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

#define MAXN 100010

using namespace std;



int n, ans, happy[MAXN][3], dp[MAXN][3];



void init(){

    for(int i=1; i<=3; i++) dp[1][i]=happy[1][i];

}



int main(){



    ios_base::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    cin>>n;



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

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

            cin>>happy[i][j];

        }

    }



    init();



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

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

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

                if(k==j) continue;

                dp[i][j]=max(dp[i][j], dp[i-1][k]+happy[i][j]);

            }

        }

    }



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



    cout<<ans;



return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.271917776557949x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

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



    return 0;

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



using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;





int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i){

	if(i == n) return 0;

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

	int& ans = dp[a][i];

	

	if(a==0){

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1){

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2){

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

}

int main(){

	cin>>n;

	for(int i=0; i<n; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

	memset(dp, -1, sizeof dp);

	cout<<max(go(0,0),max(go(1,0),go(2,0)))<<"\n";

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 11.500056436978234x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

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



    return 0;

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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9; 



int m;

const int mx = 1e5+5;

int ac[mx][3];

int dp[3][mx];

int go(int a, int i){

	if(i == m) return 0;

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

	int& ans = dp[a][i];

	

	if(a==0){

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1){

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2){

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

}

int main(){

	cin>>m;

	for(int i=0; i<m; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

	memset(dp, -1, sizeof dp);

	cout<<max(go(0,0),max(go(1,0),go(2,0)))<<"\n";

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 11.500172561382477x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

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



    return 0;

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

#define maxn (int)(2e5)

#define INF 1<<30

#define i64 long long

#define pii pair<i64,i64>

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

#define LB(x,v) lower_bound(all(x), v) - x.begin()

#define fread() freopen("in.txt", "r", stdin);

using namespace std;



int mat[maxn+2][4], ans[maxn+2][4] , n;

const int k = 3;



void init(int n){

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

        ans[i][0] = ans[i][1] = ans[i][2] = -1;

}



int solve(int day, int option){

    if(day >= n) return 0;



    if(ans[day][option] != -1) return ans[day][option];



    int res = -1;

    if(option == 0){

        res = mat[day][option] + max(solve(day+1, 1), solve(day+1, 2));

    }

    else if(option == 1){

        res = mat[day][option] + max(solve(day+1, 0), solve(day+1, 2));

    }

    else {

        res = mat[day][option] + max(solve(day+1, 0), solve(day+1, 1));

    }

    return ans[day][option] = res;

}



int main() {

    while(scanf("%d",&n)==1){

        init(n);

        for(int i=0;i<n;i++) scanf("%d %d %d",&mat[i][0], &mat[i][1], &mat[i][2]);

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

    }    

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.1498385232197466x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

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



    return 0;

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

using namespace std;

 

#define ll long long int



int main() {

    ll N;

    cin >> N;

    ll dp[N][3] = {};

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

        ll a, b, c;

        cin >> a >> b >> c;

        if(i == 0) {

            dp[i][0] = a;

            dp[i][1] = b;

            dp[i][2] = c;

        }

        else {

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

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

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

        }

    }

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

}

// Faster C++ solution for the same problem (measured speedup: 5.201968125994029x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

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



    return 0;

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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;



int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i)

{

	if(i==n) return 0;

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

	int& ans = dp[a][i];

	if(a==0)

	{

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1)

	{

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2)

	{

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

	

}

int main()

{

    cin>>n;

    for(int i=0; i<n; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

    memset(dp, -1, sizeof dp);

    cout<<max(go(0,0), max(go(1,0),go(2,0)))<<"\n";

}

	

// Faster C++ solution for the same problem (measured speedup: 11.500056436978234x):
#include <cstdio>

#include <algorithm>

using std::max;



#define MAXN 100003

int d[MAXN][3],a[MAXN][3];

int main(void)

{

    int i,N;

    scanf("%d",&N);

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

    {

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

    }

    d[0][0]=a[0][0];

    d[0][1]=a[0][1];

    d[0][2]=a[0][2];

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

    {

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

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

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

    }

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



    return 0;

}
// 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: 2.2097139438697333x):
#include<bits/stdc++.h>

using namespace std;



long long dp[100];



int main()

{

    long long n,a,b,c,d,e,f,i;

    cin>>n;

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

    {

        cin>>a>>b>>c;

        if(i==0)

        {

            dp[0]=a;

            dp[1]=b;

            dp[2]=c;

            continue;

        }

        d=dp[0];

        e=dp[1];

        f=dp[2];

        dp[0]=a+max(e,f);

        dp[1]=b+max(d,f);

        dp[2]=c+max(d,e);

    }

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

    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: 4.21991901129956x):
#include<bits/stdc++.h>

using namespace std;



long long dp[100];



int main()

{

    long long n,a,b,c,d,e,f,i;

    cin>>n;

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

    {

        cin>>a>>b>>c;

        if(i==0)

        {

            dp[0]=a;

            dp[1]=b;

            dp[2]=c;

            continue;

        }

        d=dp[0];

        e=dp[1];

        f=dp[2];

        dp[0]=a+max(e,f);

        dp[1]=b+max(d,f);

        dp[2]=c+max(d,e);

    }

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

    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: 1.3889655643039716x):
#include<bits/stdc++.h>

using namespace std;



long long dp[100];



int main()

{

    long long n,a,b,c,d,e,f,i;

    cin>>n;

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

    {

        cin>>a>>b>>c;

        if(i==0)

        {

            dp[0]=a;

            dp[1]=b;

            dp[2]=c;

            continue;

        }

        d=dp[0];

        e=dp[1];

        f=dp[2];

        dp[0]=a+max(e,f);

        dp[1]=b+max(d,f);

        dp[2]=c+max(d,e);

    }

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

    return 0;

}
// 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: 9.361807396655843x):
#include<bits/stdc++.h>

using namespace std;



long long dp[100];



int main()

{

    long long n,a,b,c,d,e,f,i;

    cin>>n;

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

    {

        cin>>a>>b>>c;

        if(i==0)

        {

            dp[0]=a;

            dp[1]=b;

            dp[2]=c;

            continue;

        }

        d=dp[0];

        e=dp[1];

        f=dp[2];

        dp[0]=a+max(e,f);

        dp[1]=b+max(d,f);

        dp[2]=c+max(d,e);

    }

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

    return 0;

}
// 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: 3.005630318316322x):
#include<bits/stdc++.h>

using namespace std;



long long dp[100];



int main()

{

    long long n,a,b,c,d,e,f,i;

    cin>>n;

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

    {

        cin>>a>>b>>c;

        if(i==0)

        {

            dp[0]=a;

            dp[1]=b;

            dp[2]=c;

            continue;

        }

        d=dp[0];

        e=dp[1];

        f=dp[2];

        dp[0]=a+max(e,f);

        dp[1]=b+max(d,f);

        dp[2]=c+max(d,e);

    }

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

    return 0;

}
// 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: 3.5197160016705342x):
#include<bits/stdc++.h>

using namespace std;



long long dp[100];



int main()

{

    long long n,a,b,c,d,e,f,i;

    cin>>n;

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

    {

        cin>>a>>b>>c;

        if(i==0)

        {

            dp[0]=a;

            dp[1]=b;

            dp[2]=c;

            continue;

        }

        d=dp[0];

        e=dp[1];

        f=dp[2];

        dp[0]=a+max(e,f);

        dp[1]=b+max(d,f);

        dp[2]=c+max(d,e);

    }

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

    return 0;

}
// 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: 2.588264755352419x):
#include<bits/stdc++.h>

using namespace std;



long long dp[100];



int main()

{

    long long n,a,b,c,d,e,f,i;

    cin>>n;

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

    {

        cin>>a>>b>>c;

        if(i==0)

        {

            dp[0]=a;

            dp[1]=b;

            dp[2]=c;

            continue;

        }

        d=dp[0];

        e=dp[1];

        f=dp[2];

        dp[0]=a+max(e,f);

        dp[1]=b+max(d,f);

        dp[2]=c+max(d,e);

    }

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

    return 0;

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



using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;





int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i){

	if(i == n) return 0;

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

	int& ans = dp[a][i];

	

	if(a==0){

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1){

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2){

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

}

int main(){

	cin>>n;

	for(int i=0; i<n; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

	memset(dp, -1, sizeof dp);

	cout<<max(go(0,0),max(go(1,0),go(2,0)))<<"\n";

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.2083681387763603x):
#include<bits/stdc++.h>

using namespace std;



long long dp[100];



int main()

{

    long long n,a,b,c,d,e,f,i;

    cin>>n;

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

    {

        cin>>a>>b>>c;

        if(i==0)

        {

            dp[0]=a;

            dp[1]=b;

            dp[2]=c;

            continue;

        }

        d=dp[0];

        e=dp[1];

        f=dp[2];

        dp[0]=a+max(e,f);

        dp[1]=b+max(d,f);

        dp[2]=c+max(d,e);

    }

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

    return 0;

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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9; 



int m;

const int mx = 1e5+5;

int ac[mx][3];

int dp[3][mx];

int go(int a, int i){

	if(i == m) return 0;

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

	int& ans = dp[a][i];

	

	if(a==0){

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1){

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2){

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

}

int main(){

	cin>>m;

	for(int i=0; i<m; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

	memset(dp, -1, sizeof dp);

	cout<<max(go(0,0),max(go(1,0),go(2,0)))<<"\n";

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.2083904382699213x):
#include<bits/stdc++.h>

using namespace std;



long long dp[100];



int main()

{

    long long n,a,b,c,d,e,f,i;

    cin>>n;

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

    {

        cin>>a>>b>>c;

        if(i==0)

        {

            dp[0]=a;

            dp[1]=b;

            dp[2]=c;

            continue;

        }

        d=dp[0];

        e=dp[1];

        f=dp[2];

        dp[0]=a+max(e,f);

        dp[1]=b+max(d,f);

        dp[2]=c+max(d,e);

    }

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

    return 0;

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

using namespace std;

typedef long long ll;

typedef long long int lli;

typedef long double ld;

const int oo = 1e9;



int n;

const int nx = 1e5+5;

int ac[nx][3];

int dp[3][nx];

int go(int a, int i)

{

	if(i==n) return 0;

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

	int& ans = dp[a][i];

	if(a==0)

	{

		ans=max(ans, go(1, i+1)+ac[i][0]);

		ans=max(ans, go(2, i+1)+ac[i][0]);

	}

	if(a==1)

	{

		ans=max(ans, go(0, i+1)+ac[i][1]);

		ans=max(ans, go(2, i+1)+ac[i][1]);

	}

	if(a==2)

	{

		ans=max(ans, go(0, i+1)+ac[i][2]);

		ans=max(ans, go(1, i+1)+ac[i][2]);

	}

	

	return ans;

	

}

int main()

{

    cin>>n;

    for(int i=0; i<n; i++) cin>>ac[i][0]>>ac[i][1]>>ac[i][2];

    memset(dp, -1, sizeof dp);

    cout<<max(go(0,0), max(go(1,0),go(2,0)))<<"\n";

}

	

// Faster C++ solution for the same problem (measured speedup: 2.2083681387763603x):
#include<bits/stdc++.h>

using namespace std;



long long dp[100];



int main()

{

    long long n,a,b,c,d,e,f,i;

    cin>>n;

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

    {

        cin>>a>>b>>c;

        if(i==0)

        {

            dp[0]=a;

            dp[1]=b;

            dp[2]=c;

            continue;

        }

        d=dp[0];

        e=dp[1];

        f=dp[2];

        dp[0]=a+max(e,f);

        dp[1]=b+max(d,f);

        dp[2]=c+max(d,e);

    }

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

    return 0;

}