Problem: 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.

Submission attempts (in order):
import sys
import numpy as np
from scipy.sparse.csgraph import bellman_ford
from scipy.sparse import csr_matrix
from collections import deque
input = sys.stdin.readline


def main():
    n, m, p = map(int, input().split())
    
    inf = 10**9
    l = np.zeros((n, n))
    tree_forth = [[] for _ in range(n)]
    tree_back = [[] for _ in range(n)]
    for i in range(m):
        a, b, c =map(int, input().split())
        tree_forth[a-1].append(b-1)
        tree_back[b-1].append(a-1)
        c = p-c
        if c == 0:
            c = 10**(-8)
        if l[a-1][b-1] == 0:
            l[a-1][b-1] = c
        else:
            l[a-1][b-1] = min(l[a-1][b-1], c)
    
    already1, already2 = [True]*n, [True]*n
    for already, tree, start in zip([already1, already2], [tree_forth, tree_back], [0, n-1]):
        not_yet = deque([start])
        already[start] = False
        while not_yet:
            key = not_yet.pop()
            for v in tree[key]:
                if already[v] == False:
                    continue
                already[v] = False
                not_yet.append(v)
    
    
    for i in range(n):
        if already1[i] or already2[i]:
            for j in range(n):
                l[i][j] = 0
                l[j][i] = 0

    print(2/0)
    
    try:
        print(max(0, int(bellman_ford(csr_matrix(l))[0][-1])*(-1)))
    except:
        print(-1)


if __name__ == "__main__":
    main()


import sys
from collections import deque

input = sys.stdin.readline

class Bellman_Ford:
    def __init__(self, v=None, e=None, lis=None, start=None, end=None, inf=float('inf')):
        self.V = v
        self.E = e
        self.lis = lis
        self.start = start if start else 0
        self.end = end if end else self.V-1
        self.inf = inf
        self.close_minus = False
    
    def getlist(self, lis):
        self.lis = lis
    
    def def_start(self, s):
        self.start = s
    
    def def_end(self, e):
        self.end = e

    def def_inf(self, inf):
        self.inf = inf
    
    def def_vertice(self, v):
        self.V = v
    
    def def_edge(self, e):
        self.E = e

    def prepare(self):
        self.cost = [self.inf]*self.V 
        self.cost[self.start] = 0 

    def search(self):
        for i in range(self.V):
            update = False 
            for x, y, z in self.lis:
                if self.cost[y] > self.cost[x] + z:
                    self.cost[y] = self.cost[x] + z
                    update = True
            if not update:
                break
            
            if i == self.V - 1:
                self.close_minus = True
                return False
        return True

    def main(self):
        self.prepare()
        self.search()
    
    def cost_all(self):
        return self.cost

def main():
    n, m, p = map(int, input().split())
    
    graph = [None]*m
    tree_forth = [[] for _ in range(n)]
    tree_back = [[] for _ in range(n)]
    
    for i in range(m):
        a, b, c =map(int, input().split())
        tree_forth[a-1].append(b-1)
        tree_back[b-1].append(a-1)
        c = p-c
        graph[i] = (a-1, b-1, c)
    
    already1, already2 = [True]*n, [True]*n
    for already, tree, start in zip([already1, already2], [tree_forth, tree_back], [0, n-1]):
        not_yet = deque([start])
        already[start] = False
        while not_yet:
            key = not_yet.pop()
            for v in tree[key]:
                if already[v] == False:
                    continue
                already[v] = False
                not_yet.append(v)
    
    graph_new = []
    for i in range(m):
        a, b, c = graph[i]
        if already1[a] or already1[b] or already2[a] or already2[b]:
            continue
        graph_new.append((a, b, c))
       
    bf = Bellman_Ford(v=n, e=m, lis=graph_new)
    bf.main()
    if bf.close_minus:
        print(-1)
    else:
        print(max(0, -bf.cost[n-1]))
    
    
    
if __name__ == "__main__":
    main()

Verdicts: Runtime Error

Accepted