std;

int p[100002][4];

int a[100002][4];

int max_sum(int c,int i,int n){

    if(c>n) return 0;

    if(p[c][i]!=-1) return p[c][i];

    int j;

    int mx=0;

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

        if(j==i) continue;

        mx=max(mx,a[c][j]+max_sum(c+1,j,n));

    }

    return p[c][i]=mx;

}

main()

{

    int n,ans;

    scanf("%d",&n);

    int i;

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

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

        p[i][0]=-1;

        p[i][1]=-1;

        p[i][2]=-1;

        p[i][3]=-1;



    }

    ans=max_sum(1,3,n);

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

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

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int N = 1e5+10 , OO = 0x3f3f3f3f ;

//const int mod = 1000000007;

 

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

double startTime;

double getCurrentTime() {

	return ((double)clock() - startTime) / CLOCKS_PER_SEC;

}

#define clr(arr,val) memset(arr , val , sizeof (arr))

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

#define rloop(i,n) for(int i = int(n)-1;i >= 0;i--)

#define xloop(i,a,b) for(int i = int(a);i <= int(b);i++)

#define range(vec) for(auto &x : vec) cin >> x;

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

#define SZ(v)  ((int)((v).size()))

int dx[]={1,0,-1,0};int dy[]={0,1,0,-1}; //4 Direction

//int dx[]={1,1,0,-1,-1,-1,0,1};int dy[]={0,1,1,1,0,-1,-1,-1};//8 direction

//int dx[]={2,1,-1,-2,-2,-1,1,2};int dy[]={1,2,2,1,-1,-2,-2,-1};//Knight Direction

//int dx[6]={2,1,-1,-2,-1,1};int dy[6]={0,1,1,0,-1,-1}; //Hexagonal Direction

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

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



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

int dp[N][4];



int solve(int idx , int lst){

	

	if(idx == n) return 0;

	int& ret = dp[idx][lst];

	if(~ret) return ret;

	ret = -OO;

	if(lst != 0) // swim

		ret = max(ret , solve(idx+1 , 0) + a[idx]);

	if(lst != 1)

		ret = max(ret , solve(idx+1 , 1) + b[idx]);

	if(lst != 2)

		ret = max(ret , solve(idx+1 , 2) + c[idx]);

		

	return ret;

}





int main()

{

    startTime = (double)clock();

    #ifndef ONLINE_JUDGE

        //freopen("in.in", "r", stdin);

        //freopen("out.in", "w", stdout);

    #endif

	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

	

	cin >> n;

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

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



	clr(dp , -1);

	cout << solve(0 , 3) << "\n";





    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.6414241956159343x):
#include<bits/stdc++.h>

#define ll long long

#define pb push_back

#define pf push_front

#define pob pop_back

#define pof pop_front

#define scn(n) scanf("%d",&n)

#define scnll(n) scanf("%lld",&n)

#define nl printf("\n")

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

using namespace std;

int p[100002][4];

int a[100002][4];

int max_sum(int c,int i,int n){

    if(c>n) return 0;

    if(p[c][i]!=-1) return p[c][i];

    int j;

    int mx=0;

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

        if(j==i) continue;

        mx=max(mx,a[c][j]+max_sum(c+1,j,n));

    }

    return p[c][i]=mx;

}

main()

{

    int n,ans;

    scanf("%d",&n);

    int i;

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

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

        p[i][0]=-1;

        p[i][1]=-1;

        p[i][2]=-1;

        p[i][3]=-1;



    }

    ans=max_sum(1,3,n);

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

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

#define F first

#define S second

#define T second.first

#define C second.second

#define FIO ios_base::sync_with_stdio(0);cin.tie(0);

using namespace std;

typedef long long ll;

typedef pair<int, int> pii;

typedef pair<int,pair<int,int>> piii;



const int N= 1e5+5, L = 14, OO = 0x3f3f3f3f;

const ll M=10 ;

const double ESP = 1e-8;



int n, k;

piii A[N];

int dp[N][5];

int solve(int idx, int prev){

    if(idx == n)

        return 0;



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

    if(~ret)

        return ret;



    ret = max(ret, (prev != 1)? solve(idx+1, 1)+A[idx].F : -1);

    ret = max(ret, (prev != 2)? solve(idx+1, 2)+A[idx].T : -1);

    return ret = max(ret, (prev != 3)? solve(idx+1, 3)+A[idx].C : -1);



}



int main(){

    FIO

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

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

    cin>>n;

    for(int i=0; i<n; i++)cin>>A[i].F>>A[i].T>>A[i].C;

    memset(dp, -1, sizeof dp);

    cout<<solve(0,4);

}


// Faster C++ solution for the same problem (measured speedup: 3.0311398808728796x):
#include<bits/stdc++.h>

#define ll long long

#define pb push_back

#define pf push_front

#define pob pop_back

#define pof pop_front

#define scn(n) scanf("%d",&n)

#define scnll(n) scanf("%lld",&n)

#define nl printf("\n")

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

using namespace std;

int p[100002][4];

int a[100002][4];

int max_sum(int c,int i,int n){

    if(c>n) return 0;

    if(p[c][i]!=-1) return p[c][i];

    int j;

    int mx=0;

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

        if(j==i) continue;

        mx=max(mx,a[c][j]+max_sum(c+1,j,n));

    }

    return p[c][i]=mx;

}

main()

{

    int n,ans;

    scanf("%d",&n);

    int i;

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

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

        p[i][0]=-1;

        p[i][1]=-1;

        p[i][2]=-1;

        p[i][3]=-1;



    }

    ans=max_sum(1,3,n);

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

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



using namespace std;



const int N=1e5;

int n;

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

int mem[N][5];



int solve(int i,int lst){

    if(i==n)

        return 0;

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

    int op1=lst!=0?solve(i+1,0)+a[i]:-1000000000;

    int op2=lst!=1?solve(i+1,1)+b[i]:-1000000000;

    int op3=lst!=2?solve(i+1,2)+c[i]:-1000000000;

    return mem[i][lst]=max(op1,max(op2,op3));

}



int main() {

    scanf("%d",&n);

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

        scanf("%d%d%d",a+i,b+i,c+i);

    memset(mem,-1,sizeof mem);

    printf("%d",max(solve(0,0),max(solve(0,1),solve(0,2))));

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 3.008849362901893x):
#include<bits/stdc++.h>

#define ll long long

#define pb push_back

#define pf push_front

#define pob pop_back

#define pof pop_front

#define scn(n) scanf("%d",&n)

#define scnll(n) scanf("%lld",&n)

#define nl printf("\n")

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

using namespace std;

int p[100002][4];

int a[100002][4];

int max_sum(int c,int i,int n){

    if(c>n) return 0;

    if(p[c][i]!=-1) return p[c][i];

    int j;

    int mx=0;

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

        if(j==i) continue;

        mx=max(mx,a[c][j]+max_sum(c+1,j,n));

    }

    return p[c][i]=mx;

}

main()

{

    int n,ans;

    scanf("%d",&n);

    int i;

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

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

        p[i][0]=-1;

        p[i][1]=-1;

        p[i][2]=-1;

        p[i][3]=-1;



    }

    ans=max_sum(1,3,n);

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

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

using namespace std;

#define ll long long

#define ull unsigned long long

#define PII pair<int,int>

#define PSI pair<string,int>

#define PIII pair< int, pair<int,int> >

const int MXN = 1e5+5;

int a[MXN][4], dp[MXN][4];

int N;



int rec(int pos, int type){



    if(pos>N) return 0;



    if(dp[pos][type]!=-1) return dp[pos][type];



    int ret = -1;

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

        if(i!=type)

            ret = max (ret, rec(pos+1,i) + a[pos+1][i]);



    return dp[pos][type] = ret;

}



void read(){



    int i,j;

    cin>>N;

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

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

            cin>>a[i][j];

}



void solve(){



    memset(dp,-1,sizeof(dp));

    read();

    cout<< rec(0,0) << "\n";

}



int main(){



    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);

    solve();

}


// Faster C++ solution for the same problem (measured speedup: 2.6295375965908376x):
#include<bits/stdc++.h>

#define ll long long

#define pb push_back

#define pf push_front

#define pob pop_back

#define pof pop_front

#define scn(n) scanf("%d",&n)

#define scnll(n) scanf("%lld",&n)

#define nl printf("\n")

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

using namespace std;

int p[100002][4];

int a[100002][4];

int max_sum(int c,int i,int n){

    if(c>n) return 0;

    if(p[c][i]!=-1) return p[c][i];

    int j;

    int mx=0;

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

        if(j==i) continue;

        mx=max(mx,a[c][j]+max_sum(c+1,j,n));

    }

    return p[c][i]=mx;

}

main()

{

    int n,ans;

    scanf("%d",&n);

    int i;

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

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

        p[i][0]=-1;

        p[i][1]=-1;

        p[i][2]=-1;

        p[i][3]=-1;



    }

    ans=max_sum(1,3,n);

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

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

#define endl '\n'

#define ll long long

using namespace std;



const int MXN = 1e5 + 5;

int n, k;

int m[MXN][3];

int dp[MXN][4];



int x[3] = {0, 1, 2};



int Dp(int d, int lst) {

    if(d == n) return 0;

    if(dp[d][lst] != -1) return dp[d][lst];

    int ans = 0;

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

        if(lst != x[i]) {

            ans = max(ans, Dp(d + 1, i) + m[d][i]);

        }

    }

    return dp[d][lst] = ans;

}



int main() {

    ios_base::sync_with_stdio(false); cin.tie(NULL);

    cin >> n;

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

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

    }

    memset(dp, -1, sizeof dp);

    cout << Dp(0, 3) << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.630389274464348x):
#include<bits/stdc++.h>

#define ll long long

#define pb push_back

#define pf push_front

#define pob pop_back

#define pof pop_front

#define scn(n) scanf("%d",&n)

#define scnll(n) scanf("%lld",&n)

#define nl printf("\n")

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

using namespace std;

int p[100002][4];

int a[100002][4];

int max_sum(int c,int i,int n){

    if(c>n) return 0;

    if(p[c][i]!=-1) return p[c][i];

    int j;

    int mx=0;

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

        if(j==i) continue;

        mx=max(mx,a[c][j]+max_sum(c+1,j,n));

    }

    return p[c][i]=mx;

}

main()

{

    int n,ans;

    scanf("%d",&n);

    int i;

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

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

        p[i][0]=-1;

        p[i][1]=-1;

        p[i][2]=-1;

        p[i][3]=-1;



    }

    ans=max_sum(1,3,n);

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

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





using namespace std;



vector<vector<int>> vec(100007);

int n;



long long meme[100007][5];





long long solve(int index, int flag){

    if(index==n){

        return 0;

    }

    

    long long &res = meme[index][flag]; 

    

    if(res != -1){

        return res;

    }

    res = 0;

    

    if(flag!=0){

        res = max(res,vec[index][0]+solve(index+1,0));

    }

    if(flag!=1){

        res =  max(res,vec[index][1]+solve(index+1,1));

    }

    if(flag!=2){

        res =  max(res,vec[index][2]+solve(index+1,2));

    }

    return res;

}



int main(){

    cin>>n;

    int num1,num2,num3;

    memset(meme,-1,sizeof(meme));

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

        cin>>num1>>num2>>num3;

        vec[i].push_back(num1);

        vec[i].push_back(num2);

        vec[i].push_back(num3);

    }

    long long sol = solve(0,3);

    cout<<sol<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 8.097865324083623x):
#include<bits/stdc++.h>

#define ll long long

#define pb push_back

#define pf push_front

#define pob pop_back

#define pof pop_front

#define scn(n) scanf("%d",&n)

#define scnll(n) scanf("%lld",&n)

#define nl printf("\n")

#define YES printf("YES\n")

#define Yes printf("Yes\n")

#define yes printf("yes\n")

#define NO printf("NO\n")

#define No printf("No\n")

#define no printf("no\n")

using namespace std;

int p[100002][4];

int a[100002][4];

int max_sum(int c,int i,int n){

    if(c>n) return 0;

    if(p[c][i]!=-1) return p[c][i];

    int j;

    int mx=0;

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

        if(j==i) continue;

        mx=max(mx,a[c][j]+max_sum(c+1,j,n));

    }

    return p[c][i]=mx;

}

main()

{

    int n,ans;

    scanf("%d",&n);

    int i;

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

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

        p[i][0]=-1;

        p[i][1]=-1;

        p[i][2]=-1;

        p[i][3]=-1;



    }

    ans=max_sum(1,3,n);

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

}