stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 2.9334879973598675x):
#include<bits/stdc++.h>

using ll = long long;



int main(){

	std::cin.tie(0);

	std::ios::sync_with_stdio(false);



	int N;

	std::cin >> N;



	std::vector<ll> Hap[3];

	std::vector<ll> dp[3];

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

		Hap[i].resize(N+1, 0);

		dp[i].resize(N+1, 0);

	}

	

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

		std::cin >> Hap[0][i] >> Hap[1][i] >> Hap[2][i];

	}





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

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

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

				if( j == k ) continue;

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

			}

		}

	}

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

	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: 2.2122153037598618x):
//下一个状态选上一个状态不与自己相同的另外两个的最大值 

#include<cstdio>

#include<cmath>

#include<cstring>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#define ll long long



using namespace std;

const int MaxN = 100005;



ll n, i;

ll a[MaxN], b[MaxN], c[MaxN];

ll dp[MaxN][5];



int max(int x, int y){

	x>y?x:y;

} 



int main(){

	cin >> n;

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

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

    dp[1][2] = b[1];

    dp[1][3] = c[1];

    for(i = 2; i <= n; 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];

    }

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

    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: 4.224695890308601x):
//下一个状态选上一个状态不与自己相同的另外两个的最大值 

#include<cstdio>

#include<cmath>

#include<cstring>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#define ll long long



using namespace std;

const int MaxN = 100005;



ll n, i;

ll a[MaxN], b[MaxN], c[MaxN];

ll dp[MaxN][5];



int max(int x, int y){

	x>y?x:y;

} 



int main(){

	cin >> n;

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

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

    dp[1][2] = b[1];

    dp[1][3] = c[1];

    for(i = 2; i <= n; 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];

    }

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

    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: 1.3905378505091421x):
//下一个状态选上一个状态不与自己相同的另外两个的最大值 

#include<cstdio>

#include<cmath>

#include<cstring>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#define ll long long



using namespace std;

const int MaxN = 100005;



ll n, i;

ll a[MaxN], b[MaxN], c[MaxN];

ll dp[MaxN][5];



int max(int x, int y){

	x>y?x:y;

} 



int main(){

	cin >> n;

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

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

    dp[1][2] = b[1];

    dp[1][3] = c[1];

    for(i = 2; i <= n; 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];

    }

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

    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: 9.3724048088621x):
//下一个状态选上一个状态不与自己相同的另外两个的最大值 

#include<cstdio>

#include<cmath>

#include<cstring>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#define ll long long



using namespace std;

const int MaxN = 100005;



ll n, i;

ll a[MaxN], b[MaxN], c[MaxN];

ll dp[MaxN][5];



int max(int x, int y){

	x>y?x:y;

} 



int main(){

	cin >> n;

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

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

    dp[1][2] = b[1];

    dp[1][3] = c[1];

    for(i = 2; i <= n; 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];

    }

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

    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: 3.0090326424694767x):
//下一个状态选上一个状态不与自己相同的另外两个的最大值 

#include<cstdio>

#include<cmath>

#include<cstring>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#define ll long long



using namespace std;

const int MaxN = 100005;



ll n, i;

ll a[MaxN], b[MaxN], c[MaxN];

ll dp[MaxN][5];



int max(int x, int y){

	x>y?x:y;

} 



int main(){

	cin >> n;

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

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

    dp[1][2] = b[1];

    dp[1][3] = c[1];

    for(i = 2; i <= n; 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];

    }

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

    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: 1.1989373600872042x):
#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;

}
// 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: 2.65981897181833x):
#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;

}
// 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: 2.1993735153046914x):
#include <stdio.h>

#include <iostream>

#include <algorithm>



using namespace std;



int main(){

    int n;

    scanf("%d", &n);

    int mat[n][3];

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

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

        int a, b, c;

        cin >> a >> b >> c;

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

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

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

    }

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

    

    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: 4.200171762472325x):
#include <stdio.h>

#include <iostream>

#include <algorithm>



using namespace std;



int main(){

    int n;

    scanf("%d", &n);

    int mat[n][3];

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

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

        int a, b, c;

        cin >> a >> b >> c;

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

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

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

    }

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

    

    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: 1.3824658545850583x):
#include <stdio.h>

#include <iostream>

#include <algorithm>



using namespace std;



int main(){

    int n;

    scanf("%d", &n);

    int mat[n][3];

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

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

        int a, b, c;

        cin >> a >> b >> c;

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

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

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

    }

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

    

    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: 9.317998513205858x):
#include <stdio.h>

#include <iostream>

#include <algorithm>



using namespace std;



int main(){

    int n;

    scanf("%d", &n);

    int mat[n][3];

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

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

        int a, b, c;

        cin >> a >> b >> c;

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

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

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

    }

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

    

    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: 2.991565373084069x):
#include <stdio.h>

#include <iostream>

#include <algorithm>



using namespace std;



int main(){

    int n;

    scanf("%d", &n);

    int mat[n][3];

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

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

        int a, b, c;

        cin >> a >> b >> c;

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

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

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

    }

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

    

    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: 3.5032453756940463x):
#include <stdio.h>

#include <iostream>

#include <algorithm>



using namespace std;



int main(){

    int n;

    scanf("%d", &n);

    int mat[n][3];

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

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

        int a, b, c;

        cin >> a >> b >> c;

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

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

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

    }

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

    

    return 0;

}