Problem:
Score : 100 points

Problem Statement

Taro's summer vacation starts tomorrow, and he has decided to make plans for it now.

The vacation consists of N days.
For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day:

A: Swim in the sea. Gain a_i points of happiness.

B: Catch bugs in the mountains. Gain b_i points of happiness.

C: Do homework at home. Gain c_i points of happiness.

As Taro gets bored easily, he cannot do the same activities for two or more consecutive days.

Find the maximum possible total points of happiness that Taro gains.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq a_i, b_i, c_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N
a_1 b_1 c_1
a_2 b_2 c_2
:
a_N b_N c_N

Output

Print the maximum possible total points of happiness that Taro gains.

Sample Input 1

3
10 40 70
20 50 80
30 60 90

Sample Output 1

210

If Taro does activities in the order C, B, C, he will gain 70 + 50 + 90 = 210 points of happiness.

Sample Input 2

1
100 10 1

Sample Output 2

100

Sample Input 3

7
6 7 8
8 8 3
2 5 2
7 8 6
4 6 8
2 3 4
7 5 1

Sample Output 3

46

Taro should do activities in the order C, A, B, A, C, B, A.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {
    static Scanner sc = new Scanner(System.in);
    public static void main(String[] args) {
        int n = sc.nextInt();
        int[] a = new int[n+1];
        a[0] = 0;
        int[] b = new int[n+1];
        b[0] = 0;
        int[] c = new int[n+1];
        c[0] = 0;


        for(int i = 1; i < n+1; i++){
            a[i] = sc.nextInt();
            b[i] = sc.nextInt();
            c[i] = sc.nextInt();

        }

        int[][] dp = new int[100005][3];
        dp[0][0] = 0;
        dp[0][1] = 0;
        dp[0][2] = 0;
        //dp的三列分别存储在第i天去a，b，c的最大值

        //さっきまでの問題と同じように
        //dp[i]=(i日目までに得られる幸福度の和の最大値)
        //としたいけど、それじゃダメなのだ。例えばa[N]=1,b[N]=1,c[N]=1000だったときには、
        // 最終日にはCを選んだ方が良さそうだけど、前の日にCを選んでたら困るのだ！
        //「2日連続で選べない」という条件は、前の日に何をしたかわかってればどうにかできるから
        //dp[i][j]=(i日目に行動jをして、i日目までに得られる幸福度の和の最大値)
        //を考えれば良さそうなのだ！
        //i日目に行動Aをするには、前日の行動はBかCじゃないとダメだから
        //dp[i][A]=max(dp[i−1][B]+a[i],dp[i−1][C]+a[i])
        //になるのだ！dp[i][B]、dp[i][C]も同様なのだ。
        //最終的な答えは、最終日の行動がどれでもいいから
        //max(dp[N][A],dp[N][B],dp[N][C])
        //なのだ！
        for (int i = 1; i < n+1; i++){
            //当这一天选择为去做A事情的时候，前一天只可能做了b事或者c事情
            //因此只要用a【i】加上前一天的b事情或者c事情就行
            dp[i][0] = Math.max(a[i] + dp[i-1][1], a[i]+dp[i-1][2]);
            dp[i][1] = Math.max(b[i] + dp[i-1][0], b[i] +dp[i-1][2]);
            dp[i][2] = Math.max(c[i] + dp[i-1][0], c[i] +dp[i-1][1]);
        }

        System.out.println(Math.max(dp[n][0], Math.max(dp[n][1], dp[n][2])));
    }

}