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):
import sys
inf = 10000000000

def BellmanFord(edges,points,start):
  
  dist=[inf for i in range(points)]
  dist[start-1]=0
  
  
  for i in range(points):
    for edge in edges:
      if dist[edge[0]-1]!=inf and  dist[edge[1]-1] > dist[edge[0]-1] + edge[2]:
        dist[edge[1]-1] = dist[edge[0]-1] + edge[2]
    dist1 = dist[N-1]
  for i in range(points):
    for edge in edges:
      if dist[edge[0]-1]!=inf and  dist[edge[1]-1] > dist[edge[0]-1] + edge[2]:
        dist[edge[1]-1] = dist[edge[0]-1] + edge[2]
    dist2 = dist[N-1]
  if dist1 != dist2:return "Loop"
  return (dist)

N,M,P = [int(i) for i in input().split()]
Edges = []
for i in range(M):
    a,b,c = [int(i) for i in input().split()]
    Edges.append((a,b,P-c))
dist = BellmanFord(Edges,N,1)
if dist =="Loop":
    print(-1)
else:
    print(max(-dist[N-1],0))

Solution (Python):
N, M, P = map(int, input().split())
edge = []

for i in range(M):
    A, B, C = map(int, input().split())
    edge.append((A, B, P-C))

inf = float('inf')
cost = [inf] * (N+1)
cost[1] = 0
for i in range(N-1):
    for e in edge:
        a, b, c = e
        if cost[a] + c < cost[b]:
            cost[b] = cost[a] + c

for i in range(N):
    for e in edge:
        a, b, c = e
        if cost[a] + c < cost[b]:
            if b == N:
                print(-1)
                exit()
            else:
                cost[b] = cost[a] + c
        
print(max(-cost[N], 0))

Solution (Python):
N, M, P = map(int, input().split())
E = [list(map(int, input().split())) for _ in range(M)]
A = [-float("inf")] * (N+1)
A[1] = 0
for v in range(N):
    for a, b, c in E:
        t = A[a] + c - P
        if t > A[b]:
            A[b] = t
ans = A[N]
for v in range(N):
    for a, b, c in E:
        t = A[a] + c - P
        if t > A[b]:
            A[b] = t
if ans!=A[N]:
    print(-1)
else:
    print(max(0, ans))

Solution (Python):
n,m,p = map(int, input().split())
G = [[] for i in range(n)]
for i in range(m):
    a,b,c = map(int, input().split())
    a,b = a-1, b-1
    G[a].append((b, p-c))


INF = 10**18
dist = [INF] * n
dist[0] = 0
for i in range(2*n):
    for v, e in enumerate(G):
        if dist[v] == INF:
            continue
        for t, cost in e:
            if dist[v] + cost < dist[t]:
                dist[t] = dist[v] + cost
                
                if i >= n-1 and t == n-1:
                    print(-1)
                    exit()
print(max(0, -dist[-1]))

Solution (Python):
import sys

stdin = sys.stdin

ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip()  

n, m, p = na()

def bf(es, n, start):
    d = [999999999999] * n
    d[start] = 0

    for r in range(n-1):
        updated = False
        for e in es:
            if d[e[0]] == 999999999999:
                continue
            if d[e[0]] + e[2] < d[e[1]]:
                d[e[1]] = d[e[0]] + e[2]
                updated = True
        if not updated:
            break

    for e in es:
        if d[e[0]] < 999999999999 and d[e[0]] + e[2] < d[e[1]]:
            return None

    return d

es = []
for i in range(m):
    a, b, c = na()
    es.append((a-1, b-1, -(c-p)))

import collections
q = collections.deque()
q.append(n-1)

valid = [False] * n
valid[n-1] = True
while len(q) > 0:
    cur = q.popleft()
    for e in es:
        if e[1] == cur and not valid[e[0]]:
            valid[e[0]] = True
            q.append(e[0])

nes = []
for e in es:
    if valid[e[0]] and valid[e[1]]:
        nes.append(e)
es = nes

d = bf(es, n, 0)
if not d:
    print(-1)
else:
    print(max(0, -d[n-1]))

Solution (Python):



def Bellman_Ford(g,start):
    n = len(g)    
    dist = [float("inf")]*n
    dist[start] = 0
    a0 = float("inf")
    
    for i in range(n<<1):
        updated = False
        for v in range(n):
            for to, cost in g[v]:
                if dist[to] <= dist[v] + cost: continue
                dist[to] = dist[v] + cost
                updated = True
        if not updated: break
        if i <= n: a0 = dist[-1]
    if a0 != dist[-1]: return None, False
    return dist, True


import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline 

n,m,p = [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+p))


dist, TF = Bellman_Ford(g, 0)

if TF: print(max(-dist[-1],0))
else: print(-1)