Problem:
 Score : 500 points    Problem Statement There is a directed graph with N vertices numbered 1 to N and M edges.
The i -th edge is directed from Vertex A_i to Vertex B_i , and there are C_i coins placed along that edge.
Additionally, there is a button on Vertex N .  We will play a game on this graph.
You start the game on Vertex 1 with zero coins, and head for Vertex N by traversing the edges while collecting coins.
It takes one minute to traverse an edge, and you can collect the coins placed along the edge each time you traverse it.
As usual in games, even if you traverse an edge once and collect the coins, the same number of coins will reappear next time you traverse that edge, which you can collect again.  When you reach Vertex N , you can end the game by pressing the button. (You can also choose to leave Vertex N without pressing the button and continue traveling.)
However, when you end the game, you will be asked to pay T \times P coins, where T is the number of minutes elapsed since the start of the game. If you have less than T \times P coins, you will have to pay all of your coins instead.  Your score will be the number of coins you have after this payment.
Determine if there exists a maximum value of the score that can be obtained. If the answer is yes, find that maximum value.      Constraints  2 \leq N \leq 2500  1 \leq M \leq 5000  1 \leq A_i, B_i \leq N  1 \leq C_i \leq 10^5  0 \leq P \leq 10^5  All values in input are integers.  Vertex N can be reached from Vertex 1 .         Input Input is given from Standard Input in the following format:  N  M  P  A_1  B_1  C_1  :  A_M  B_M  C_M       Output If there exists a maximum value of the score that can be obtained, print that maximum value; otherwise, print -1 .        Sample Input 1 3 3 10
1 2 20
2 3 30
1 3 45      Sample Output 1 35   There are two ways to travel from Vertex 1 to Vertex 3 :   Vertex 1 \rightarrow 2 \rightarrow 3 : You collect 20 + 30 = 50 coins on the way. After two minutes from the start of the game, you press the button, pay 2 \times 10 = 20 coins, and you have 50 - 20 = 30 coins left.  Vertex 1 \rightarrow 2 : You collect 45 coins on the way. After one minute from the start of the game, you press the button, pay 1 \times 10 = 10 coins, and you have 45 - 10 = 35 coins left.   Thus, the maximum score that can be obtained is 35 .       Sample Input 2 2 2 10
1 2 100
2 2 100      Sample Output 2 -1   The edge extending from Vertex 1 takes you to Vertex 2 . If you then traverse the edge extending from Vertex 2 to itself t times and press the button, your score will be 90 + 90t . Thus, you can infinitely increase your score, which means there is no maximum value of the score that can be obtained.       Sample Input 3 4 5 10
1 2 1
1 4 1
3 4 1
2 2 100
3 3 100      Sample Output 3 0   There is no way to travel from Vertex 1 to Vertex 4 other than traversing the edge leading from Vertex 1 to Vertex 4 directly. You will pick up one coin along this edge, but after being asked to paying 10 coins, your score will be 0 .  Note that you can collect an infinite number of coins if you traverse the edge leading from Vertex 1 to Vertex 2 , but this is pointless since you can no longer reach Vertex 4 and end the game.

Solution (Python):

from collections import deque
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline

def inpl(): return list(map(int, input().split()))

N, M, P = inpl()
INF = 10**12
G = [[] for _ in range(N)]
E = []
G2 = [[] for _ in range(N)]
rG2 = [[] for _ in range(N)]
for _ in range(M):
    a, b, c = inpl()
    G[a-1].append([b-1, -c+P])
    G2[a-1].append(b-1)
    rG2[b-1].append(a-1)
    E.append([a-1, b-1, -c+P])


def SCC(G, rG):
    N = len(G)
    def dfs(i):
        nonlocal t, rorder, searched
        searched[i] = True
        for j in G[i]:
            if not searched[j]:
                dfs(j)
        rorder[t] = i
        t += 1

    def rdfs(i):
        nonlocal t, group, g
        group[i] = g
        for j in rG[i]:
            if group[j] == -1:
                rdfs(j)

    t = 0
    rorder = [-1]*N
    searched = [0]*N
    group = [-1]*N

    for i in range(N):
        if not searched[i]:
            dfs(i)
    g = 0
    for i in range(N-1, -1, -1):
        if group[rorder[i]] == -1:
            rdfs(rorder[i])
            g += 1
    return group, g


def belman_ford(V, edges, r):
    "If glaph contains negative cycles then return True"
    D = [INF]*(V)
    D[r] = 0
    if V == 1:
        return False, [0]
    OK = True
    for _ in range(V-1):
        update = False
        for s, t, d in edges:
            if D[t] > D[s] + d:
                D[t] = D[s] + d
                update = True
        if not update:
            OK = False
    return OK, D

group, gsize = SCC(G2, rG2)
groups = [[] for _ in range(gsize)]
for i in range(N):
    groups[group[i]].append(i)

score = [10**12 for _ in range(N)]
negative_groups = [0]*N


for g in range(gsize):
    ggs = set(groups[g])
    s = groups[g][0]
    score[s] = 0
    Q = deque([s])
    while Q:
        p = Q.popleft()
        for q, c in G[p]:
            if not q in ggs:
                continue
            if score[p] + c < score[q]:
                score[q] = score[p] + c
                Q.append(q)
        if score[s] < 0:
            for g in ggs:
                negative_groups[g] = 1
            break

s = 0
Q = deque([s])
reach1 = [0]*N
reach1[0] = 1

while Q:
    p = Q.popleft()
    for q in G2[p]:
        if not reach1[q]:
            reach1[q] = 1
            Q.append(q)

s = N-1
Q = deque([s])
reachN = [0]*N
reachN[-1] = 1

while Q:
    p = Q.popleft()
    for q in rG2[p]:
        if not reachN[q]:
            reachN[q] = 1
            Q.append(q)

NEGATIVE, D = belman_ford(N, E, 0)

if any([reach1[i]&reachN[i]&negative_groups[i] for i in range(N)]) or reach1[-1]==0:
    print(-1)
else:
    print(max(0, -D[-1]))

Solution (Python):
INF = pow(10, 15)

def main():
  n, m, p = map(int, input().split())
  edges = []
  for _ in range(m):
    a, b, c = map(int, input().split())
    edges.append((a-1, b-1, p-c))  

  dist = [INF for _ in range(n)]
  frome = [-1 for _ in range(n)]
  dist[0] = 0
  for _ in range(n):
    for edge in edges:
      if dist[edge[0]] == INF:
        continue
      if dist[edge[1]] > dist[edge[0]] + edge[2]:
        dist[edge[1]] = dist[edge[0]] + edge[2]
        frome[edge[1]] = edge[0]

  for i, edge in enumerate(edges):
    if dist[edge[1]] > dist[edge[0]] + edge[2]:
      edges[i] = (edge[0], edge[1], -INF)

  dist = [INF for _ in range(n)]
  frome = [-1 for _ in range(n)]
  dist[0] = 0
  for _ in range(n):
    for edge in edges:
      if dist[edge[0]] == INF:
        continue
      if dist[edge[1]] > dist[edge[0]] + edge[2]:
        dist[edge[1]] = dist[edge[0]] + edge[2]
        frome[edge[1]] = edge[0]

  fromd = {}
  f = n-1
  while frome[f] != -1:
    if f in fromd:
      break
    fromd[f] = frome[f]
    f = frome[f]

  is_loop = False
  for edge in edges:
    if dist[edge[1]] > dist[edge[0]] + edge[2]:
      if edge[0] in fromd:
        is_loop = True
        break

  if is_loop:
    print(-1)
  else:
    if dist[-1] > 0:
      print(0)
    else:
      print(-dist[-1])


if __name__ == '__main__':
  main()

Solution (Python):
N, M, P = map(int,input().split())
dist = [0] + [float('inf')] * (N - 1)

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

for i in range(M):
    i,j,k = map(int,input().split())
    a.append(i)
    b.append(j)
    c.append(-1*(k - P))

INF = float('inf')

for i in range(N-1):
    for j in range(M):
        if dist[a[j] - 1] == INF: continue
        if dist[b[j] - 1] > dist[a[j] - 1] + c[j]:
            dist[b[j] - 1] = dist[a[j] - 1] + c[j]
        
ans = dist[N -1]

negative = [False] * N

for i in range(N):
    for j in range(M):
        if dist[a[j] - 1] == INF: continue
        if dist[b[j] - 1] > dist[a[j] - 1] + c[j]:
            dist[b[j] - 1] = dist[a[j] - 1] + c[j]
            negative[b[j] - 1] = True

        if negative[a[j] - 1] == True:
            negative[b[j] - 1] = True

if negative[N - 1]:
    print(-1)
else:
    print(max(-1 * ans, 0))

Solution (Python):
import os
import sys

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7

N, M, P = list(map(int, sys.stdin.readline().split()))
ABC = [list(map(int, sys.stdin.readline().split())) for _ in range(M)]

graph = [[] for _ in range(N)]
for a, b, c in ABC:
    graph[a - 1].append((b - 1, P - c))


def bellman_ford(graph, from_v, to_v):
    """
    到達できないなら INF、負閉路があったら -INF
    :param list of (list of (int, int)) graph: (to, cap) の隣接リスト
    :param int from_v:
    :param int to_v:
    :rtype: int
    """

    def reachable(graph, from_v):
        """
        graph 上で from_v から到達できるかどうかのリスト
        :param graph:
        :param from_v:
        """
        ret = [False] * len(graph)
        ret[from_v] = True
        stack = [from_v]
        while stack:
            v = stack.pop()
            for u, _ in graph[v]:
                if not ret[u]:
                    ret[u] = True
                    stack.append(u)
        return ret

    
    
    vertices = []
    revs = [[] for _ in range(len(graph))]
    for v, ud in enumerate(graph):
        for u, d in ud:
            revs[u].append((v, d))
    for v, (r1, r2) in enumerate(zip(reachable(graph, from_v), reachable(revs, to_v))):
        if r1 & r2:
            vertices.append(v)

    dist = [float('inf')] * len(graph)
    dist[from_v] = 0
    updated = True
    for _ in range(len(graph) + 1):
        updated = False
        for v in vertices:
            for u, d in graph[v]:
                if dist[v] + d < dist[u]:
                    dist[u] = dist[v] + d
                    updated = True
        if not updated:
            break
    return -float('inf') if updated else dist[to_v]


ans = -bellman_ford(graph, 0, N - 1)
if ans == INF or ans == -INF:
    print(-1)
else:
    print(max(0, ans))

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

    """ convenient functions
    
    
    
    
    
    """

    """convenient decorator
    
    
        
        
        
        
        
        
        
        
        
        
    """
    
    
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product 
    
    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 = []  
        b, e = 2, 0  
        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
        """
        
        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 
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest 
    from copy import deepcopy, copy 
    import operator
    from operator import itemgetter 
    
    
    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 
    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)
    
    
    
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) 
    def modinv(a, p = MOD):
        
        
        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):
        
        
        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]

    
    
    
    
    

    

    """ initialize variables and set inputs
    
        
        
    
        
        
        
        
        
        
        
    
        
        
        
        
    
        
        
    """

    
    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
    
    
    
    
    
    
    
    """

    
    if sample_file:
        sys.stdin = open(sample_file)
        
    
    

    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))
                
    
    
    

    """memo: how to use defaultdict of list
    
    Dic = defaultdict(list)
    
    Dic[x].append(y)
    
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    
    >>> a = 5
    >>> bin_str_a = format(a, '
    >>> print(bin_str_a)
    0b0101 
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

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

    """ convenient functions
    
    
    
    
    
    """

    """convenient decorator
    
    
        
        
        
        
        
        
        
        
        
        
    """
    
    
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product 
    
    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 = []  
        b, e = 2, 0  
        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
        """
        
        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 
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest 
    from copy import deepcopy, copy 
    import operator
    from operator import itemgetter 
    
    
    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 
    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)
    
    
    
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) 
    def modinv(a, p = MOD):
        
        
        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):
        
        
        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]

    
    
    
    
    

    

    """ initialize variables and set inputs
    
        
        
    
        
        
        
        
        
        
        
    
        
        
        
        
    
        
        
    """

    
    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
    
    
    
    
    
    
    
    """

    
    if sample_file:
        sys.stdin = open(sample_file)
        
    
    

    N, M, P = R()
    ABC = [R() for _ in range(M)]
    ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
    G = [[INF]*N for _ in range(N)]
    
    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:
        G[a][b] = c
        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):
        for a, b, c in ABC:
            if reachable_from_both_sides[a]:
                if dist[a] != INF and dist[a] + c > dist[b]:
                    dist[b] = dist[a] + c
                    if i == N-1:
                        print(-1)
                        return
    print(max(dist[vertex_end], 0))
                
    
    
    

    """memo: how to use defaultdict of list
    
    Dic = defaultdict(list)
    
    Dic[x].append(y)
    
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    
    >>> a = 5
    >>> bin_str_a = format(a, '
    >>> print(bin_str_a)
    0b0101 
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Solution (Python):

from collections import defaultdict
import sys

sys.setrecursionlimit(10 ** 6)

def dfs(u):
    ok[u] = 1
    for v in to[u]:
        if ok[v]: continue
        dfs(v)

def rdfs(u):
    ok[u] += 2
    for v in fr[u]:
        if ok[v] > 1: continue
        rdfs(v)

inf = 10 ** 9
to = defaultdict(list)
fr = defaultdict(list)
abc = []
n, m, p = map(int, input().split())
ok = [0] * n
for _ in range(m):
    a, b, c = map(int, input().split())
    a -= 1
    b -= 1
    c = -(c - p)
    to[a].append(b)
    fr[b].append(a)
    abc.append([a, b, c])
dfs(0)
rdfs(n - 1)
score = [inf] * n
score[0] = 0
for i in range(n):
    stop = True
    for a, b, c in abc:
        if ok[a] < 3 or ok[b] < 3: continue
        tmp = score[a] + c
        if tmp < score[b]:
            score[b] = tmp
            stop = False
    if stop:
        print(max(0, -score[n - 1]))
        break
else:
    print(-1)