stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 8.050159653555877x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int main()

{

    int n;

    cin>>n;

    int a[n+5][3];

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

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



    int dp[n+5][3];

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

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

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

    }



    cout<< max(dp[n-1][0], max(dp[n-1][1],dp[n-1][2]))<<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: 2.2318280767467464x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int main()

{

    int n;

    cin>>n;

    int a[n+5][3];

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

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



    int dp[n+5][3];

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

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

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

    }



    cout<< max(dp[n-1][0], max(dp[n-1][1],dp[n-1][2]))<<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: 4.279138663642478x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int main()

{

    int n;

    cin>>n;

    int a[n+5][3];

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

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



    int dp[n+5][3];

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

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

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

    }



    cout<< max(dp[n-1][0], max(dp[n-1][1],dp[n-1][2]))<<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: 2.2037109766021135x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int main()

{

    int n;

    cin>>n;

    int a[n+5][3];

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

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



    int dp[n+5][3];

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

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

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

    }



    cout<< max(dp[n-1][0], max(dp[n-1][1],dp[n-1][2]))<<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: 2.207670229308755x):
#include <bits/stdc++.h>

#define ll long long

using namespace std;



int main()

{

    int n;

    cin>>n;

    int a[n+5][3];

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

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



    int dp[n+5][3];

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

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

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

    }



    cout<< max(dp[n-1][0], max(dp[n-1][1],dp[n-1][2]))<<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: 2.220224836986749x):
#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 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: 4.23999179846429x):
#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 <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: 1.3955724233637683x):
#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 N (int)2e5

int n,a[N],b[N],c[N],memo[N][10];



// a = 0, b = 1, c = 2

int dp(int curr , int last) {

    if (curr > n) return 0;

    int &sol = memo[curr][last];

    if (sol != -1) return sol;

    if (last == 0) {

        sol = max(

            dp(curr + 1 , 1) + b[curr] ,

            dp(curr + 1 , 2) + c[curr]

        );

    } else if (last == 1) {

        sol = max(

            dp(curr + 1 , 0) + a[curr] ,

            dp(curr + 1 , 2) + c[curr]

        );

    } else {

        sol = max(

            dp(curr + 1 , 0) + a[curr] ,

            dp(curr + 1 , 1) + b[curr]

        );

    }

    return sol;

}



int main() {

    scanf("%d" , &n);

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

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

    memset (memo , -1 , sizeof memo);

    printf("%d\n" , max(dp(2 , 0) + a[1] , max(dp(2 , 1) + b[1] , dp(2 , 2) + c[1])));

}


// Faster C++ solution for the same problem (measured speedup: 9.406338480510081x):
#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;



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

int dp[100007][5];



int Solve(int idx, int n, int prev){

	

	if(idx == n)return 0;

	

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

	

	int ans = -1;

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

		if(i == prev)continue;

		if(i == 1)ans = max(ans, a[idx] + Solve(idx+1, n, i));

		else if(i == 2)ans = max(ans, b[idx] + Solve(idx+1, n, i));

		else ans = max(ans, c[idx] + Solve(idx+1, n, i));		

	}

	return dp[idx][prev] = ans;

}



int main(){

	int n; cin>>n;

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

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

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

	}

	int ret = -1;

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

		ret = max(ret, Solve(0,n,i));

	}

	cout<<ret<<endl;



//	cout<<Solve(0,n,0)<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.019927127689648x):
#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 <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <math.h>

#include <queue>

#include <stack>

#include <vector>

#include <algorithm>

#include <iostream>



using namespace std;



const int MAXN = 1e5 + 5;

int n,m;

int a,b,c;

int d,e,f;

vector <int> A;

vector <int> B;

vector <int> C;

vector <int> dp[MAXN];



int main()

{

	scanf("%d",&n);

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

	{

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

		A.push_back(a);//第一天的a现在是a[0]

		B.push_back(b);

		C.push_back(c);

	}

	dp[0].push_back(A[0]);

	dp[0].push_back(B[0]);

	dp[0].push_back(C[0]);

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

	{

		d = max(dp[i - 1][1],dp[i - 1][2]);

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

		f = max(dp[i - 1][0],dp[i - 1][1]);

		dp[i].push_back(d + A[i]);

		dp[i].push_back(e + B[i]);

		dp[i].push_back(f + C[i]);

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

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

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

	}

	m = max(max(dp[n - 1][1],dp[n - 1][2]),dp[n - 1][0]);

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.536458150037043x):
#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 signed long long ll;

const int MX = 1e5 + 5;

const int INF = 1e9 + 5;

int n, A[MX][3], dp[MX][4];



int solve(int pos, int prev) {

    if(pos >= n) return 0;

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

    int ans = 0;

    if(prev == 1) 

        ans = max(ans, max(A[pos][1] + solve(pos + 1, 2), A[pos][2] + solve(pos + 1, 3)));

    else if(prev == 2) 

        ans = max(ans, max(A[pos][0] + solve(pos + 1, 1), A[pos][2] + solve(pos + 1, 3)));

    else if(prev == 3) 

        ans = max(ans, max(A[pos][1] + solve(pos + 1, 2), A[pos][0] + solve(pos + 1, 1)));

    else

        ans = max(A[pos][0] + solve(pos + 1, 1), max(A[pos][1] + solve(pos + 1, 2), A[pos][2] + solve(pos + 1, 3)));

    return dp[pos][prev] = ans;

}



int main() {

    scanf("%d", &n);



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

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

    }

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

    int ans = solve(0, 0);

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

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.6005762920006457x):
#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);

}