stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 3.0136275070902143x):
#include<bits/stdc++.h>

using namespace std;

int dp[100001][3];

int a[100001][3];

int solve(int n)

{

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

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];

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

    {

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

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

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

    }

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

    return max(mx,dp[n-1][2]);

}

int main()

{

        int n;

        cin >> n;

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

        {

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

            cin >> a[j][k];

        }

        cout << solve(n) << endl;

}
// 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: 2.9914657723156024x):
#include<bits/stdc++.h>

using namespace std;

int dp[100001][3];

int a[100001][3];

int solve(int n)

{

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

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];

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

    {

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

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

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

    }

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

    return max(mx,dp[n-1][2]);

}

int main()

{

        int n;

        cin >> n;

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

        {

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

            cin >> a[j][k];

        }

        cout << solve(n) << endl;

}
// 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: 2.614345475119423x):
#include<bits/stdc++.h>

using namespace std;

int dp[100001][3];

int a[100001][3];

int solve(int n)

{

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

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];

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

    {

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

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

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

    }

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

    return max(mx,dp[n-1][2]);

}

int main()

{

        int n;

        cin >> n;

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

        {

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

            cin >> a[j][k];

        }

        cout << solve(n) << endl;

}
// 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: 2.615192232434381x):
#include<bits/stdc++.h>

using namespace std;

int dp[100001][3];

int a[100001][3];

int solve(int n)

{

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

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];

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

    {

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

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

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

    }

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

    return max(mx,dp[n-1][2]);

}

int main()

{

        int n;

        cin >> n;

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

        {

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

            cin >> a[j][k];

        }

        cout << solve(n) << endl;

}
// 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: 8.051080005698315x):
#include<bits/stdc++.h>

using namespace std;

int dp[100001][3];

int a[100001][3];

int solve(int n)

{

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

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];

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

    {

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

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

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

    }

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

    return max(mx,dp[n-1][2]);

}

int main()

{

        int n;

        cin >> n;

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

        {

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

            cin >> a[j][k];

        }

        cout << solve(n) << endl;

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

//#define int long long

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

using namespace std;

#define endl '\n'

#define fori(s,n) for(int i=s;i<n;i++) 

#define forj(s,n) for(int j=s;j<n;j++) 

#define vii vector<int>

#define pii pair<int,int>

int n;

vector<vii> v;

int maxi = 0;

int dp[100005][3];

int solve(int x, int y) {//{0,0} intially

	

	if (x == n) {

		return 0;

	}

	int& ret = dp[x][y];

	if (ret != -1) return ret;

	if (y != 0 || x == 0)

		ret = max(ret, v[x][0] + solve(x + 1, 0));

	if (y != 1 || x == 0)

		ret = max(ret, v[x][1] + solve(x + 1, 1));

	if (y != 2 || x == 0)

		ret = max(ret, v[x][2] + solve(x + 1, 2));

	return ret;

}

signed main()

{

	fast;

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

	cin >> n;

	v = vector<vii>(n, vii(3));

	fori(0, n)

		forj(0, 3)

		cin >> v[i][j];

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

}

// Faster C++ solution for the same problem (measured speedup: 2.2320832353821505x):
#include<bits/stdc++.h>

using namespace std;

int dp[100001][3];

int a[100001][3];

int solve(int n)

{

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

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];

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

    {

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

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

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

    }

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

    return max(mx,dp[n-1][2]);

}

int main()

{

        int n;

        cin >> n;

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

        {

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

            cin >> a[j][k];

        }

        cout << solve(n) << endl;

}
// Slower C++ solution:
/**

 * @author      : Luis Miguel Baez <es.luismiguelbaez@gmail.com>

 * @created     : Thursday August 13, 2020

 */



#include <bits/stdc++.h>



using namespace std;



#define endl '\n'

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

#define rall(x) (x).rbegin(), (x).rend()

#define ms(arr, value) memset(arr, value, sizeof arr)

#define ll    int64_t



#define pii   pair<int, int>

#define piii  tuple<int, int, int>

#define pll   pair<ll, ll>

#define plll  tuple<ll, ll, ll>



#define F  first

#define S  second



const int oo  = (int)(1e9);

const ll lloo  = (ll)(1e18);



const int N = int(1e5) + 10;



class Task {

public:



    int n;

    vector<piii> A;



    ll dp[N][4];





    ll go(int index, int last) {

        if(index > n) return 0LL;

        ll &ans = dp[index][last];

        if(ans != -1) return ans;

        ll x=0, y=0;

        if(last == 1) {

            x = get<1>(A[index]) + go(index+1, 2);

            y = get<2>(A[index]) + go(index+1, 3);

        } else if(last == 2) {

            x = get<0>(A[index]) + go(index+1, 1);

            y = get<2>(A[index]) + go(index+1, 3);

        } else if(last == 3) {

            x = get<0>(A[index]) + go(index+1, 1);

            y = get<1>(A[index]) + go(index+1, 2);

        } else {

            ll z = 0;

            x = get<0>(A[index]) + go(index+1, 1);

            y = get<1>(A[index]) + go(index+1, 2);

            z = get<2>(A[index]) + go(index+1, 3);

            x = max(x, z);

        }

        ans = max(x, y);

        return ans;

    }



    ll solveOne(istream &in, ostream &out) {

        int a, b, c;

        in >> n;

        A.resize(n+1);

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

            in >> a >> b >> c;

            A[i] = {a, b, c};

        }

        memset(dp, -1, sizeof dp);

        ll ans = go(1, -1);

        return ans;  

    }

    

    void solve(istream &in, ostream &out) {

        int t = 1;

        while(t--) {

            out << solveOne(in, out) << endl;

        }

    }

};



auto main() -> int {

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    

    Task solver;

    istream &in(cin);

    ostream &out(cout);

    solver.solve(in, out);

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.279627885546932x):
#include<bits/stdc++.h>

using namespace std;

int dp[100001][3];

int a[100001][3];

int solve(int n)

{

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

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];

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

    {

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

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

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

    }

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

    return max(mx,dp[n-1][2]);

}

int main()

{

        int n;

        cin >> n;

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

        {

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

            cin >> a[j][k];

        }

        cout << solve(n) << endl;

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

using namespace std;

int k;

const int N=1e5+10;

int f[N][3];

int memo[N][3];

 

int dp(int pos, int now){

    if(pos==k) return 0;

    if(memo[pos][now]!=-1) return memo[pos][now];

    else{

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

            if(now==i) continue;

            else{

            memo[pos][now]=max(memo[pos][now],dp(pos+1,i)+f[pos][i]);

            }

        }

    }

    return memo[pos][now];

    

}

 

int main(){

    cin>>k;

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

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

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

            cin>>f[i][j];

        }

    }

    int ans=0;

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

        ans=max(ans,dp(0,i));

    }

    cout<<ans<<'\n';

}

// Faster C++ solution for the same problem (measured speedup: 2.2039629206884315x):
#include<bits/stdc++.h>

using namespace std;

int dp[100001][3];

int a[100001][3];

int solve(int n)

{

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

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];

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

    {

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

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

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

    }

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

    return max(mx,dp[n-1][2]);

}

int main()

{

        int n;

        cin >> n;

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

        {

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

            cin >> a[j][k];

        }

        cout << solve(n) << endl;

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

using namespace std;

int ara[100005][3];

int dp[100005][3];



int func(int pos,int des,int idx){

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

    int res;

    if(pos>des) res=0;

    else{

        int t1,t2;

        if(idx==0){

            t1=ara[pos][0]+func(pos+1,des,1);

            t2=ara[pos][0]+func(pos+1,des,2);

        }

        else if(idx==1){

             t1=ara[pos][1]+func(pos+1,des,0);

            t2=ara[pos][1]+func(pos+1,des,2);

        }

        else{

             t1=ara[pos][2]+func(pos+1,des,1);

            t2=ara[pos][2]+func(pos+1,des,0);

        }

        res=max(t1,t2);

    }

    dp[pos][idx]=res;

    return res;

}

int main(){

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

    int n,x,y,z;

    cin >> n;

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

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

    }

    x=func(1,n,0);

    y=func(1,n,1);

    z=func(1,n,2);

    int ans;

    

    ans=max(x,y);

    ans=max(ans,z);

    

    cout << ans << endl;

    

}

// Faster C++ solution for the same problem (measured speedup: 2.2079226260453146x):
#include<bits/stdc++.h>

using namespace std;

int dp[100001][3];

int a[100001][3];

int solve(int n)

{

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

    dp[0][1]=a[0][1];

    dp[0][2]=a[0][2];

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

    {

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

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

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

    }

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

    return max(mx,dp[n-1][2]);

}

int main()

{

        int n;

        cin >> n;

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

        {

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

            cin >> a[j][k];

        }

        cout << solve(n) << endl;

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

using namespace std;

typedef long long int ll;

const int MAX=1e5+10;

int N,A[MAX],B[MAX],C[MAX],dp[MAX][3];

int solve(int idx,int pr)

{

    if(idx>N)return 0;

    if(dp[idx][pr]!=-1)return dp[idx][pr];

    int res=0;

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

        if(i==pr)continue;

        int val=0;

        if(i==1)val=A[idx];

        if(i==2)val=B[idx];

        if(i==3)val=C[idx];

        res = max(res,solve(idx+1,i)+val);

    }

    return dp[idx][pr]=res;

}

int  main ()

{

    memset(dp,-1,sizeof dp);

    cin >> N;

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

        cin >> A[i] >> B[i] >> C[i];

    }

    cout<<solve(1,0)<<endl;

    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 2.2379344058340735x):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<vector>

#include<cstring>

#include<set>

#include<queue>

#include<map>

#include<math.h>

#include<climits>



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

#define ll long long

#define dd double

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

#define scd(n)        scanf("%lf",&n)

#define scc(n)        scanf("%c",&n)

#define scs(n)        scanf("%s",&n)

#define ptl(n)        printf("%lld",n)

#define ptd(n)        printf("%f",n)

#define ptc(n)        printf("%c",n)

#define pts(n)        printf("%s",n)

#define nline         printf("\n")

#define pb            push_back

#define FOR(i,a,b)    for(ll i=a;i<=b;i++)

#define FORr(i,a,b)   for(ll i=a;i>=b;i--)

#define pi            2*acos(0.0)

#define pii           pair<ll,ll>

#define uu            first

#define vv            second



using namespace std;

ll n,a[100005],b[100005],c[100005],dp[100010][3],op0,op1,op2;

int main()

{

    scl(n);

    FOR(i,1,n)

    {

        scl(a[i]);

        scl(b[i]);

        scl(c[i]);

    }

    dp[n][0]=a[n];

    dp[n][1]=b[n];

    dp[n][2]=c[n];

    FORr(i,n-1,1)

    {

        if(i+1<=n)

        {

            op0=dp[i+1][0];

            op1=dp[i+1][1];

            op2=dp[i+1][2];

        }

        else

        {

            op0=LONG_MIN;

            op1=LONG_MIN;

            op2=LONG_MIN;

        }

        dp[i][0]=max(op1+a[i],op2+a[i]);

        dp[i][1]=max(op0+b[i],op2+b[i]);

        dp[i][2]=max(op0+c[i],op1+c[i]);

       // cout<<dp[i][0]<<endl;

    }

    ll ans=max(dp[1][0],dp[1][1]);

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

    ptl(ans);

    return 0;

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

using namespace std;

#define print printf("==================\n")

#define ll long long

#define pi acos(-1.0)

#define eps 1e-16

#define max3(a,b,c) max(a,max(b,c))

const ll INF = 1 << 30;

typedef pair<ll, ll  > payar;

typedef struct

{

    ll  x, y;

} point;

priority_queue<payar, vector<payar>,  greater<payar> > pq; ///accending

vector< pair<ll,payar  > > vpp;

vector < payar >  vp;



ll n ;

int ara[100005][4];

ll dp[100005][4];

ll cal(int day, int itm)

{

    if(day>n)

    {

        return 0;

    }



    if(dp[day][itm]!=-1)

        return dp[day][itm];

    ll maxi = -1;

    if(day==1)

    {

   return   maxi = dp[day][itm]=max3( cal(day+1,1)+ara[day][1], cal(day+1,2)+ara[day][2], cal(day+1,3)+ara[day][3] );



    }

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

    {

            if(i==itm)

                continue;

            maxi=max( cal(day+1,i)+ara[day][i],maxi);





    }

    return dp[day][itm]=maxi;

}



int main()

{



    cin>>n;

    memset(dp,-1,sizeof dp);

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

    {

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

        {

            cin>>ara[i][j];

        }

    }

    cout<<cal(1,1);







    return  0 ;

}


// Faster C++ solution for the same problem (measured speedup: 4.273811988841452x):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<vector>

#include<cstring>

#include<set>

#include<queue>

#include<map>

#include<math.h>

#include<climits>



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

#define ll long long

#define dd double

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

#define scd(n)        scanf("%lf",&n)

#define scc(n)        scanf("%c",&n)

#define scs(n)        scanf("%s",&n)

#define ptl(n)        printf("%lld",n)

#define ptd(n)        printf("%f",n)

#define ptc(n)        printf("%c",n)

#define pts(n)        printf("%s",n)

#define nline         printf("\n")

#define pb            push_back

#define FOR(i,a,b)    for(ll i=a;i<=b;i++)

#define FORr(i,a,b)   for(ll i=a;i>=b;i--)

#define pi            2*acos(0.0)

#define pii           pair<ll,ll>

#define uu            first

#define vv            second



using namespace std;

ll n,a[100005],b[100005],c[100005],dp[100010][3],op0,op1,op2;

int main()

{

    scl(n);

    FOR(i,1,n)

    {

        scl(a[i]);

        scl(b[i]);

        scl(c[i]);

    }

    dp[n][0]=a[n];

    dp[n][1]=b[n];

    dp[n][2]=c[n];

    FORr(i,n-1,1)

    {

        if(i+1<=n)

        {

            op0=dp[i+1][0];

            op1=dp[i+1][1];

            op2=dp[i+1][2];

        }

        else

        {

            op0=LONG_MIN;

            op1=LONG_MIN;

            op2=LONG_MIN;

        }

        dp[i][0]=max(op1+a[i],op2+a[i]);

        dp[i][1]=max(op0+b[i],op2+b[i]);

        dp[i][2]=max(op0+c[i],op1+c[i]);

       // cout<<dp[i][0]<<endl;

    }

    ll ans=max(dp[1][0],dp[1][1]);

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

    ptl(ans);

    return 0;

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

#include <cstdint>

#include <cinttypes>

#include <cstring>

#include <algorithm>



using i32 = std::int_fast32_t;



const int MaxN = 100001;



int A[MaxN][3], dp[MaxN][3];



signed main(signed argc, char **argv) {

	i32 n;

	std::scanf("%" SCNdFAST32, &n);

	

	for (i32 i = 0; i < n; ++ i) {

		std::scanf("%" SCNdFAST32 "%" SCNdFAST32 "%" SCNdFAST32, A[i], A[i] + 1, A[i] + 2);

	}

	

	memset(dp, 0, sizeof dp);

	

	dp[0][0] = A[0][0];

	dp[0][1] = A[0][1];

	dp[0][2] = A[0][2];

	

	for (i32 i = 1; i < n; ++ i) {

		for (i32 j = 0; j < 3; ++ j) {

			for (i32 a = 0; a < 3; ++ a) {

				if (a == j) {

					continue;

				}

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

			}

		}

	}

	

	std::printf("%" PRIdFAST32, std::max(dp[n - 1][0], std::max(dp[n - 1][1], dp[n - 1][2])));

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4067041725007299x):
#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<vector>

#include<cstring>

#include<set>

#include<queue>

#include<map>

#include<math.h>

#include<climits>



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

#define ll long long

#define dd double

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

#define scd(n)        scanf("%lf",&n)

#define scc(n)        scanf("%c",&n)

#define scs(n)        scanf("%s",&n)

#define ptl(n)        printf("%lld",n)

#define ptd(n)        printf("%f",n)

#define ptc(n)        printf("%c",n)

#define pts(n)        printf("%s",n)

#define nline         printf("\n")

#define pb            push_back

#define FOR(i,a,b)    for(ll i=a;i<=b;i++)

#define FORr(i,a,b)   for(ll i=a;i>=b;i--)

#define pi            2*acos(0.0)

#define pii           pair<ll,ll>

#define uu            first

#define vv            second



using namespace std;

ll n,a[100005],b[100005],c[100005],dp[100010][3],op0,op1,op2;

int main()

{

    scl(n);

    FOR(i,1,n)

    {

        scl(a[i]);

        scl(b[i]);

        scl(c[i]);

    }

    dp[n][0]=a[n];

    dp[n][1]=b[n];

    dp[n][2]=c[n];

    FORr(i,n-1,1)

    {

        if(i+1<=n)

        {

            op0=dp[i+1][0];

            op1=dp[i+1][1];

            op2=dp[i+1][2];

        }

        else

        {

            op0=LONG_MIN;

            op1=LONG_MIN;

            op2=LONG_MIN;

        }

        dp[i][0]=max(op1+a[i],op2+a[i]);

        dp[i][1]=max(op0+b[i],op2+b[i]);

        dp[i][2]=max(op0+c[i],op1+c[i]);

       // cout<<dp[i][0]<<endl;

    }

    ll ans=max(dp[1][0],dp[1][1]);

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

    ptl(ans);

    return 0;

}