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.



Sample input:
3
10 40 70
20 50 80
30 60 90


Sample output:
210

Submission status: Accepted

Solution code:
import cmath
import sys

INF=10**18
MOD=10**9+7
MAX=10**5+7

def main():
    N=int(input())
    dp=[0, 0, 0]
    for line in sys.stdin.readlines():
        a,b,c=map(int, line.split())
        ndp=[0, 0, 0]
        ndp[0]=max(dp[1], dp[2])+a
        ndp[1]=max(dp[2], dp[0])+b
        ndp[2]=max(dp[0], dp[1])+c
        dp=ndp
    print(max(dp))

main()

Submission status: Accepted

Solution code:
import sys

INF=10**18
MOD=10**9+7
MAX=10**5+7

def main():
    N=int(input())
    dp=[0, 0, 0]
    for line in sys.stdin.readlines():
        a,b,c=map(int, line.split())
        ndp=[0, 0, 0]
        ndp[0]=max(dp[1], dp[2])+a
        ndp[1]=max(dp[2], dp[0])+b
        ndp[2]=max(dp[0], dp[1])+c
        dp=ndp
    print(max(dp))

main()

Submission status: Accepted

Solution code:
import cmath
import numpy as np
import sys

INF=10**18
MOD=10**9+7
MAX=10**5+7

def main():
    N=int(input())
    dp=[0, 0, 0]
    for _ in range(N):
        a,b,c=map(int, input().split())
        ndp=[0, 0, 0]
        ndp[0]=max(dp[1], dp[2])+a
        ndp[1]=max(dp[2], dp[0])+b
        ndp[2]=max(dp[0], dp[1])+c
        dp=ndp
    print(max(dp))

if __name__=='__main__':
    main()

Submission status: Accepted

Solution code:
N = int(input())
happy = []
result = [[0 for i in range(3)] for j in range(N)]

for i in range(N):
    t = list( map( int, input().split() ) )
    happy.append(t)

for i in range(3):
    result[0][i] = happy[0][i] 

for i in range(1,N):
    for j in range(3):
        result[i][j] = max( [ result[i-1][k] + happy[i][j] for k in range(3) if k != j ] )

print( max( result[N-1] ) )

Submission status: Accepted

Solution code:
import sys
sys.setrecursionlimit(10**6)

memo = {}

def solve(i, last_option):
	global a, b, c
	if (i, last_option) not in memo:
		if i == len(a):
			result = 0
		elif i == 0:
			result = max(a[i] + solve(i + 1, 1), b[i] + solve(i + 1, 2), c[i] + solve(i + 1, 3))
		else:
			if last_option == 1:
				result = max(b[i] + solve(i + 1, 2), c[i] + solve(i + 1, 3))
			elif last_option == 2:
				result = max(a[i] + solve(i + 1, 1), c[i] + solve(i + 1, 3))
			else:
				result = max(a[i] + solve(i + 1, 1), b[i] + solve(i + 1, 2))
		memo[(i, last_option)] = result
		return result
	else:
		return memo[(i, last_option)]

n = int(input())
a = [None for i in range(n)]
b = [None for i in range(n)]
c = [None for i in range(n)]
for i in range(n):
	a[i], b[i], c[i] = map(int, input().split())
print(solve(0, 4))

Submission status: Accepted

Solution code:
N = int(input())
dp = [list(map(int, input().split())) for i in range(N)]

for i in range(N-1):
  for j in range(3):
    maxN = 0
    for k in range(3):
      if j==k: continue
      maxN = max(maxN, dp[i][k])
    dp[i+1][j] += maxN

print(max(dp[-1]))

Submission status: Accepted

Solution code:
n = int(input())
a,b,c = [],[],[]
for i in range(n):
    line = input()
    line = line.strip("\n").split(" ")
    a.append(int(line[0]))
    b.append(int(line[1]))
    c.append(int(line[2]))

def bottomup(a, b, c):
    dp = [[0 for _ in range(3)] for _ in range(n)]
    
    dp[(n-1)] = [a[n-1], b[n-1], c[n-1]]

    for i in range(n-2,-1,-1):
        dp[i][0] = a[i] + max(dp[i+1][1:])
        dp[i][1] = b[i] + max(dp[i+1][0], dp[i+1][2])
        dp[i][2] = c[i] + max(dp[i+1][:2])
    
    return max(dp[0])
    
print(bottomup(a,b,c))

Submission status: Accepted

Solution code:
def main():
    N = int(input())
    A = [[int(i) for i in input().split()] for _ in range(N)]
    dp = [[0]*3 for _ in range(N)] # dp[i][j]:i日目に行動jをするときの最大値
    dp[0] = A[0]
    for i in range(1, N):
        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][2], dp[i-1][0])
        dp[i][2] = A[i][2] + max(dp[i-1][0], dp[i-1][1])
    print(max(dp[N-1]))
main()

Submission status: Accepted

Solution code:
N = int(input())
ABC = [list(map(int, input().split())) for i in range(N)]

DP = [[0] * 3 for i in range(N)]
DP[0][0], DP[0][1], DP[0][2] = ABC[0][0], ABC[0][1], ABC[0][2]

for i in range(1, N):
  for j in range(3):
    for k in range(3):
      if j != k:
        DP[i][j] = max(DP[i - 1][k] + ABC[i][j], DP[i][j])
        
ans = max(DP[-1])
print(ans)

Submission status: Accepted

Solution code:
import numpy as np
import sys
def solve():
    N=int(sys.stdin.readline())
    dp=np.zeros((3,N)).astype('int64')
    for i in range(N):
        a,b,c=list(map(int,sys.stdin.readline().split()))
        if i:
            dp[0,i]=max(dp[1,i-1]+a,dp[2,i-1]+a)
            dp[1,i]=max(dp[0,i-1]+b,dp[2,i-1]+b)
            dp[2,i]=max(dp[1,i-1]+c,dp[0,i-1]+c)
        else:
            dp[0,i]=a
            dp[1,i]=b
            dp[2,i]=c

    return np.max(dp[:,N-1])
    
if __name__ == '__main__':
    print(solve())

Submission status: Accepted

Solution code:
n = int(input())
abc = [list(map(int , input().split())) for _ in range(n)]
dp = [[0,0,0] for _ in range(n)]
dp[0] = abc[0]
for i in range(1, n):
    for j in range(3):
        maxprev = max(dp[i-1][(j + 1)%3], dp[i-1][(j + 2)%3])
        dp[i][j] = abc[i][j] + maxprev
#print(abc)
#print(dp)
print(max(dp[n-1]))

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10 ** 7)

N = int(input())
a = [0] * N
b = [0] * N
c = [0] * N

for i in range(N):
    a[i], b[i], c[i] = map(int, input().split())
    if i > 0:
        a[i] += max(b[i-1], c[i-1])
        b[i] += max(a[i-1], c[i-1])
        c[i] += max(a[i-1], b[i-1])

print(max(a[N-1], b[N-1], c[N-1]))

Submission status: Accepted

Solution code:
n = int(input())
h = [tuple(map(int, input().split())) for _ in range(n)]

# i日目の活動を３パターン考える
dp = [[-1]*3 for _ in range(n+1)]
dp[0][0] = dp[0][1] = dp[0][2] = 0
dp[1][0], dp[1][1], dp[1][2] = h[0][0], h[0][1], h[0][2]

for i in range(2, n+1):
    for j in range(3):
        h_max = 0
        for k in range(3):
            if j != k and h_max < dp[i-1][k] + h[i-1][j]:
                h_max = dp[i-1][k] + h[i-1][j]
        dp[i][j] = h_max

print(max(dp[n]))

Submission status: Accepted

Solution code:
N = int(input())

a = []
b = []
c = []

tmpa, tmpb, tmpc = map(int, input().split())
a.append(tmpa)
b.append(tmpb)
c.append(tmpc)

for _ in range(N-1):
    tmpa, tmpb, tmpc = map(int, input().split())
    tmpa += max(b[-1], c[-1])
    tmpb += max(a[-1], c[-1])
    tmpc += max(b[-1], a[-1])
    
    a.append(tmpa)
    b.append(tmpb)
    c.append(tmpc)

    
print(max(a[-1], b[-1], c[-1]))

Submission status: Accepted

Solution code:
n = int(input())
list1 = [list(map(int,input().split())) for i in range(n)]

dpa = [0]*n
dpb = [0]*n
dpc = [0]*n
dpa[0],dpb[0],dpc[0] = list1[0][0],list1[0][1],list1[0][2]


for i in range(1,n):
	dpa[i] = max(dpb[i-1]+list1[i][0],dpc[i-1]+list1[i][0])
	dpb[i] = max(dpa[i-1]+list1[i][1],dpc[i-1]+list1[i][1])
	dpc[i] = max(dpa[i-1]+list1[i][2],dpb[i-1]+list1[i][2])

print(max(dpa[-1],dpb[-1],dpc[-1]))

Submission status: Accepted

Solution code:
import numpy as np

N = int(input())
dp = np.zeros((3,N),dtype=int)

for i in range(N):
    dp[:,i] = list(int(x) for x in input().split())
    if i > 0:
        dp[0,i] += max(dp[1,i-1], dp[2,i-1])
        dp[1,i] += max(dp[2,i-1], dp[0,i-1])
        dp[2,i] += max(dp[0,i-1], dp[1,i-1])


print(max(dp[:,N-1]))

Submission status: Accepted

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array, use [[0] * N for _ in range(N)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    R = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main

    N = r_int()
    ABC = [R() for _ in range(N)]
    la = [0] * (N+5)
    lb = [0] * (N+5)
    lc = [0] * (N+5)
    for i, (a, b, c) in enumerate(ABC):
        la[i+1] = max(lb[i], lc[i]) + a
        lb[i+1] = max(la[i], lc[i]) + b
        lc[i+1] = max(la[i], lb[i]) + c
    print(max(la[N],lb[N],lc[N]))

            # dp[t][n] = max(dp[])
    
    # to finish code, use return instead of exit()
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
N=int(input())
a=[0]*N
b=[0]*N
c=[0]*N
a[0],b[0],c[0]=map(int,input().split())
for i in range(1,N):
  a[i],b[i],c[i]=map(int,input().split())
  a[i] += max(b[i-1],c[i-1])
  b[i] += max(a[i-1],c[i-1])
  c[i] += max(a[i-1],b[i-1])

print(max(a[-1],b[-1],c[-1]))

Submission status: Accepted

Solution code:
n = int(input())
abc = [list(map(int, input().split())) for _ in range(n)]

dp = [[0] * 3 for _ in range(n)]
dp[0] = abc[0]

for i in range(1, n):
    for j in range(3):
        for k in range(3):
            if j != k:
                dp[i][j] = max(dp[i][j], dp[i-1][k] + abc[i][j])
print(max(dp[n - 1]))

Submission status: Accepted

Solution code:
from pprint import pprint
import sys
import copy

n = int(sys.stdin.readline().strip())

events = []
for _ in range(n):
    a, b, c = map(int, sys.stdin.readline().strip().split(" "))
    events.append((a, b, c))
# pprint(events)

pre_states = events[0]
for day in range(1, n):
    states = [0] * 3
    for i in range(3):
        tmp = []
        for j in range(3):
            if i == j:
                continue

            tmp.append(pre_states[j] + events[day][i])
        states[i] = max(tmp)
    # pprint(states)
    pre_states = copy.deepcopy(states)

pprint(max(pre_states))

Submission status: Accepted

Solution code:
N = int(input())
a = [0] * N
b = [0] * N
c = [0] * N

for i in range(N):
    a[i], b[i], c[i] = map(int, input().split())
    if i > 0:
        a[i] += max(b[i-1], c[i-1])
        b[i] += max(a[i-1], c[i-1])
        c[i] += max(a[i-1], b[i-1])

print(max(a[N-1],b[N-1],c[N-1]))

Submission status: Accepted

Solution code:
N = int(input())

cur, nxt = [0] * 3, [0] * 3
for i in range(N):
    abc = [int(c) for c in input().split()]
    for k in range(3):
        for last in range(3):
            if k != last:
                nxt[k] = max(nxt[k], cur[last] + abc[k])

    cur = nxt
    nxt = [0] * 3

ans = max(cur[k] for k in range(3))
print(ans)

Submission status: Accepted

Solution code:
N = int(input())
abc = []
for _ in range(N):
    abc.append(tuple(map(int, input().split())))

dp = [[0] * 3 for _ in range(N+1)]
for i in range(1, N+1):
    for k in range(3):
        for last in range(3):
            if k != last:
                dp[i][k] = max(dp[i][k], dp[i-1][last] + abc[i-1][last])

ans = max(dp[N][k] for k in range(3))
print(ans)

Submission status: Accepted

Solution code:
import numpy as np
n = int(input())
ha = []
hb = []
hc = []
for i in range(n):
  t1, t2, t3 = map(int, input().split())
  ha.append(t1)
  hb.append(t2)
  hc.append(t3)
dp = np.zeros((n,3))

dp[0,0] = ha[0]
dp[0,1] = hb[0]
dp[0,2] = hc[0]
for i in range(1, n):
  for c in range(3):
    if c == 0:
      tmp = ha[i]
    if c == 1:
      tmp = hb[i]    
    if c == 2:
      tmp = hc[i]
    dp[i,c] = max([dp[i-1,j] for j in range(3) if j!=c] ) + tmp
print(int(max(dp[n-1,0],dp[n-1,1],dp[n-1,2])))

Submission status: Accepted

Solution code:
import math
import numpy as np


def inputIntList():
    return [int(s) for s in input().split()]


def inputInt():
    return int(input())


def main():
    N = inputInt()
    act = [inputIntList() for _ in range(N)]
    act.append([0, 0, 0])
    dp = [[0 for _ in range(3)] for _ in range(N + 1)]
    dp[0] = act[0]
    for i in range(1, N + 1):
        for j in range(3):
            dp[i][j] = max(dp[i - 1][j - 1], dp[i - 1][j - 2]) + act[i][j]
            # print(dp)

    return max(dp[-1])


if __name__ == "__main__":
    print(main())

Submission status: Accepted

Solution code:
N = int(input())
abc = [[int(i) for i in input().split()] for i in range(N)]

for i in range(1,N):
    for g in range(3):
        abc[i][g] += max(abc[i-1][g-1], abc[i-1][g-2])

print(max(abc[-1]))

Submission status: Accepted

Solution code:
def main():
    N = int(input())
    
    array = []
    
    for i in range(N):
        array2 = list(map(int,input().split()))
        array.append(array2)

    num = [[0,0,0] for z in range(N)]
    for m in range(3):
        num[0][m] = array[0][m]

    

    for j in range(1,N):
        
        for z in range(3):
            for y in range(3):
                if y == z:
                    continue
                else:
                    num[j][z] = max(num[j][z],array[j][z]+num[j-1][y])
            
    
    print(max(num[-1]))
    
main()

Submission status: Accepted

Solution code:
def main():
    N = int(input())
    
    array = []
    
    for i in range(N):
        array2 = list(map(int,input().split()))
        array.append(array2)

    num = [[0,0,0] for z in range(N)]
    for m in range(3):
        num[0][m] = array[0][m]

    

    for j in range(1,N):
        
        for z in range(3):
            if z == 0:
                num[j][z] = array[j][z]+max(num[j-1][1],num[j-1][2])
                    
            if z == 1:
                num[j][z] =array[j][z]+max(num[j-1][0],num[j-1][2])
            if z == 2:
                num[j][z] =array[j][z]+max(num[j-1][1],num[j-1][0])
        
    
    print(max(num[-1]))
    
main()

Submission status: Accepted

Solution code:
import sys


def solve():
    readline = sys.stdin.buffer.readline
    mod = 10 ** 9 + 7
    n = int(readline())
    a = [0] * n
    b = [0] * n
    c = [0] * n
    for i in range(n):
        av, bv, cv = list(map(int, readline().split()))
        if i == 0:
            a[i] = av
            b[i] = bv
            c[i] = cv
        else:
            a[i] = max(b[i-1] + av, c[i-1] + av)
            b[i] = max(a[i-1] + bv, c[i-1] + bv)
            c[i] = max(a[i-1] + cv, b[i-1] + cv)
    print(max(a[-1], b[-1], c[-1]))


if __name__ == '__main__':
    solve()

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**8)

def chmax(a, b):
    if a < b:
        a = b
        return b
    return a

N = int(input())
a = [list(map(int, input().split())) for _ in range(N)]
dp = [list([0] * 3) for _ in range(N + 1)]
for i in range(N):
    for j in range(3):
        for k in range(3):
            if j == k:
                continue
            dp[i + 1][k] = chmax(dp[i + 1][k], dp[i][j] + a[i][k])

ans = 0
for i in range(3):
    ans = max(ans, dp[N][i])

print(ans)

Submission status: Accepted

Solution code:
tdlist = lambda i, j, value=0: [[value] * j for _ in range(i)]
n = int(input())
abc = [tuple(map(int, input().split())) for _ in range(n)]
dp = tdlist(n, 3)
dp[0] = list(abc[0])

for i in range(1, n):
    dp[i][0] = max(dp[i-1][1], dp[i-1][2]) + abc[i][0]
    dp[i][1] = max(dp[i-1][0], dp[i-1][2]) + abc[i][1]
    dp[i][2] = max(dp[i-1][0], dp[i-1][1]) + abc[i][2]

print(max(dp[n-1]))

Submission status: Accepted

Solution code:
#動的計画法
#C - Vacation
#aのところをdp[i][0]とするとなぜか事故る、そういう仕様なのか？、それどもdp = [[0,0,0]]*Nが影響しているのか
N = int(input())
H = [list(map(int,input().split())) for i in range(N)]
dp = [[0,0,0]]*N
dp[0] = H[0]
for i in range(1,N):
    a = max(dp[i-1][1],dp[i-1][2])+H[i][0]
    b = max(dp[i-1][0],dp[i-1][2])+H[i][1]
    c = max(dp[i-1][0],dp[i-1][1])+H[i][2]
    dp[i] = [a,b,c]
print(max(dp[N-1]))

Submission status: Accepted

Solution code:
n = int(input())
x = [list(map(int, input().split())) for i in range(n)]

dp = [[-float("inf")] * 3 for i in range(n + 1)]

for i in range(3):
    dp[0][i] = 0

for i in range(n):
    for j in range(3):
        for k in range(3):
            if j != k:
                dp[i + 1][k] = max(dp[i + 1][k], dp[i][j] + x[i][k])

print(max(dp[n]))