Problem:


Score : 400 points

Problem Statement

There is a directed graph with N vertices and M edges.
The i-th edge (1≤i≤M) points from vertex a_i to vertex b_i, and has a weight c_i.
We will play the following single-player game using this graph and a piece.

Initially, the piece is placed at vertex 1, and the score of the player is set to 0.
The player can move the piece as follows:

When the piece is placed at vertex a_i, move the piece along the i-th edge to vertex b_i. After this move, the score of the player is increased by c_i.

The player can end the game only when the piece is placed at vertex N.
The given graph guarantees that it is possible to traverse from vertex 1 to vertex N.

When the player acts optimally to maximize the score at the end of the game, what will the score be?
If it is possible to increase the score indefinitely, print inf.

Constraints

2≤N≤1000 

1≤M≤min(N(N-1),2000) 

1≤a_i,b_i≤N (1≤i≤M) 

a_i≠b_i (1≤i≤M) 

a_i≠a_j or b_i≠b_j (1≤i<j≤M) 

-10^9≤c_i≤10^9 (1≤i≤M)

c_i is an integer.

In the given graph, there exists a path from vertex 1 to vertex N.

Input

Input is given from Standard Input in the following format:  

N M  
a_1 b_1 c_1  
a_2 b_2 c_2
:  
a_M b_M c_M  

Output

Print the maximum possible score at the end of the game, if it is finite. If it is possible to increase the score indefinitely, print inf.  

Sample Input 1

3 3
1 2 4
2 3 3
1 3 5

Sample Output 1

7

There are two ways to move the piece to vertex N=3:

vertex 1 → vertex 2 → vertex 3 : score 4+3=7

vertex 1 → vertex 3 : score 5

Thus, the maximum possible score at the end of the game is 7.

Sample Input 2

2 2
1 2 1
2 1 1

Sample Output 2

inf

It is possible to increase the score indefinitely by alternating between vertex 1 and 2.

Sample Input 3

6 5
1 2 -1000000000
2 3 -1000000000
3 4 -1000000000
4 5 -1000000000
5 6 -1000000000

Sample Output 3

-5000000000



Sample input:
3 3
1 2 4
2 3 3
1 3 5


Sample output:
7

Submission status: Accepted

Solution code:
def d_score_attack(N, M, R, INF=float('inf')):  # ベルマン-フォード法による
    edges = [(a - 1, b - 1, -c) for a, b, c in R]
    dist = [INF] * N
    dist[0] = 0

    for _ in range(N - 1):
        for u, v, w in edges:
            if dist[u] == INF:
                continue
            if dist[v] > dist[u] + w:
                dist[v] = dist[u] + w
    ans_tmp = dist[N - 1]  # 負閉路がなければ、これが解

    # 負閉路がなければここからのループで距離は変わらない
    is_negative = [False] * N
    for _ in range(N):
        for u, v, w in edges:
            if dist[u] == INF:
                continue
            if dist[v] > dist[u] + w:
                dist[v] = dist[u] + w
                is_negative[v] = True

            # ある頂点が負閉路の一部なら、それに隣接する頂点も負閉路の一部
            if is_negative[u]:
                is_negative[v] = True
    # 頂点Nが負閉路に含まれるなら、元のグラフでのスコアは無限大
    return 'inf' if is_negative[N - 1] else -ans_tmp

N, M = [int(i) for i in input().split()]
R = [[int(i) for i in input().split()] for j in range(M)]
print(d_score_attack(N, M, R))

Submission status: Accepted

Solution code:
N, M = map(int, input().split())

v = [[] for _ in range(N)]
abc = []
for _ in range(M) :
    a, b, c = map(int, input().split())
    v[b-1].append((a-1, -c))
    
visited = [False] * N
visited[N-1] = True
queue = [N-1]
while queue :
    cur = queue.pop()
    for nex, cost in v[cur] :
        if not visited[nex] :
            queue.append(nex)
            visited[nex] = True
        abc.append((nex, cur, cost))
        
dist = [float('inf')] * N
dist[0] = 0

def bellmanford() :
    for i in range(N) :
        for a, b, c in abc :
            if dist[a] + c < dist[b] :
                dist[b] = dist[a] + c
                if i == N-1 :
                    return True
    return False
    
if bellmanford() :
    print('inf')
else :
    print(-dist[N-1])

Submission status: Accepted

Solution code:
n, m = [ int(v) for v in input().split() ]
node_list = []
for i in range(m):
    a, b, c = [ int(v) for v in input().split() ]
    node_list.append((a-1,b-1,-c))
distance_list = [ 10**15 for i in range(n) ]
parent_list = [ -1 for i in range(n) ]

distance_list[0] = 0
parent_list[0] = -1



for i in range(n-1):
    new_distanece_list = distance_list
    for j in node_list:
        a, b, c = j
        if distance_list[a] != 10**15:
            if new_distanece_list[b] > distance_list[a] + c:
                new_distanece_list[b] = distance_list[a] + c
                parent_list[b] = a
                change_log = 1
    distance_list = new_distanece_list
distance_list_1 = [ i for i in distance_list]

for i in range(n):
    new_distanece_list = distance_list
    for j in node_list:
        a, b, c = j
        if distance_list[a] != 10**15:
            if new_distanece_list[b] > distance_list[a] + c:
                new_distanece_list[b] = distance_list[a] + c
                parent_list[b] = a
                change_log = 1
    distance_list = new_distanece_list
distance_list_2 = distance_list



if distance_list_1[-1] == distance_list_2[-1]:
    print(-distance_list_1[-1])
else:
    print("inf")

Submission status: Accepted

Solution code:
#単一始点最短経路問題
#負の辺があっても動作する

#最短経路を求める
#計算量O(N * M)

INF = 10 ** 15
#入力
N, M = map(int, input().split())

# 入力は1-index
# 内部で0-indexにして処理
G = []
for _ in range(M): #M個の辺の情報を受け取る
    l, r, s = map(int, input().split()) #lからrへ重みsの辺が存在
    G += [[l-1, r-1, -s]] #有向グラフのときはここだけ


#メイン文
d = [INF] * N #最短距離を記録しておくリスト
# prev = [-1] * N
negative = [False] * N
#最短経路復元
def restore(g): #gまでの最短経路を復元
    path = []
    g -= 1 #0-indexに直す
    path.append(g)
    g = prev[g]
    count = 0
    while g != -1:
        # print ('C')
        path.append(g)
        g = prev[g]
        count += 1
        if count == N + 2:
            break
    path.reverse()
    return path

def shortest_path(s): #S番目の頂点から各頂点への最短距離を求める
    s -= 1
    for i in range(N):
        d[i] = INF
    d[s] = 0
    count = 0
    while count < N -1:
        count += 1
        # print ('A')
        # update = False
        for i in range(M): #無向グラフのときは辺は実質的には2倍
            e = G[i]
            if d[e[0]] != INF and d[e[1]] > d[e[0]] + e[2]:
                d[e[1]] = d[e[0]] + e[2]
                # update = True
                # prev[e[1]] = e[0]
        # if not update:
        #     return True

    count = 0
    while count < N:
        count += 1
        # print ('A')
        for i in range(M): #無向グラフのときは辺は実質的には2倍
            e = G[i]
            if d[e[0]] == INF:
                continue
            if d[e[0]] != INF and d[e[1]] > d[e[0]] + e[2]:
                d[e[1]] = d[e[0]] + e[2]
                negative[e[1]] = True
            if negative[e[0]]:
                negative[e[1]] = True


#呼び出しスタート地点を入れる
shortest_path(1)
if not negative[N - 1]:
    print (-d[N - 1])
else:
    print ('inf')

# print (d)

Submission status: Accepted

Solution code:
inf = float('inf')
V,E = map(int,input().split())
edge = []
edge_rev = []
for i in range(E):
    A,B,C = map(int,input().split())
    edge_rev.append([B-1,A-1,-C])
    edge.append([A-1,B-1,-C])

d = [inf]*V
d[0] = 0
for i in range(V):
    update = False
    for j in range(E):
        From,To,cost = edge[j]
        if d[From] != inf and d[To]>d[From]+cost:
            d[To]=d[From]+cost
            update = True
    if not update:
        break

d_rev = [inf]*V
d_rev[V-1] = 0
for i in range(V):
    update = False
    for j in range(E):
        From,To,cost = edge_rev[j]
        if d_rev[From] != inf and d_rev[To]>d_rev[From]+cost:
            d_rev[To]=d_rev[From]+cost
            update = True
    if not update:
        break

d_ = [x for x in d]
d_rev_ = [x for x in d_rev]



for j in range(E):
    From,To,cost = edge[j]
    if d[From] != inf and d_rev[From] != inf and d[To]>d[From]+cost:
        d[To]=d[From]+cost


for j in range(E):
    From,To,cost = edge_rev[j]
    if d[From] != inf and d_rev[From] != inf and d_rev[To]>d_rev[From]+cost:
        d_rev[To]=d_rev[From]+cost


loop = False
for i in range(V):
    if d[i] != 0 and d_rev[i] != 0:
        loop = (d[i]+d_rev[i] != d_[i]+d_rev_[i]) or loop

if loop:
    print('inf')

else:
    ans = -d[-1]
    print(ans)

Submission status: Accepted

Solution code:
#単一始点最短経路問題
#負の辺があっても動作する

#最短経路を求める
#計算量O(N * M)

INF = 10 ** 15
#入力
N, M = map(int, input().split())

# 入力は1-index
# 内部で0-indexにして処理
G = []
for _ in range(M): #M個の辺の情報を受け取る
    l, r, s = map(int, input().split()) #lからrへ重みsの辺が存在
    G += [[l-1, r-1, -s]] #有向グラフのときはここだけ


#メイン文
d = [INF] * N #最短距離を記録しておくリスト
prev = [-1] * N

#最短経路復元
def restore(g): #gまでの最短経路を復元
    path = []
    g -= 1 #0-indexに直す
    path.append(g)
    g = prev[g]
    count = 0
    while g != -1:
        # print ('C')
        path.append(g)
        g = prev[g]
        count += 1
        if count == N + 2:
            break
    path.reverse()
    return path

def shortest_path(s): #S番目の頂点から各頂点への最短距離を求める
    s -= 1
    for i in range(N):
        d[i] = INF
    d[s] = 0
    count = 0
    while True:
        if count == N:
            # print ('B')
            path = restore(N)
            # print (prev)
            # print (path)
            update = False
            for i in range(M): #無向グラフのときは辺は実質的には2倍
                e = G[i]
                if d[e[0]] != INF and d[e[1]] > d[e[0]] + e[2]:
                    d[e[1]] = d[e[0]] + e[2]
                    if e[1] in path:
                        return False
                    update = True
                    prev[e[1]] = e[0]
            return True

        count += 1
        # print ('A')
        update = False
        for i in range(M): #無向グラフのときは辺は実質的には2倍
            e = G[i]
            if d[e[0]] != INF and d[e[1]] > d[e[0]] + e[2]:
                d[e[1]] = d[e[0]] + e[2]
                update = True
                prev[e[1]] = e[0]
        if not update:
            return True

#呼び出しスタート地点を入れる
if shortest_path(1):
    print (-d[N - 1])
else:
    print ('inf')

# print (d)

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

    # 

    """ 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, M = R()
    ABC = [R() for _ in range(M)]
    ABC = [(a-1, b-1, c) for a, b, c in ABC]

    def BellmanFord(N, M, ABC, vertex_start=0, vertex_end = N-1, return_if_indefinite = -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, optional): start vertex. Defaults to 0.
            vertex_end (int, optional): end vertex. Defaults to N-1.
            return_if_indefinite (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 return_if_indefinite to str)
        """

        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 return_if_indefinite
            if not updated:
                break
        return dist[vertex_end]

    print(BellmanFord(N, M, ABC, return_if_indefinite = 'inf'))

    # N, M, P = R()
    # ABC = [R() for _ in range(M)]
    # ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
    
    # vertex_start, vertex_end = 0, N-1
    # 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 = []
    # reachable_to_direct = defaultdict(list)
    # reachable_from_direct = defaultdict(list)

    # 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]
    
    # 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
    #         if dist[a] != INF and dist[a] + c > dist[b]:
    #             dist[b] = dist[a] + c
    #             updated = True
    #             if i == N-1:
    #                 print(-1)
    #                 return
    #     if not updated:
    #         break
    # print(max(dist[vertex_end], 0))
                
    # 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:
import sys
inf = float('inf')

def solve():
    N, M = map(int, input().split())
    edges = [None] * M

    for i in range(M):
        a, b, c = map(int, sys.stdin.readline().split())
        a, b = a-1, b-1
        edges[i] = (a, b, -c)

    ans = BellmanFord(N, M, edges)
    if ans is None:
        print('inf')
    else:
        print(-ans)

def BellmanFord(N, M, edges):
    d = [inf] * N
    d[0] = 0

    for i in range(N-1):
        for (u, v, c) in edges:
            if d[u] + c < d[v]:
                d[v] = d[u] + c

    for (u, v, c) in edges:
        if d[u] + c < d[v]:
            d[v] = -inf

    for i in range(N-1):
        for (u, v, c) in edges:
            if d[u] + c < d[v]:
                d[v] = -inf
            if v == N-1 and d[v] == -inf:
                return None
    return d[N-1]

if __name__ == '__main__':
    solve()

Submission status: Accepted

Solution code:
"""
Bellman-Ford
始点と終点が決められているとき、
経由する閉路のみを確認する必要がある。
"""
def Bellman_Ford2(g,start,goal):
    n = len(g)    
    dist = [float("inf")]*n
    dist[start] = 0
    a0 = float("inf")
    for i in range(2*n):
        updated = False
        for v in range(n):
            for to, cost in g[v]:
                if dist[to] <= dist[v] + cost: continue
                if i < n:
                    dist[to] = dist[v] + cost
                else:
                    dist[to] = -float("inf")
                updated = True
        if not updated: break
    if dist[goal] == -float("inf"): return None, False
#    if negative[-1]: return None, False
    return dist, True
 

import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline #文字列入力のときは注意

n,m = [int(i) for i in readline().split()]
g = [[] for _ in range(n)]
for i in range(m):
    a,b,c = [int(i) for i in readline().split()]
    g[a-1].append((b-1,-c))
#    g[b-1].append((a-1,c))

dist, TF = Bellman_Ford2(g,0,-1)

if TF: print(-dist[-1])
else: print("inf")

Submission status: Accepted

Solution code:
import sys
sys.setrecursionlimit(10**9)

def BF(edges,n,s):
    inf=float("inf")
    d=[inf for i in range(n+1)]
    d[s]=0
    update=True
    for i in range(n+1):
      update=False
      for v,u,cost in edges:
        if cand[v]!=1 or cand[u]!=1:
          continue
        if d[v]!=inf and d[u]>d[v]+cost:
          d[u]=d[v]+cost
          update=True
      if update==False:
        break
      if i==n:
        return inf
    return -d[-1]

def DFS_1toN(v):
  for u in to[v]:
    if toflag[u]==0:
      toflag[u]=1
      DFS_1toN(u)
      
def DFS_Nto1(v):
  for u in ot[v]:
    if otflag[u]==0:
      otflag[u]=1
      DFS_Nto1(u)
    
n,m=map(int,input().split())
g=[]
to=[[] for _ in range(n+1)]
ot=[[] for _ in range(n+1)]
for _ in range(m):
  a,b,c=map(int,input().split())
  g.append([a,b,-c])
  to[a].append(b)
  ot[b].append(a)
toflag=[0]*(n+1)
toflag[1]=1
DFS_1toN(1)
otflag=[0]*(n+1)
otflag[n]=1
DFS_Nto1(n)
cand=[0]*(n+1)
for i in range(1,n+1):
  cand[i]=toflag[i]*otflag[i]
ans=BF(g,n,1)
if ans==float('inf'):
  print('inf')
else:
  print(ans)

Submission status: Accepted

Solution code:
import sys

inf = 1 << 60

def solve():
    N, M = map(int, input().split())
    edges = [None] * M

    for i in range(M):
        a, b, c = map(int, sys.stdin.readline().split())
        a, b = a-1, b-1
        edges[i] = (a, b, -c)

    ans = BellmanFord(N, M, edges)
    if ans is None:
        print('inf')
    else:
        print(-ans)

def BellmanFord(N, M, edges):
    d = [inf] * N
    d[0] = 0

    for i in range(N-1):
        for (u, v, c) in edges:
            if d[u] + c < d[v]:
                d[v] = d[u] + c

    #for i in range(N):
    for (u, v, c) in edges:
        if d[u] + c < d[v]:
            if v == N-1:
                return None
            
            d[v] = d[u] + c
    return d[N-1]

if __name__ == '__main__':
    solve()