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



using namespace std;



    const int n = 1e5+5;

    int num;

    int act[n][3];

    int dyn[3][n];

    

    int init(int a, int i){

        

    	if(i == num) return 0;

    	

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

    	int& res = dyn[a][i];

    	

    	if(a==0){

    	    

    		res=max(res, init(1, i+1)+act[i][0]);

    		res=max(res, init(2, i+1)+act[i][0]);

    	}

    	

    	if(a==1){

    	    

    		res=max(res, init(0, i+1)+act[i][1]);

    		res=max(res, init(2, i+1)+act[i][1]);

    	}

    	

    	if(a==2){

    	    

    		res=max(res, init(0, i+1)+act[i][2]);

    		res=max(res, init(1, i+1)+act[i][2]);

    	}

    	

    	

    	return res;

    }

    

    int main(){

        

    	cin>>num;

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

    	memset(dyn, -1, sizeof dyn);

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

    	return 0;

    	

    }

// Faster C++ solution for the same problem (measured speedup: 11.490605618211816x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

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

using namespace std;



int dp[3];



int main(){

    int n,a,b,c,x,y,z;

    cin>>n;



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

        cin>>a>>b>>c;

        if(i==0){

            dp[0]=a;

            dp[1]=b;

            dp[2]=c;

            continue;

        }



        x=dp[0];

        y=dp[1];

        z=dp[2];



        dp[0]=a+max(y,z);

        dp[1]=b+max(x,z);

        dp[2]=c+max(x,y);







    }



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





    return 0;







}


// Faster C++ solution for the same problem (measured speedup: 5.208067738601987x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

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

#include<cstdio>

#include<cstring> 

#include<string>

#include<map>

#include<vector>

#include<cmath>

#include<string>

#include<algorithm>

using namespace std;

int dp[100000][3];

int main()

{

	  int a,b,c,d;

	  cin>>a;

	  dp[0][0]=0;

	  dp[0][1]=0;

	  dp[0][2]=0;

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

	  {

	  	cin>>b;

	  	cin>>c;

	  	cin>>d;	 

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

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

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

	  }	  

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

	  return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.206750852998071x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

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

#include <string.h>

#define MAXN 100003



int array_a[MAXN][3];

int array_d[MAXN][3];



int max(int a,int b){

	return a>b? a:b;

}



int main(){

	memset(array_a,-1,sizeof(int)*MAXN*3);

	memset(array_d,-1,sizeof(int)*MAXN*3);

	int N;

	scanf("%d",&N);



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

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

	}



	array_d[0][0]=array_a[0][0];

	array_d[0][1]=array_a[0][1];

	array_d[0][2]=array_a[0][2];

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

		array_d[i][0]=array_a[i][0]+max(array_d[i-1][1],array_d[i-1][2]);

		array_d[i][1]=array_a[i][1]+max(array_d[i-1][0],array_d[i-1][2]);

		array_d[i][2]=array_a[i][2]+max(array_d[i-1][0],array_d[i-1][1]);

	}

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

	return 0;



}

// Faster C++ solution for the same problem (measured speedup: 13.486339629105531x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

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

#include<string>

#include<cmath>

#include<algorithm>

#include<vector>

#include<map>

#include<stack>

#include<queue>

using namespace std;

#define MAXN 100001

int dp[MAXN*3];

int main(){

	int N;

	cin>>N;

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

		cin>>dp[i];

	}

	//dp[i]对应与本次选择i%3(A,B,C)的最大值，等于前面不选择i%3的最大值加上选择i%3的值

	for(int i=3;i<N*3;i++){

		if(i%3==0){

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

		}else if(i%3==1){

			dp[i]=dp[i]+max(dp[i-4],dp[i-2]);

		}else{

			dp[i]=dp[i]+max(dp[i-4],dp[i-5]);

		}

	}

	int x;

	if(dp[3*N-1]>=dp[3*N-2]){

		x=dp[3*N-1];

	}else{

		x=dp[3*N-2];

	}

   cout<<(x>dp[3*N-3]?x:dp[3*N-3]);

   return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 5.20073644597813x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

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

#include<cstring>

#include<vector>

#include<stack>

#include<cstdio>

using namespace std;

int main()

{

	int n,a,b,c,flag=1,ans=0;

	int x[2][3];

	scanf("%d",&n);

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

	{

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

		if(i==0)

		{

			x[i][0]=a;

		    x[i][1]=b;

		    x[i][2]=c;

		    continue;

		}

		x[flag][0]=a+max(x[1-flag][1],x[1-flag][2]);

		x[flag][1]=b+max(x[1-flag][2],x[1-flag][0]);

		x[flag][2]=c+max(x[1-flag][1],x[1-flag][0]);

		flag=1-flag;

	}

	ans=max(x[1-flag][0],max(x[1-flag][1],x[1-flag][2]));

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

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.1382133651285855x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

}
// Slower C++ solution:
//       Author: Mostafa Mounir Shehab

 

#include <bits/stdc++.h>

using namespace std;

 

#define PI  3.14159265  // PI = acos(-1)

#define EPS (1e-10)

#define endl "\n"

#define SZ(v) (int)(v.size())

typedef long long ll;

typedef long double ld;

//template <class Type>

 

 

void Fast(){

	std::ios_base::sync_with_stdio(0);

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

}

 

inline int D(){

	int t;

	scanf("%d",&t);

	return t;

}

inline ll llD(){

	ll t;

	scanf("%lld",&t);

	return t;

}

 

 

ll MOD=1e9 + 7; 

const int N=1e5 + 5,M=50 + 1;

int arr[N][4];

ll dp[N][4];

ll OO=1e12; 

 

ll n,m,k,cnt,SU,idx,len,MN,MX;

ll l,r,mid;

ll a,b,c,d;



ll Solve(int idx,int row){

	

		

	if(idx == n)

		return 0;

	

	ll &ret = dp[idx][row];

	if(~ret)

		return ret;

	

	ret = -OO;

	

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

		if(row != i)

			ret = max(ret,arr[idx][i] + Solve(idx+1,i));

	

	return ret;

}







 

 

int main(){

	

	memset(dp, -1 ,sizeof dp);

	cin>>n;

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

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



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

	



	





}

// Faster C++ solution for the same problem (measured speedup: 22.00683085367636x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

}
// 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: 22.05422778508199x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

}
// 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: 17.717334826334437x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

}
// 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: 26.437736292644693x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

}
// 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: 21.934523892496447x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

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

using namespace std;

int main()

{

    int n,ans=0;

    cin>>n;

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

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

    {

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

    }

    int dp[n-1][3];

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

    dp[0][1]=b[0];

    dp[0][2]=c[0];

    ans=max(dp[0][0],max(dp[0][1],dp[0][2]));

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

    {

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

        {

            if(j==0)

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

            if(j==1)

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

            if(j==2)

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

            if(i==n-1)

                ans=max(dp[i][j],ans);

        }

    }

    cout<<ans;

   return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.203529298829334x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

}
// 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: 13.558430701981163x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

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



using namespace std;

const int N = 1e5 + 5;

long long dp[3][N], ans;



int main()

{

	int n;

	cin >> n;

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

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

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

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

	dp[1][0] = B[0];

	dp[2][0] = C[0];

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

	{

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

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

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

	}

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

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

	cout << ans << "\n";



	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.213587857256487x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

}
// 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: 13.619556882177985x):
#include <stdio.h>



int max(int a, int b)

{

    return a > b ? a : b;

}



int main()

{

    int n;

    scanf("%d", &n);

    int a[n][3];

    int dp[n][3];

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

    {

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

    }

    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]);

    }

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

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

}