stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#define _CRT_SECURE_NO_WARNINGS

#include <bits/stdc++.h>

#include <unordered_map>

#define stop(x) {cout<<x<<endl;exit(0);}

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

#define szz(s) int(s.size())

#define endl '\n'

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

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

typedef long long ll;

using namespace std;



void file() {

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

#ifndef ONLINE_JUDGE

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

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

#endif

}





ll n;

ll mem[100004][4];

vector < vector < int > > v;



ll solve(int idx , int last) {

    if (idx == n)return 0;



    ll& ret = mem[idx][last];

    if (~ret)return ret;

    ret = 0;

    if (last != 0)ret = max(ret, solve(idx + 1, 0) + v[idx][0]);

    if (last != 1)ret = max(ret, solve(idx + 1, 1) + v[idx][1]);

    if (last != 2)ret = max(ret, solve(idx + 1, 2) + v[idx][2]);

    return ret;

}



int main() {

    //file();



    memset(mem, -1, sizeof mem);

    cin >> n;

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

    for (auto& e : v)

        for (auto& f : e)

            cin >> f;



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



}

// Faster C++ solution for the same problem (measured speedup: 4.243929502543706x):
#include<bits/stdc++.h>

using namespace std;

const int maxv=1000010;

typedef long long ll;

ll n;

ll a[maxv];

ll b[maxv];

ll c[maxv];

ll ans[maxv][3];



int main()

{



    cin>>n;





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

    {

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

    }



    ans[0][0]=a[0];

    ans[0][1]=b[0];

    ans[0][2]=c[0];

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

    {

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

        ans[i][1]+=max(ans[i-1][0],ans[i-1][2])+b[i];

        ans[i][2]+=max(ans[i-1][0],ans[i-1][1])+c[i];





    }

    ll result=max(ans[n-1][2],max(ans[n-1][0],ans[n-1][1]));

    cout<<result;



    return 0;



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

author :- Mahmoud Tarek.

*******/



#include<bits/stdc++.h>



using namespace std ;



#define rg register

#define dbg(v) cerr << #v << " " << v ;

typedef long long ll ;



int n;

struct doW{

int w[5] ;

};



vector<doW> a ;

int mem[100002][6] ;



int solve(int i, int ch){

if(i == n) return 0 ;

int & ret = mem[i][ch] ;

if(~ret) return ret ;

ret = 0 ;

for(int j = 2 ; j <= 4 ; j++){

    if(ch != j){

        ret = max(ret, a[i].w[j] + solve(i + 1, j)) ;

    }

}

return ret ;

}



int main() {

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

    cin >> n ;

    a.resize(n) ;

    memset(mem, -1, sizeof mem) ;

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

        for(int j = 2 ; j <= 4 ; j++) cin >> a[i].w[j] ;

    }

    cout << solve(0, 1) ;



}



/*

  Empty set of troubles.

*/


// Faster C++ solution for the same problem (measured speedup: 3.4093744160376747x):
#include<bits/stdc++.h>

using namespace std;

const int maxv=1000010;

typedef long long ll;

ll n;

ll a[maxv];

ll b[maxv];

ll c[maxv];

ll ans[maxv][3];



int main()

{



    cin>>n;





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

    {

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

    }



    ans[0][0]=a[0];

    ans[0][1]=b[0];

    ans[0][2]=c[0];

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

    {

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

        ans[i][1]+=max(ans[i-1][0],ans[i-1][2])+b[i];

        ans[i][2]+=max(ans[i-1][0],ans[i-1][1])+c[i];





    }

    ll result=max(ans[n-1][2],max(ans[n-1][0],ans[n-1][1]));

    cout<<result;



    return 0;



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

using namespace std;



const int N=1e5+30;

typedef long long ll;

int n;

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

ll dp[N][5];

ll solve(int i,int last){

    if(i==n)return 0;

    int path1=-1 ,path2=-1,path3=-1;

    ll &ret=dp[i][last];

    if(~ret)return ret;

    if(last!=1){

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

    }

    if(last!=2){

        path2=b[i]+solve(i+1,2);

    }

    if(last!=3){

        path3=c[i]+solve(i+1,3);

    }

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



}

int main()

{

    memset(dp,-1,sizeof dp);

    cin>>n;

    for(int i=0;i<n;i++)cin>>a[i]>>b[i]>>c[i];

    cout<<solve(0,4);



}




// Faster C++ solution for the same problem (measured speedup: 5.0874548919241676x):
#include<bits/stdc++.h>

using namespace std;

const int maxv=1000010;

typedef long long ll;

ll n;

ll a[maxv];

ll b[maxv];

ll c[maxv];

ll ans[maxv][3];



int main()

{



    cin>>n;





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

    {

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

    }



    ans[0][0]=a[0];

    ans[0][1]=b[0];

    ans[0][2]=c[0];

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

    {

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

        ans[i][1]+=max(ans[i-1][0],ans[i-1][2])+b[i];

        ans[i][2]+=max(ans[i-1][0],ans[i-1][1])+c[i];





    }

    ll result=max(ans[n-1][2],max(ans[n-1][0],ans[n-1][1]));

    cout<<result;



    return 0;



}
// Slower C++ solution:
/* MaTb3aa */

#include<bits/stdc++.h>

#define ll long long

using namespace std;

int n;

const int N = 100005;

vector<int>v;

ll dp[100005][4];

int A[N], B[N], C[N];

ll solve(int i,int lst) {

	if (i == n)return 0;

	ll& ret = dp[i][lst];

	if (~ret)return ret;

	if (lst != 1)ret = max(ret, solve(i + 1, 1) + A[i]);

	if (lst != 2)ret = max(ret, solve(i + 1, 2) + B[i]);

	if (lst != 3)ret = max(ret, solve(i + 1, 3) + C[i]);

	return ret;

}

int main() {

	cin >> n;

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

	memset(dp, -1, sizeof dp);

	cout << solve(0, 0);

	

		 

}

// Faster C++ solution for the same problem (measured speedup: 4.2208946955097115x):
#include<bits/stdc++.h>

using namespace std;

const int maxv=1000010;

typedef long long ll;

ll n;

ll a[maxv];

ll b[maxv];

ll c[maxv];

ll ans[maxv][3];



int main()

{



    cin>>n;





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

    {

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

    }



    ans[0][0]=a[0];

    ans[0][1]=b[0];

    ans[0][2]=c[0];

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

    {

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

        ans[i][1]+=max(ans[i-1][0],ans[i-1][2])+b[i];

        ans[i][2]+=max(ans[i-1][0],ans[i-1][1])+c[i];





    }

    ll result=max(ans[n-1][2],max(ans[n-1][0],ans[n-1][1]));

    cout<<result;



    return 0;



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

using namespace std;

int n;

int arr[100005][3];



int dp[100005][4];



int solve(int i, int prev){

	if(i == n) return 0;

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



	dp[i][prev] = -1e7;

	int a = 0, b = 0, c = 0;

	if(prev != 0)

		a = solve(i+1, 0) + arr[i][0];

	if(prev != 1)

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

	if(prev != 2)

		c = solve(i+1, 2) + arr[i][2];



	dp[i][prev] = max(a, max(b, c));

	return dp[i][prev];

}



int main(){

	cin>>n;

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

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



	memset(dp, -1, sizeof dp);

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.609069998962029x):
#include<bits/stdc++.h>

using namespace std;

const int maxv=1000010;

typedef long long ll;

ll n;

ll a[maxv];

ll b[maxv];

ll c[maxv];

ll ans[maxv][3];



int main()

{



    cin>>n;





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

    {

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

    }



    ans[0][0]=a[0];

    ans[0][1]=b[0];

    ans[0][2]=c[0];

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

    {

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

        ans[i][1]+=max(ans[i-1][0],ans[i-1][2])+b[i];

        ans[i][2]+=max(ans[i-1][0],ans[i-1][1])+c[i];





    }

    ll result=max(ans[n-1][2],max(ans[n-1][0],ans[n-1][1]));

    cout<<result;



    return 0;



}
// Slower C++ solution:
#define _CRT_SECURE_NO_WARNINGS

#include<iostream>

#include<functional>

#include <bitset>

#include<algorithm>

#include<fstream>

#include<string>

#include <cstring>

#include<stdio.h>

#include<iomanip>

#include<sstream>

#include<cmath>

#include<vector>

#include<assert.h>

#include<queue>

#include<deque>

#include<stack>

#include<set>

#include<map>

// never lose hope ..

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

#define eps 1e-8

using namespace std;

double const  PI = 3.141592653;

const int OO = 1000000000 + 9;



void fast() {

	ios_base::sync_with_stdio(0);

	cin.tie(NULL), cout.tie(NULL);

}

struct s {

	int x, y, z;

};



int n;

int v[(int)1e5][4];

int dp[(int)1e5][4];



int solve(int i = 0, int pre = 4) {

	if (i >= n)return 0;

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

	int ch1, ch2, ch3;

	ch1 = ch2 = ch3 = 0;

	if (pre != 1) {

		ch1 += solve(i + 1, 1) + v[i][1];

	}

	if (pre != 2) {

		ch2 += solve(i + 1, 2) + v[i][2];

	}

	if (pre != 3) {

		ch3 += solve(i + 1, 3) + v[i][3];

	}

	return dp[i][pre] = max({ ch1,ch2,ch3 });

}





int main()

{

	fast();



	cin >> n;

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

		cin >> v[i][1] >> v[i][2] >> v[i][3];

	}

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

	cout << solve();



	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.6208326052996025x):
#include<bits/stdc++.h>

using namespace std;

const int maxv=1000010;

typedef long long ll;

ll n;

ll a[maxv];

ll b[maxv];

ll c[maxv];

ll ans[maxv][3];



int main()

{



    cin>>n;





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

    {

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

    }



    ans[0][0]=a[0];

    ans[0][1]=b[0];

    ans[0][2]=c[0];

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

    {

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

        ans[i][1]+=max(ans[i-1][0],ans[i-1][2])+b[i];

        ans[i][2]+=max(ans[i-1][0],ans[i-1][1])+c[i];





    }

    ll result=max(ans[n-1][2],max(ans[n-1][0],ans[n-1][1]));

    cout<<result;



    return 0;



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

using namespace std;

const int MAX_n=1e5+10;

int n;

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

int dp[MAX_n][3];

int main()

{

    cin>>n;

    for(int i=1;i<=n;i++)    cin>>a[i]>>b[i]>>c[i];

    dp[1][0]=max(b[1],c[1]);

    dp[1][1]=max(a[1],c[1]);

    dp[1][2]=max(a[1],b[1]);

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

    {

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

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

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

    }

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

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

    cout<<ans;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.213136612136461x):
#include <cstdio>

#include <algorithm>

using std::max;



int main(void)

{

    int dp[2][3],a[3],i,N;

    i=0; 

    dp[0][0]=0;

    dp[0][1]=0;

    dp[0][2]=0;

 

    scanf("%d",&N);

    while(N--)

    {

        scanf("%d%d%d",a,a+1,a+2);

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

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

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

        i=1-i;

    }

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

    return 0;

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

#include <vector>

#include <algorithm>

#include <numeric>

#include <cmath>

#include <set>

#include <tuple>

#include <functional>

#include <map>

#include <unordered_map>

#include <queue>

#include <tuple>



using namespace std;



#define endl '\n'



using ull = unsigned long int;

using ll = long long;



int main() {

  ios_base::sync_with_stdio(false);

  cin.tie(nullptr);

  const int precision = 13;

  cout.precision(precision);

  cout.setf(ios_base::fixed);



  int n; cin >> n;

  vector<vector<int>> dp(2, vector<int>(3, 0));



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

    int a, b, c; cin >> a >> b >> c;

    int cur = i%2;

    int prev = (i-1)%2;

    dp[cur][0] = a + max(dp[prev][1], dp[prev][2]);

    dp[cur][1] = b + max(dp[prev][0], dp[prev][2]);

    dp[cur][2] = c + max(dp[prev][0], dp[prev][1]);

  }



  cout << *max_element(dp[n%2].begin(), dp[n%2].end()) << endl;





  return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.2910203266304245x):
#include <cstdio>

#include <algorithm>

using std::max;



int main(void)

{

    int dp[2][3],a[3],i,N;

    i=0; 

    dp[0][0]=0;

    dp[0][1]=0;

    dp[0][2]=0;

 

    scanf("%d",&N);

    while(N--)

    {

        scanf("%d%d%d",a,a+1,a+2);

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

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

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

        i=1-i;

    }

    printf("%d\n",max(max(dp[i][0],dp[i][1]),dp[i][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: 26.376949643100602x):
#include <cstdio>

#include <algorithm>

using std::max;



int main(void)

{

    int dp[2][3],a[3],i,N;

    i=0; 

    dp[0][0]=0;

    dp[0][1]=0;

    dp[0][2]=0;

 

    scanf("%d",&N);

    while(N--)

    {

        scanf("%d%d%d",a,a+1,a+2);

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

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

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

        i=1-i;

    }

    printf("%d\n",max(max(dp[i][0],dp[i][1]),dp[i][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: 17.69490316483919x):
#include <cstdio>

#include <algorithm>

using std::max;



int main(void)

{

    int dp[2][3],a[3],i,N;

    i=0; 

    dp[0][0]=0;

    dp[0][1]=0;

    dp[0][2]=0;

 

    scanf("%d",&N);

    while(N--)

    {

        scanf("%d%d%d",a,a+1,a+2);

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

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

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

        i=1-i;

    }

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

    return 0;

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

using namespace std;



int main(){

    int n, i, j, k;

    scanf("%d",&n);

    int a[n+5], b[n+5], c[n+5];

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

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

    }

    int dp[n+5][3];

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

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

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

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

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

    }

    int ans  = 0;

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

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

    printf("%d", ans);



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.190603856375076x):
#include <cstdio>

#include <algorithm>

using std::max;



int main(void)

{

    int dp[2][3],a[3],i,N;

    i=0; 

    dp[0][0]=0;

    dp[0][1]=0;

    dp[0][2]=0;

 

    scanf("%d",&N);

    while(N--)

    {

        scanf("%d%d%d",a,a+1,a+2);

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

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

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

        i=1-i;

    }

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

    return 0;

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

using namespace std;



int main()

{

    int i,n;

    cin>>n;

    int a[n+2], b[n+2], c[n+2], x=0, y=0, z=0, p, q, r;

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

    {

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

        if(i)

        {

            p=x,q=y,r=z;

            x=a[i]+max(q,r);

            y=b[i]+max(p,r);

            z=c[i]+max(q,p);

        }

        else

        {

            x=a[i], y=b[i], z=c[i];

        }

    }

    cout<<max(x, max(y, z))<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.221232181785649x):
#include <cstdio>

#include <algorithm>

using std::max;



int main(void)

{

    int dp[2][3],a[3],i,N;

    i=0; 

    dp[0][0]=0;

    dp[0][1]=0;

    dp[0][2]=0;

 

    scanf("%d",&N);

    while(N--)

    {

        scanf("%d%d%d",a,a+1,a+2);

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

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

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

        i=1-i;

    }

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

    return 0;

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

using namespace std;

#define ll long long

#define MX 100005



const ll INF = 1e12;



int A[MX][4];

ll dp[MX][4];



ll recur(int i, int j){

    if (i == 1) return dp[i][j] = A[i][j];

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



    ll mx = 0;

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

        if (x != j) mx = max(mx, recur(i-1, x));

    }

    return dp[i][j] = (mx + A[i][j]);

}



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]);

            dp[i][j] = -1;

        }

    }



    ll ans = 0;

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

        ans = max(ans, recur(n, j));

    }



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



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.154165453004849x):
#include <cstdio>

#include <algorithm>

using std::max;



int main(void)

{

    int dp[2][3],a[3],i,N;

    i=0; 

    dp[0][0]=0;

    dp[0][1]=0;

    dp[0][2]=0;

 

    scanf("%d",&N);

    while(N--)

    {

        scanf("%d%d%d",a,a+1,a+2);

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

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

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

        i=1-i;

    }

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

    return 0;

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

using namespace std;



#define pb push_back

#define mp make_pair

#define fi first

#define se second

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

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

#define MAX 1000000

//3

typedef long long ll;

typedef pair<int,int> pii;

priority_queue< int > pq;

priority_queue< int , vector<int> , greater<int> > rpq;

set<int>s;

set <int > :: iterator itr; 

vector<int> vec[1001];

int r,n,m,inp,tc;

int arr[1000007]= {};

map<int , int> mymap;





int main()

{

	int n;

	int arr[3];

	int dp[3];

	cin>>n;

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

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

		if(i == 0 ) {

			dp[0] = arr[0];

			dp[1] = arr[1];

			dp[2] = arr[2];

		}

		else{

			int a = dp[0];

			int b = dp[1];

			int c = dp[2];

			dp[0] = max(arr[0] + b , arr[0] + c);

			dp[1] = max(arr[1] + c , arr[1] + a);

			dp[2] = max(arr[2] + a , arr[2] + b);

		}

		//cout<<dp[0]<<" "<<dp[1]<<" "<<dp[2]<<endl;

	}

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

    return 0;

}



//65 900 17 13

// Faster C++ solution for the same problem (measured speedup: 5.348343964151414x):
#include <cstdio>

#include <algorithm>

using std::max;



int main(void)

{

    int dp[2][3],a[3],i,N;

    i=0; 

    dp[0][0]=0;

    dp[0][1]=0;

    dp[0][2]=0;

 

    scanf("%d",&N);

    while(N--)

    {

        scanf("%d%d%d",a,a+1,a+2);

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

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

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

        i=1-i;

    }

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

    return 0;

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

#define ll long long

using namespace std;



ll arr[100007][3];

ll dp[100007][3];

bool vis[100007][3];

int n;



ll rec( int i, int prev  )

{

    if(i>=n)

        return 0;

    if(vis[i][prev])

        return dp[i][prev];



    ll ans= 0;

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

    {

        if(prev==j)

            continue;

        ans= max( ans, arr[i][j]+rec(i+1,j) );

    }



    vis[i][prev]= 1;



    return dp[i][prev]= ans;

}



int main()

{

    cin>>n;



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

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

            cin>>arr[i][j];



    cout<< rec( 0, -1 ) <<endl;

}


// Faster C++ solution for the same problem (measured speedup: 5.233170667359828x):
#include <cstdio>

#include <algorithm>

using std::max;



int main(void)

{

    int dp[2][3],a[3],i,N;

    i=0; 

    dp[0][0]=0;

    dp[0][1]=0;

    dp[0][2]=0;

 

    scanf("%d",&N);

    while(N--)

    {

        scanf("%d%d%d",a,a+1,a+2);

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

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

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

        i=1-i;

    }

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

    return 0;

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

#include <algorithm>



using namespace std;



int main() {

	int n;

	cin >> n;

	int activities[n][3];

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

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

			cin >> activities[i][j];

		}

	}



	int escolhaA = 0, escolhaB = 0, escolhaC = 0;

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

		int auxA = escolhaA, auxB = escolhaB, auxC = escolhaC;

		escolhaA = max(auxB,auxC) + activities[i][0];

		escolhaB = max(auxA,auxC) + activities[i][1];

		escolhaC = max(auxA,auxB) + activities[i][2];

	}



	cout << max(escolhaA,max(escolhaB,escolhaC)) << endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.221347371027732x):
#include <cstdio>

#include <algorithm>

using std::max;



int main(void)

{

    int dp[2][3],a[3],i,N;

    i=0; 

    dp[0][0]=0;

    dp[0][1]=0;

    dp[0][2]=0;

 

    scanf("%d",&N);

    while(N--)

    {

        scanf("%d%d%d",a,a+1,a+2);

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

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

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

        i=1-i;

    }

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

    return 0;

}