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

using namespace std;

#define INF 100005

#define maxx 1000000000000



int n;

long long a[INF][3];

long long mem[INF][3];



long long rec(int level,int pre)

{

    if(level==n) return 0;

    if(mem[level][pre]!=-1) return mem[level][pre];

    long long now=0;

    if(pre==0)

    {

        now=max(now,max(a[level][0]+rec(level+1,1),a[level][0]+rec(level+1,2)));

    }else

    if(pre==1)

    {

        now=max(now,max(a[level][1]+rec(level+1,0),a[level][1]+rec(level+1,2)));

    }else

    if(pre==2)

    {

        now=max(now,max(a[level][2]+rec(level+1,0),a[level][2]+rec(level+1,1)));

    }

    return mem[level][pre]=now;

}



int main()

{

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

    scanf("%d",&n);

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

    {

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

    }

    long long ans=max(rec(0,0),max(rec(0,1),rec(0,2)));

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

}


// Faster C++ solution for the same problem (measured speedup: 3.4274015251572623x):
#include <bits/stdc++.h> 



using namespace std;

 

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;

 

#define nl '\n'

 

#define y1 aza_ibagaliev

 

#define f first

#define s second

	

#define ed end()

#define bg begin()

#define rbg rbegin()

 

#define rv reverse

#define mp make_pair

#define pb push_back

#define ppb pop_back

#define ppf pop_front

 

#define sz size()

#define all(x) x.bg, x.ed

 

#define pll pair < ll, ll >

#define pii pair < int, int >

 

#define sqr(x) ((x) * 1ll * (x))

#define sqrd(x) ((x) * 1.0 * (x))

 

#define speed ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)

 

#define rnl cout << "----------------------------------------" << nl

 

#define bit __builtin_popcount

 

inline ll bip(ll x, ll n, ll mod){ll res=1;while(n){if(n & 1){res=(res*x)%mod;}x=(x*x)%mod;n>>=1;}return res;}

        	

const int ppr = 257;

const ll INF = 2e18;

const int inf = 2e9;

const int mod = 1e9 + 7;

const int N = 1e5 + 123;

const ld pi = 3.1741592653589793238462643;

 

const int dx[] = {1, 0, -1, 0};

const int dy[] = {0, 1, 0, -1};



int n, x[5][N], dp[5][N];



main() {             	

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

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



//	srand(time(NULL));



	scanf("%d", &n);



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

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

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



		if(i == 1) {

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

				dp[j][i] = x[j][i];

		} else {

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

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

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

		}

	}



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



    return 0;                                        

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

#include<algorithm>

#include<cstring>

using namespace std;

const int maxn = 1e5 + 5;



int a[maxn][4];

int dp[maxn][4];//d[i][j] i day choose j thing 's max happy



int main(){

	int n;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] ,dp[i - 1][3]) + a[i][1];

		dp[i][2] = max(dp[i - 1][1] ,dp[i - 1][3]) + a[i][2];

		dp[i][3] = max(dp[i - 1][2] ,dp[i - 1][1]) + a[i][3];

	}



	int ans = 0;

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

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

}

// Faster C++ solution for the same problem (measured speedup: 1.807065853939665x):
#include <bits/stdc++.h> 



using namespace std;

 

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;

 

#define nl '\n'

 

#define y1 aza_ibagaliev

 

#define f first

#define s second

	

#define ed end()

#define bg begin()

#define rbg rbegin()

 

#define rv reverse

#define mp make_pair

#define pb push_back

#define ppb pop_back

#define ppf pop_front

 

#define sz size()

#define all(x) x.bg, x.ed

 

#define pll pair < ll, ll >

#define pii pair < int, int >

 

#define sqr(x) ((x) * 1ll * (x))

#define sqrd(x) ((x) * 1.0 * (x))

 

#define speed ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)

 

#define rnl cout << "----------------------------------------" << nl

 

#define bit __builtin_popcount

 

inline ll bip(ll x, ll n, ll mod){ll res=1;while(n){if(n & 1){res=(res*x)%mod;}x=(x*x)%mod;n>>=1;}return res;}

        	

const int ppr = 257;

const ll INF = 2e18;

const int inf = 2e9;

const int mod = 1e9 + 7;

const int N = 1e5 + 123;

const ld pi = 3.1741592653589793238462643;

 

const int dx[] = {1, 0, -1, 0};

const int dy[] = {0, 1, 0, -1};



int n, x[5][N], dp[5][N];



main() {             	

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

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



//	srand(time(NULL));



	scanf("%d", &n);



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

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

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



		if(i == 1) {

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

				dp[j][i] = x[j][i];

		} else {

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

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

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

		}

	}



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



    return 0;                                        

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

using namespace std;



long long w1,w2,hap=0,maxi,aa,bb,cc,N,i,j,k,a[100005][5];

long long M[100005][5];





int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cin>>N;

    for(i=0; i<100003; i++)

        for(j=0; j<4; j++)

            M[i][j]=0;



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

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

    M[N-1][0]=a[N-1][0];

    M[N-1][1]=a[N-1][1];

    M[N-1][2]=a[N-1][2];



    for(i=N-2; i>=0; i--)

    {

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

        {

            if(j==0)

            {

                w1=1;

                w2=2;

            }

            else if(j==1)

            {

                w1=0;

                w2=2;

            }

            else

            {

                w1=0;

                w2=1;

            }

        M[i][j]=a[i][j]+max(M[i+1][w1],M[i+1][w2]);

        }

    }

    aa=max(M[0][0],M[0][1]);

    aa=max(aa,M[0][2]);

    cout<<aa<<endl;

    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 4.996032135782925x):
#include <iostream>

#include <iomanip>

#include <cmath>

#include <cstdio>

#include <cstring>

#include <ctime>

#include <string>

#include <stdio.h>

#include <set>

#include <map>

#include <algorithm>

#include <queue>

#include <vector>

#define ll long long

#define pb push_back

#define mp make_pair

#define f first

#define s second



using namespace std;



const int inf = 1e9 + 7;

const long long INF = 1e18 + 7;

const int maxn = 2e5 + 7;



int n, dp[maxn][4];

pair < pair < int, int >, int > a[maxn];



int main(){

	ios_base::sync_with_stdio(NULL);

	cin.tie(0);

	cout.tie(0);

	cin >> n;

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

		cin >> a[i].f.f >> a[i].f.s >> a[i].s;

	dp[1][1] = a[1].f.f;

	dp[1][2] = a[1].f.s;

	dp[1][3] = a[1].s;

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

		dp[i][1] = max(dp[i - 1][3], dp[i - 1][2]) + a[i].f.f;

		dp[i][2] = max(dp[i - 1][3], dp[i - 1][1]) + a[i].f.s;

		dp[i][3] = max(dp[i - 1][1], dp[i - 1][2]) + a[i].s;

	}

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

	return 0;

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

using namespace std;

#define INF 100005

#define maxx 1000000000000



int n;

long long a[INF][3];

long long mem[INF][3];



long long rec(int level,int pre)

{

    if(level==n) return 0;

    if(mem[level][pre]!=-1) return mem[level][pre];

    long long now=0;

    if(pre==0)

    {

        now=max(now,max(a[level][0]+rec(level+1,1),a[level][0]+rec(level+1,2)));

    }else

    if(pre==1)

    {

        now=max(now,max(a[level][1]+rec(level+1,0),a[level][1]+rec(level+1,2)));

    }else

    if(pre==2)

    {

        now=max(now,max(a[level][2]+rec(level+1,0),a[level][2]+rec(level+1,1)));

    }

    return mem[level][pre]=now;

}



int main()

{

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

    scanf("%d",&n);

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

    {

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

    }

    long long ans=max(rec(0,0),max(rec(0,1),rec(0,2)));

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

}


// Faster C++ solution for the same problem (measured speedup: 3.351574236114426x):
#include <iostream>

#include <iomanip>

#include <cmath>

#include <cstdio>

#include <cstring>

#include <ctime>

#include <string>

#include <stdio.h>

#include <set>

#include <map>

#include <algorithm>

#include <queue>

#include <vector>

#define ll long long

#define pb push_back

#define mp make_pair

#define f first

#define s second



using namespace std;



const int inf = 1e9 + 7;

const long long INF = 1e18 + 7;

const int maxn = 2e5 + 7;



int n, dp[maxn][4];

pair < pair < int, int >, int > a[maxn];



int main(){

	ios_base::sync_with_stdio(NULL);

	cin.tie(0);

	cout.tie(0);

	cin >> n;

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

		cin >> a[i].f.f >> a[i].f.s >> a[i].s;

	dp[1][1] = a[1].f.f;

	dp[1][2] = a[1].f.s;

	dp[1][3] = a[1].s;

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

		dp[i][1] = max(dp[i - 1][3], dp[i - 1][2]) + a[i].f.f;

		dp[i][2] = max(dp[i - 1][3], dp[i - 1][1]) + a[i].f.s;

		dp[i][3] = max(dp[i - 1][1], dp[i - 1][2]) + a[i].s;

	}

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

	return 0;

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

#include<algorithm>

#include<cstring>

using namespace std;

const int maxn = 1e5 + 5;



int a[maxn][4];

int dp[maxn][4];//d[i][j] i day choose j thing 's max happy



int main(){

	int n;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] ,dp[i - 1][3]) + a[i][1];

		dp[i][2] = max(dp[i - 1][1] ,dp[i - 1][3]) + a[i][2];

		dp[i][3] = max(dp[i - 1][2] ,dp[i - 1][1]) + a[i][3];

	}



	int ans = 0;

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

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

}

// Faster C++ solution for the same problem (measured speedup: 1.7670866149096434x):
#include <iostream>

#include <iomanip>

#include <cmath>

#include <cstdio>

#include <cstring>

#include <ctime>

#include <string>

#include <stdio.h>

#include <set>

#include <map>

#include <algorithm>

#include <queue>

#include <vector>

#define ll long long

#define pb push_back

#define mp make_pair

#define f first

#define s second



using namespace std;



const int inf = 1e9 + 7;

const long long INF = 1e18 + 7;

const int maxn = 2e5 + 7;



int n, dp[maxn][4];

pair < pair < int, int >, int > a[maxn];



int main(){

	ios_base::sync_with_stdio(NULL);

	cin.tie(0);

	cout.tie(0);

	cin >> n;

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

		cin >> a[i].f.f >> a[i].f.s >> a[i].s;

	dp[1][1] = a[1].f.f;

	dp[1][2] = a[1].f.s;

	dp[1][3] = a[1].s;

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

		dp[i][1] = max(dp[i - 1][3], dp[i - 1][2]) + a[i].f.f;

		dp[i][2] = max(dp[i - 1][3], dp[i - 1][1]) + a[i].f.s;

		dp[i][3] = max(dp[i - 1][1], dp[i - 1][2]) + a[i].s;

	}

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

	return 0;

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

using namespace std;



long long w1,w2,hap=0,maxi,aa,bb,cc,N,i,j,k,a[100005][5];

long long M[100005][5];





int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cin>>N;

    for(i=0; i<100003; i++)

        for(j=0; j<4; j++)

            M[i][j]=0;



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

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

    M[N-1][0]=a[N-1][0];

    M[N-1][1]=a[N-1][1];

    M[N-1][2]=a[N-1][2];



    for(i=N-2; i>=0; i--)

    {

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

        {

            if(j==0)

            {

                w1=1;

                w2=2;

            }

            else if(j==1)

            {

                w1=0;

                w2=2;

            }

            else

            {

                w1=0;

                w2=1;

            }

        M[i][j]=a[i][j]+max(M[i+1][w1],M[i+1][w2]);

        }

    }

    aa=max(M[0][0],M[0][1]);

    aa=max(aa,M[0][2]);

    cout<<aa<<endl;

    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 5.083954407662018x):
#include<iostream>

#include<cstdio>

using namespace std;

const int N=100010;

int a[N][4],f[N][4];

int main()

{

	int n;

	scanf("%d",&n);

	for(int i=1;i<=n;i++) scanf("%d%d%d",&a[i][1],&a[i][2],&a[i][3]);

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

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

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

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

			}

		}

	}

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

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

using namespace std;

#define INF 100005

#define maxx 1000000000000



int n;

long long a[INF][3];

long long mem[INF][3];



long long rec(int level,int pre)

{

    if(level==n) return 0;

    if(mem[level][pre]!=-1) return mem[level][pre];

    long long now=0;

    if(pre==0)

    {

        now=max(now,max(a[level][0]+rec(level+1,1),a[level][0]+rec(level+1,2)));

    }else

    if(pre==1)

    {

        now=max(now,max(a[level][1]+rec(level+1,0),a[level][1]+rec(level+1,2)));

    }else

    if(pre==2)

    {

        now=max(now,max(a[level][2]+rec(level+1,0),a[level][2]+rec(level+1,1)));

    }

    return mem[level][pre]=now;

}



int main()

{

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

    scanf("%d",&n);

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

    {

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

    }

    long long ans=max(rec(0,0),max(rec(0,1),rec(0,2)));

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

}


// Faster C++ solution for the same problem (measured speedup: 3.4105566471962225x):
#include<iostream>

#include<cstdio>

using namespace std;

const int N=100010;

int a[N][4],f[N][4];

int main()

{

	int n;

	scanf("%d",&n);

	for(int i=1;i<=n;i++) scanf("%d%d%d",&a[i][1],&a[i][2],&a[i][3]);

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

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

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

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

			}

		}

	}

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

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

#include<algorithm>

#include<cstring>

using namespace std;

const int maxn = 1e5 + 5;



int a[maxn][4];

int dp[maxn][4];//d[i][j] i day choose j thing 's max happy



int main(){

	int n;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] ,dp[i - 1][3]) + a[i][1];

		dp[i][2] = max(dp[i - 1][1] ,dp[i - 1][3]) + a[i][2];

		dp[i][3] = max(dp[i - 1][2] ,dp[i - 1][1]) + a[i][3];

	}



	int ans = 0;

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

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

}

// Faster C++ solution for the same problem (measured speedup: 1.7981845473423064x):
#include<iostream>

#include<cstdio>

using namespace std;

const int N=100010;

int a[N][4],f[N][4];

int main()

{

	int n;

	scanf("%d",&n);

	for(int i=1;i<=n;i++) scanf("%d%d%d",&a[i][1],&a[i][2],&a[i][3]);

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

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

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

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

			}

		}

	}

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

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

using namespace std;



long long w1,w2,hap=0,maxi,aa,bb,cc,N,i,j,k,a[100005][5];

long long M[100005][5];





int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cin>>N;

    for(i=0; i<100003; i++)

        for(j=0; j<4; j++)

            M[i][j]=0;



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

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

    M[N-1][0]=a[N-1][0];

    M[N-1][1]=a[N-1][1];

    M[N-1][2]=a[N-1][2];



    for(i=N-2; i>=0; i--)

    {

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

        {

            if(j==0)

            {

                w1=1;

                w2=2;

            }

            else if(j==1)

            {

                w1=0;

                w2=2;

            }

            else

            {

                w1=0;

                w2=1;

            }

        M[i][j]=a[i][j]+max(M[i+1][w1],M[i+1][w2]);

        }

    }

    aa=max(M[0][0],M[0][1]);

    aa=max(aa,M[0][2]);

    cout<<aa<<endl;

    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 5.123274654673822x):
#include<iostream>

#include<cstdio>

using namespace std;

int a[100001],b[100001],c[100001],n,dp[100001][3];

int main(){

	scanf("%d",&n);

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

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

	dp[1][0]=a[1];dp[1][1]=b[1];dp[1][2]=c[1];

	for(int i=2;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];

	}

	printf("%d",max(max(dp[n][0],dp[n][1]),dp[n][2]));

	return 0;

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

using namespace std;

#define INF 100005

#define maxx 1000000000000



int n;

long long a[INF][3];

long long mem[INF][3];



long long rec(int level,int pre)

{

    if(level==n) return 0;

    if(mem[level][pre]!=-1) return mem[level][pre];

    long long now=0;

    if(pre==0)

    {

        now=max(now,max(a[level][0]+rec(level+1,1),a[level][0]+rec(level+1,2)));

    }else

    if(pre==1)

    {

        now=max(now,max(a[level][1]+rec(level+1,0),a[level][1]+rec(level+1,2)));

    }else

    if(pre==2)

    {

        now=max(now,max(a[level][2]+rec(level+1,0),a[level][2]+rec(level+1,1)));

    }

    return mem[level][pre]=now;

}



int main()

{

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

    scanf("%d",&n);

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

    {

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

    }

    long long ans=max(rec(0,0),max(rec(0,1),rec(0,2)));

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

}


// Faster C++ solution for the same problem (measured speedup: 3.4369345253324624x):
#include<iostream>

#include<cstdio>

using namespace std;

int a[100001],b[100001],c[100001],n,dp[100001][3];

int main(){

	scanf("%d",&n);

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

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

	dp[1][0]=a[1];dp[1][1]=b[1];dp[1][2]=c[1];

	for(int i=2;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];

	}

	printf("%d",max(max(dp[n][0],dp[n][1]),dp[n][2]));

	return 0;

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

#include<algorithm>

#include<cstring>

using namespace std;

const int maxn = 1e5 + 5;



int a[maxn][4];

int dp[maxn][4];//d[i][j] i day choose j thing 's max happy



int main(){

	int n;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] ,dp[i - 1][3]) + a[i][1];

		dp[i][2] = max(dp[i - 1][1] ,dp[i - 1][3]) + a[i][2];

		dp[i][3] = max(dp[i - 1][2] ,dp[i - 1][1]) + a[i][3];

	}



	int ans = 0;

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

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

}

// Faster C++ solution for the same problem (measured speedup: 1.8120920403890084x):
#include<iostream>

#include<cstdio>

using namespace std;

int a[100001],b[100001],c[100001],n,dp[100001][3];

int main(){

	scanf("%d",&n);

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

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

	dp[1][0]=a[1];dp[1][1]=b[1];dp[1][2]=c[1];

	for(int i=2;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];

	}

	printf("%d",max(max(dp[n][0],dp[n][1]),dp[n][2]));

	return 0;

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

using namespace std;



long long w1,w2,hap=0,maxi,aa,bb,cc,N,i,j,k,a[100005][5];

long long M[100005][5];





int main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cin>>N;

    for(i=0; i<100003; i++)

        for(j=0; j<4; j++)

            M[i][j]=0;



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

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

    M[N-1][0]=a[N-1][0];

    M[N-1][1]=a[N-1][1];

    M[N-1][2]=a[N-1][2];



    for(i=N-2; i>=0; i--)

    {

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

        {

            if(j==0)

            {

                w1=1;

                w2=2;

            }

            else if(j==1)

            {

                w1=0;

                w2=2;

            }

            else

            {

                w1=0;

                w2=1;

            }

        M[i][j]=a[i][j]+max(M[i+1][w1],M[i+1][w2]);

        }

    }

    aa=max(M[0][0],M[0][1]);

    aa=max(aa,M[0][2]);

    cout<<aa<<endl;

    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 5.055527365867471x):
#include <iostream>

#include <algorithm>



const int maxn = 1e5+7;



int n;

int a[maxn][3];

int dp[maxn][3];



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

	std::cin >> n;

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

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

			std::cin >> a[i][j];

		}

	}

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

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

	}

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

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

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

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

	}

	std::cout << std::max(std::max(dp[n-1][0], dp[n-1][1]), dp[n-1][2]) << '\n';

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

using namespace std;

#define INF 100005

#define maxx 1000000000000



int n;

long long a[INF][3];

long long mem[INF][3];



long long rec(int level,int pre)

{

    if(level==n) return 0;

    if(mem[level][pre]!=-1) return mem[level][pre];

    long long now=0;

    if(pre==0)

    {

        now=max(now,max(a[level][0]+rec(level+1,1),a[level][0]+rec(level+1,2)));

    }else

    if(pre==1)

    {

        now=max(now,max(a[level][1]+rec(level+1,0),a[level][1]+rec(level+1,2)));

    }else

    if(pre==2)

    {

        now=max(now,max(a[level][2]+rec(level+1,0),a[level][2]+rec(level+1,1)));

    }

    return mem[level][pre]=now;

}



int main()

{

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

    scanf("%d",&n);

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

    {

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

    }

    long long ans=max(rec(0,0),max(rec(0,1),rec(0,2)));

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

}


// Faster C++ solution for the same problem (measured speedup: 3.391486445424468x):
#include <iostream>

#include <algorithm>



const int maxn = 1e5+7;



int n;

int a[maxn][3];

int dp[maxn][3];



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

	std::cin >> n;

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

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

			std::cin >> a[i][j];

		}

	}

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

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

	}

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

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

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

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

	}

	std::cout << std::max(std::max(dp[n-1][0], dp[n-1][1]), dp[n-1][2]) << '\n';

}