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.



Sample input:
3 3 10
1 2 20
2 3 30
1 3 45


Sample output:
35

Submission status: Wrong Answer

Solution code:
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")


n,m,p = map(int, input().split())
ns = [[] for _ in range(n)]
mc = -1
for _ in range(m):
    u,v,c = map(int, input().split())
    u -= 1
    v -= 1
    ns[u].append((p-c,v))
### ダイクストラ dijkstra
def dijkstra(start):
    import heapq
    vals = [None] * n
    h = [(0, start)] # (距離, ノード番号)
    vals[start] = 0
    while h:
        val, u = heapq.heappop(h)
        if val>vals[u]:
            continue
        for d, v in ns[u]:
            if vals[v] is None or vals[v]>val+d:
                vals[v] = val+d
                heapq.heappush(h, (vals[v], v))
    return vals
def bf(start):
    dist = [float("inf")]*n
    dist[start] = 0
    seen = [False] * n
    seen[start] = True
    for i in range(2*n):
        for u in range(n):
            if not seen[u]:
                continue
            for c,v in ns[u]:
                if dist[v] > dist[u] + c:
                    seen[v] = True
                    dist[v] = dist[u] + c
                    if i>=n-1 and v==n-1:
                        return -1
    return dist

dist = bf(0)
if dist==-1:
    ans = -1
else:
    ans = max(0, dist[n-1]*-1)
print(ans)

Submission status: Wrong Answer

Solution code:
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")


n,m,p = map(int, input().split())
ns = [[] for _ in range(n)]
mc = -1
for _ in range(m):
    u,v,c = map(int, input().split())
    u -= 1
    v -= 1
    ns[u].append((p-c,v))
### ダイクストラ dijkstra
def dijkstra(start):
    import heapq
    vals = [None] * n
    h = [(0, start)] # (距離, ノード番号)
    vals[start] = 0
    while h:
        val, u = heapq.heappop(h)
        if val>vals[u]:
            continue
        for d, v in ns[u]:
            if vals[v] is None or vals[v]>val+d:
                vals[v] = val+d
                heapq.heappush(h, (vals[v], v))
    return vals
def bf(start):
    dist = [float("inf")]*n
    dist[start] = 0
    seen = [False] * n
    seen[start] = True
    for i in range(n):
        for u in range(n):
            if not seen[u]:
                continue
            for c,v in ns[u]:
                if dist[v] > dist[u] + c:
                    seen[v] = True
                    dist[v] = dist[u] + c
                    if i==n-1 and v==n-1:
                        return -1
    return dist

dist = bf(0)
if dist==-1:
    ans = -1
else:
    ans = max(0, dist[n-1]*-1)
print(ans)

Submission status: Wrong Answer

Solution code:
import math
from collections import deque

def BellmanFord(edges, N, M, start):
    dist = [math.inf]*N
    dist[start] = 0

    for i in range(N):
        for j in range(M):
            if is_valid(edges[j][0], edges[j][1]):
                if dist[edges[j][1]] > dist[edges[j][0]] + edges[j][2]:
                    dist[edges[j][1]] = dist[edges[j][0]] + edges[j][2]
                    if i == N - 1:
                        return False

    return dist

def dfs(graph, N, start):
    visited = [False]*N
    visited[start] = True
    q = deque([start])

    while q:
        v = q.pop()
        for n_v in graph[v]:
            if not visited[n_v]:
                visited[n_v] = True
                q.append(n_v)
    
    return visited

def is_valid(f, t):
    return can_from_start[f] and can_from_start[t] and can_to_goal[f] and can_to_goal[t]


N, M, P = map(int, input().split())
graph = [[] for _ in range(N)]
graph_inv = [[] for _ in range(N)]
edges = []

for _ in range(M):
    A, B, C = map(int, input().split())
    graph[A - 1].append(B - 1)
    graph_inv[B - 1].append(A - 1)
    edges.append((A - 1, B - 1, - max(0, C - P)))

can_from_start = dfs(graph, N, 0)
can_to_goal = dfs(graph_inv, N, N - 1)

dist = BellmanFord(edges, N, M, 0)

if dist:
    print(-dist[N - 1])
else:
    print(-1)

Submission status: Wrong Answer

Solution code:
import math

def BellmanFord(edges, N, M, start, goal):
    dist = [math.inf]*N
    dist[start] = 0

    for i in range(2*N):
        for j in range(M):
            if dist[edges[j][1]] > dist[edges[j][0]] + edges[j][2] and dist[edges[j][0]] < math.inf:
                if i >= N - 1 and edges[j][1] == goal:
                    return False
                elif i >= N - 1:
                    dist[edges[j][1]] = -math.inf
                else:
                    dist[edges[j][1]] = dist[edges[j][0]] + edges[j][2]

    return dist

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, - max(0, C - P)))

dist = BellmanFord(edges, N, M, 0, N - 1)

if dist:
    print(-dist[N - 1])
else:
    print(-1)

Submission status: Wrong Answer

Solution code:
import math

def BellmanFord(edges, N, M, start, goal):
    dist = [math.inf]*N
    dist[start] = 0

    for i in range(2*N):
        for j in range(M):
            if dist[edges[j][1]] > dist[edges[j][0]] + edges[j][2]:
                if i >= N - 1 and edges[j][1] == goal:
                    return False
                elif i >= N - 1:
                    dist[edges[j][1]] = -math.inf
                else:
                    dist[edges[j][1]] = dist[edges[j][0]] + edges[j][2]

    return dist

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, - max(0, C - P)))

dist = BellmanFord(edges, N, M, 0, N - 1)

if dist:
    print(-dist[N - 1])
else:
    print(-1)

Submission status: Wrong Answer

Solution code:
from collections import deque
N,M,P = map(int,input().split())
INF = 10**18
edge = []
rev_edge = [[] for _ in range(N+1)]
dis = [INF]*(N+1)
dis[1] = 0
for i in range(M):
    A,B,C = map(int,input().split())
    edge.append((A,B,P-C))
    rev_edge[B].append(A)
d = deque()
can_arrive = set()
d.append(N)
for i in range(2500):
    if d:
        x = d.popleft()
        can_arrive.add(x)
        for y in rev_edge[x]:
            if not y in can_arrive:
                d.append(y)

for i in range(N):
    update = False
    for j in range(M):
        A,B,cost = edge[j]
        if dis[A] != INF and dis[B] > dis[A]+cost:
            dis[B] = dis[A]+cost
            if B in can_arrive:
                update = True
ans = dis[N]
if update:
    print(-1)
else:
    print(max(0,-ans))

Submission status: Wrong Answer

Solution code:
from collections import deque
N,M,P = map(int,input().split())
INF = 10**18
edge = []
rev_edge = [[] for _ in range(N+1)]
dis = [INF]*(N+1)
dis[1] = 0
for i in range(M):
    A,B,C = map(int,input().split())
    edge.append((A,B,P-C))
d = deque()
can_arrive = set()
d.append(N)
while d:
    x = d.popleft()
    can_arrive.add(x)
    for y in rev_edge[x]:
        if not y in can_arrive:
            d.append(y)

for i in range(N):
    update = False
    for j in range(M):
        A,B,cost = edge[j]
        if dis[A] != INF and dis[B] > dis[A]+cost:
            dis[B] = dis[A]+cost
            if B in can_arrive:
                update = True
ans = dis[N]
if update:
    print(-1)
else:
    print(max(0,-ans))

Submission status: Wrong Answer

Solution code:
N,M,P = map(int,input().split())
INF = 10**18
edge = []
dis = [INF]*(N+1)
dis[1] = 0
for i in range(M):
    A,B,C = map(int,input().split())
    edge.append((A,B,P-C))
for i in range(N):
    update = False
    for j in range(M):
        A,B,cost = edge[j]
        if dis[A] != INF and dis[B] > dis[A]+cost:
            dis[B] = dis[A]+cost
            update = True
ans = dis[N]
if update:
    ans = dis[N]
    for i in range(N):
        for j in range(M):
            A,B,cost = edge[j]
            if dis[A] != INF and dis[B] > dis[A]+cost:
                dis[B] = dis[A]+cost
    if dis[N] == ans:
        print(max(0,-ans))
    else:
        print(-1)
else:
    print(max(0,-ans))

Submission status: Wrong Answer

Solution code:
import bisect
import collections
import copy
import functools
import heapq
import math
import sys
from collections import deque
from collections import defaultdict
input = sys.stdin.readline
MOD = 10**9+7

N,M,P = map(int,(input().split()))
line = [[] for i in range(N)]
for i in range(M):
    a,b,c = map(int,(input().split()))
    line[a-1].append([b-1,-c+P])

min_distance = [float("inf")]*N # min_distance[i] は始点から街iまでの最短距離
min_distance[0] = 0 # 始点から始点への最短距離は0とする
exit()
def Bellman_Ford(S,V,E,distance):
    update = False
    flag = 0
    for _ in range(V):
      for s in range(N):
        for g,d in line[s]: # [s,g,d] = [道路の始点,道路の終点,道路の長さ] について、すべての道路を見る
            if min_distance[s] != float("inf") and min_distance[s] + d < min_distance[g]: # 最短経路が確立されている街から伸びる道路を通ることで、距離を短くできるならば、
                min_distance[g] = min_distance[s] + d # 最短距離を上書きする
                update = True
        if not update:
            flag = 1
            break
    return min_distance,flag

def Bellman_Ford2(S,V,E,distance):
    for _ in range(V):
      for s in range(N):
        for g,d in line[s]: # [s,g,d] = [道路の始点,道路の終点,道路の長さ] について、すべての道路を見る
            if min_distance[s] != float("inf") and min_distance[s] + d < min_distance[g]: # 最短経路が確立されている街から伸びる道路を通ることで、距離を短くできるならば、
                min_distance[g] = -float("inf")
                if g == N-1:
                    break
        else:
            continue
        break
    return min_distance

min_distance,flag = Bellman_Ford(1,N,M,line)

if flag == 1:
    ans = min_distance[N-1]
    print(max(-ans,0))
else:
    min_distance = Bellman_Ford2(1,N,M,line)
    ans = min_distance[N-1]
    if ans == -float("inf"):
        print(-1)
    else:
        print(max(-ans,0))

#print(min_distance)

Submission status: Wrong Answer

Solution code:
import bisect
import collections
import copy
import functools
import heapq
import math
import sys
from collections import deque
from collections import defaultdict
input = sys.stdin.readline
MOD = 10**9+7

N,M,P = map(int,(input().split()))
line = []
for i in range(M):
    a,b,c = map(int,(input().split()))
    line.append([a-1,b-1,-c+P])

min_distance = [float("inf")]*N # min_distance[i] は始点から街iまでの最短距離
min_distance[0] = 0 # 始点から始点への最短距離は0とする

def Bellman_Ford(S,V,E,distance):
    update = False
    flag = 0
    for _ in range(V):
        for s,g,d in distance: # [s,g,d] = [道路の始点,道路の終点,道路の長さ] について、すべての道路を見る
            if min_distance[s] != float("inf") and min_distance[s] + d < min_distance[g]: # 最短経路が確立されている街から伸びる道路を通ることで、距離を短くできるならば、
                min_distance[g] = min_distance[s] + d # 最短距離を上書きする
                update = True
        if not update:
            flag = 1
            break
    return min_distance,flag

def Bellman_Ford2(S,V,E,distance):
    for _ in range(V):
        for s,g,d in distance: # [s,g,d] = [道路の始点,道路の終点,道路の長さ] について、すべての道路を見る
            if min_distance[s] != float("inf") and min_distance[s] + d < min_distance[g]: # 最短経路が確立されている街から伸びる道路を通ることで、距離を短くできるならば、
                min_distance[g] = -float("inf")
                if g == N-1:
                    print(-1)
                    exit()
                    break
        else:
            continue
        break
    return min_distance

min_distance,flag = Bellman_Ford(1,N,M,line)
exit()

if flag == 1:
    ans = min_distance[N-1]
    print(max(-ans,0))
else:
    min_distance = Bellman_Ford2(1,N,M,line)
    ans = min_distance[N-1]
    if ans == -float("inf"):
        print(-1)
    else:
        print(max(-ans,0))

#print(min_distance)

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-

#############
# Libraries #
#############

import sys
input = sys.stdin.readline

import math
#from math import gcd
import bisect
import heapq
from collections import defaultdict
from collections import deque
from collections import Counter
from functools import lru_cache

#############
# Constants #
#############

MOD = 10**9+7
INF = float('inf')
AZ = "abcdefghijklmnopqrstuvwxyz"

#############
# Functions #
#############

######INPUT######
def I(): return int(input().strip())
def S(): return input().strip()
def IL(): return list(map(int,input().split()))
def SL(): return list(map(str,input().split()))
def ILs(n): return list(int(input()) for _ in range(n))
def SLs(n): return list(input().strip() for _ in range(n))
def ILL(n): return [list(map(int, input().split())) for _ in range(n)]
def SLL(n): return [list(map(str, input().split())) for _ in range(n)]

######OUTPUT######
def P(arg): print(arg); return
def Y(): print("Yes"); return
def N(): print("No"); return
def E(): exit()
def PE(arg): print(arg); exit()
def YE(): print("Yes"); exit()
def NE(): print("No"); exit()

#####Shorten#####
def DD(arg): return defaultdict(arg)

#####Inverse#####
def inv(n): return pow(n, MOD-2, MOD)

######Combination######
kaijo_memo = []
def kaijo(n):
  if(len(kaijo_memo) > n): return kaijo_memo[n]
  if(len(kaijo_memo) == 0): kaijo_memo.append(1)
  while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
  return kaijo_memo[n]

gyaku_kaijo_memo = []
def gyaku_kaijo(n):
  if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
  if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
  while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
  return gyaku_kaijo_memo[n]

def nCr(n,r):
  if n == r: return 1
  if n < r or r < 0: return 0
  ret = 1
  ret = ret * kaijo(n) % MOD
  ret = ret * gyaku_kaijo(r) % MOD
  ret = ret * gyaku_kaijo(n-r) % MOD
  return ret

######Factorization######
def factorization(n):
  arr = []
  temp = n
  for i in range(2, int(-(-n**0.5//1))+1):
    if temp%i==0:
      cnt=0
      while temp%i==0: 
        cnt+=1 
        temp //= i
      arr.append([i, cnt])
  if temp!=1: arr.append([temp, 1])
  if arr==[]: arr.append([n, 1])
  return arr

#####MakeDivisors######
def make_divisors(n):
  divisors = []
  for i in range(1, int(n**0.5)+1):
    if n % i == 0:
      divisors.append(i)
      if i != n // i: 
        divisors.append(n//i)
  return divisors

#####MakePrimes######
def make_primes(N):
  max = int(math.sqrt(N))
  seachList = [i for i in range(2,N+1)]
  primeNum = []
  while seachList[0] <= max:
    primeNum.append(seachList[0])
    tmp = seachList[0]
    seachList = [i for i in seachList if i % tmp != 0]
  primeNum.extend(seachList)
  return primeNum

#####GCD#####
def gcd(a, b):
    while b: a, b = b, a % b
    return a

#####LCM#####
def lcm(a, b):
    return a * b // gcd (a, b)

#####BitCount#####
def count_bit(n):
  count = 0
  while n:
    n &= n-1
    count += 1
  return count

#####ChangeBase#####
def base_10_to_n(X, n):
  if X//n: return base_10_to_n(X//n, n)+[X%n]
  return [X%n]

def base_n_to_10(X, n):
  return sum(int(str(X)[-i-1])*n**i for i in range(len(str(X))))

def base_10_to_n_without_0(X, n):
  X -= 1
  if X//n: return base_10_to_n_without_0(X//n, n)+[X%n]
  return [X%n]

#####IntLog#####
def int_log(n, a):
  count = 0
  while n>=a:
    n //= a
    count += 1
  return count

#############
# Main Code #
#############

N,M,P = IL()

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

s = 0
d = [INF] * N
d[s] = 0
update = True
for _ in range(N):
  update = False
  for v, e in enumerate(graph):
    for t, cost in e:
      if d[v] != INF and d[v] + cost < d[t]:
        d[t] = d[v] + cost
        update = True
  if not update:
    break
else:
  for _ in range(3*N):
    for v, e in enumerate(graph):
      for t, cost in e:
        if d[v] != INF and d[v] + cost < d[t]:
          d[t] = d[v] + cost
          if t==N-1:
            print(-1)
            exit()

print(max(-d[-1],0))

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-

#############
# Libraries #
#############

import sys
input = sys.stdin.readline

import math
#from math import gcd
import bisect
import heapq
from collections import defaultdict
from collections import deque
from collections import Counter
from functools import lru_cache

#############
# Constants #
#############

MOD = 10**9+7
INF = float('inf')
AZ = "abcdefghijklmnopqrstuvwxyz"

#############
# Functions #
#############

######INPUT######
def I(): return int(input().strip())
def S(): return input().strip()
def IL(): return list(map(int,input().split()))
def SL(): return list(map(str,input().split()))
def ILs(n): return list(int(input()) for _ in range(n))
def SLs(n): return list(input().strip() for _ in range(n))
def ILL(n): return [list(map(int, input().split())) for _ in range(n)]
def SLL(n): return [list(map(str, input().split())) for _ in range(n)]

######OUTPUT######
def P(arg): print(arg); return
def Y(): print("Yes"); return
def N(): print("No"); return
def E(): exit()
def PE(arg): print(arg); exit()
def YE(): print("Yes"); exit()
def NE(): print("No"); exit()

#####Shorten#####
def DD(arg): return defaultdict(arg)

#####Inverse#####
def inv(n): return pow(n, MOD-2, MOD)

######Combination######
kaijo_memo = []
def kaijo(n):
  if(len(kaijo_memo) > n): return kaijo_memo[n]
  if(len(kaijo_memo) == 0): kaijo_memo.append(1)
  while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
  return kaijo_memo[n]

gyaku_kaijo_memo = []
def gyaku_kaijo(n):
  if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
  if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
  while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
  return gyaku_kaijo_memo[n]

def nCr(n,r):
  if n == r: return 1
  if n < r or r < 0: return 0
  ret = 1
  ret = ret * kaijo(n) % MOD
  ret = ret * gyaku_kaijo(r) % MOD
  ret = ret * gyaku_kaijo(n-r) % MOD
  return ret

######Factorization######
def factorization(n):
  arr = []
  temp = n
  for i in range(2, int(-(-n**0.5//1))+1):
    if temp%i==0:
      cnt=0
      while temp%i==0: 
        cnt+=1 
        temp //= i
      arr.append([i, cnt])
  if temp!=1: arr.append([temp, 1])
  if arr==[]: arr.append([n, 1])
  return arr

#####MakeDivisors######
def make_divisors(n):
  divisors = []
  for i in range(1, int(n**0.5)+1):
    if n % i == 0:
      divisors.append(i)
      if i != n // i: 
        divisors.append(n//i)
  return divisors

#####MakePrimes######
def make_primes(N):
  max = int(math.sqrt(N))
  seachList = [i for i in range(2,N+1)]
  primeNum = []
  while seachList[0] <= max:
    primeNum.append(seachList[0])
    tmp = seachList[0]
    seachList = [i for i in seachList if i % tmp != 0]
  primeNum.extend(seachList)
  return primeNum

#####GCD#####
def gcd(a, b):
    while b: a, b = b, a % b
    return a

#####LCM#####
def lcm(a, b):
    return a * b // gcd (a, b)

#####BitCount#####
def count_bit(n):
  count = 0
  while n:
    n &= n-1
    count += 1
  return count

#####ChangeBase#####
def base_10_to_n(X, n):
  if X//n: return base_10_to_n(X//n, n)+[X%n]
  return [X%n]

def base_n_to_10(X, n):
  return sum(int(str(X)[-i-1])*n**i for i in range(len(str(X))))

def base_10_to_n_without_0(X, n):
  X -= 1
  if X//n: return base_10_to_n_without_0(X//n, n)+[X%n]
  return [X%n]

#####IntLog#####
def int_log(n, a):
  count = 0
  while n>=a:
    n //= a
    count += 1
  return count

#############
# Main Code #
#############

N,M,P = IL()

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

s = 0
d = [INF] * N
d[s] = 0
update = True
for _ in range(N):
  update = False
  for v, e in enumerate(graph):
    for t, cost in e:
      if d[v] != INF and d[v] + cost < d[t]:
        d[t] = d[v] + cost
        update = True
  if not update:
    break
else:
  for _ in range(N):
    for v, e in enumerate(graph):
      for t, cost in e:
        if d[v] != INF and d[v] + cost < d[t]:
          d[t] = d[v] + cost
          if t==N-1:
            print(-1)
            exit()

print(max(-d[-1],0))

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-

#############
# Libraries #
#############

import sys
input = sys.stdin.readline

import math
#from math import gcd
import bisect
import heapq
from collections import defaultdict
from collections import deque
from collections import Counter
from functools import lru_cache

#############
# Constants #
#############

MOD = 10**9+7
INF = float('inf')
AZ = "abcdefghijklmnopqrstuvwxyz"

#############
# Functions #
#############

######INPUT######
def I(): return int(input().strip())
def S(): return input().strip()
def IL(): return list(map(int,input().split()))
def SL(): return list(map(str,input().split()))
def ILs(n): return list(int(input()) for _ in range(n))
def SLs(n): return list(input().strip() for _ in range(n))
def ILL(n): return [list(map(int, input().split())) for _ in range(n)]
def SLL(n): return [list(map(str, input().split())) for _ in range(n)]

######OUTPUT######
def P(arg): print(arg); return
def Y(): print("Yes"); return
def N(): print("No"); return
def E(): exit()
def PE(arg): print(arg); exit()
def YE(): print("Yes"); exit()
def NE(): print("No"); exit()

#####Shorten#####
def DD(arg): return defaultdict(arg)

#####Inverse#####
def inv(n): return pow(n, MOD-2, MOD)

######Combination######
kaijo_memo = []
def kaijo(n):
  if(len(kaijo_memo) > n): return kaijo_memo[n]
  if(len(kaijo_memo) == 0): kaijo_memo.append(1)
  while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
  return kaijo_memo[n]

gyaku_kaijo_memo = []
def gyaku_kaijo(n):
  if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
  if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
  while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
  return gyaku_kaijo_memo[n]

def nCr(n,r):
  if n == r: return 1
  if n < r or r < 0: return 0
  ret = 1
  ret = ret * kaijo(n) % MOD
  ret = ret * gyaku_kaijo(r) % MOD
  ret = ret * gyaku_kaijo(n-r) % MOD
  return ret

######Factorization######
def factorization(n):
  arr = []
  temp = n
  for i in range(2, int(-(-n**0.5//1))+1):
    if temp%i==0:
      cnt=0
      while temp%i==0: 
        cnt+=1 
        temp //= i
      arr.append([i, cnt])
  if temp!=1: arr.append([temp, 1])
  if arr==[]: arr.append([n, 1])
  return arr

#####MakeDivisors######
def make_divisors(n):
  divisors = []
  for i in range(1, int(n**0.5)+1):
    if n % i == 0:
      divisors.append(i)
      if i != n // i: 
        divisors.append(n//i)
  return divisors

#####MakePrimes######
def make_primes(N):
  max = int(math.sqrt(N))
  seachList = [i for i in range(2,N+1)]
  primeNum = []
  while seachList[0] <= max:
    primeNum.append(seachList[0])
    tmp = seachList[0]
    seachList = [i for i in seachList if i % tmp != 0]
  primeNum.extend(seachList)
  return primeNum

#####GCD#####
def gcd(a, b):
    while b: a, b = b, a % b
    return a

#####LCM#####
def lcm(a, b):
    return a * b // gcd (a, b)

#####BitCount#####
def count_bit(n):
  count = 0
  while n:
    n &= n-1
    count += 1
  return count

#####ChangeBase#####
def base_10_to_n(X, n):
  if X//n: return base_10_to_n(X//n, n)+[X%n]
  return [X%n]

def base_n_to_10(X, n):
  return sum(int(str(X)[-i-1])*n**i for i in range(len(str(X))))

def base_10_to_n_without_0(X, n):
  X -= 1
  if X//n: return base_10_to_n_without_0(X//n, n)+[X%n]
  return [X%n]

#####IntLog#####
def int_log(n, a):
  count = 0
  while n>=a:
    n //= a
    count += 1
  return count

#############
# Main Code #
#############

N,M,P = IL()

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

s = 0
d = [INF] * N
d[s] = 0
update = True
for _ in range(N):
  update = [0 for i in range(N)]
  for v, e in enumerate(graph):
    for t, cost in e:
      if d[v] != INF and d[v] + cost < d[t]:
        d[t] = d[v] + cost
        update[t] = 1
  if not max(update):
    break
else:
  if update[-1]:
    print(-1)
    exit()
  else:
    for _ in range(N):
      for v, e in enumerate(graph):
        for t, cost in e:
          if d[v] != INF and d[v] + cost < d[t]:
            d[t] = d[v] + cost
            if t==N-1:
              print(-1)
              exit()

print(max(-d[-1],0))

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-

#############
# Libraries #
#############

import sys
input = sys.stdin.readline

import math
#from math import gcd
import bisect
import heapq
from collections import defaultdict
from collections import deque
from collections import Counter
from functools import lru_cache

#############
# Constants #
#############

MOD = 10**9+7
INF = float('inf')
AZ = "abcdefghijklmnopqrstuvwxyz"

#############
# Functions #
#############

######INPUT######
def I(): return int(input().strip())
def S(): return input().strip()
def IL(): return list(map(int,input().split()))
def SL(): return list(map(str,input().split()))
def ILs(n): return list(int(input()) for _ in range(n))
def SLs(n): return list(input().strip() for _ in range(n))
def ILL(n): return [list(map(int, input().split())) for _ in range(n)]
def SLL(n): return [list(map(str, input().split())) for _ in range(n)]

######OUTPUT######
def P(arg): print(arg); return
def Y(): print("Yes"); return
def N(): print("No"); return
def E(): exit()
def PE(arg): print(arg); exit()
def YE(): print("Yes"); exit()
def NE(): print("No"); exit()

#####Shorten#####
def DD(arg): return defaultdict(arg)

#####Inverse#####
def inv(n): return pow(n, MOD-2, MOD)

######Combination######
kaijo_memo = []
def kaijo(n):
  if(len(kaijo_memo) > n): return kaijo_memo[n]
  if(len(kaijo_memo) == 0): kaijo_memo.append(1)
  while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
  return kaijo_memo[n]

gyaku_kaijo_memo = []
def gyaku_kaijo(n):
  if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
  if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
  while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
  return gyaku_kaijo_memo[n]

def nCr(n,r):
  if n == r: return 1
  if n < r or r < 0: return 0
  ret = 1
  ret = ret * kaijo(n) % MOD
  ret = ret * gyaku_kaijo(r) % MOD
  ret = ret * gyaku_kaijo(n-r) % MOD
  return ret

######Factorization######
def factorization(n):
  arr = []
  temp = n
  for i in range(2, int(-(-n**0.5//1))+1):
    if temp%i==0:
      cnt=0
      while temp%i==0: 
        cnt+=1 
        temp //= i
      arr.append([i, cnt])
  if temp!=1: arr.append([temp, 1])
  if arr==[]: arr.append([n, 1])
  return arr

#####MakeDivisors######
def make_divisors(n):
  divisors = []
  for i in range(1, int(n**0.5)+1):
    if n % i == 0:
      divisors.append(i)
      if i != n // i: 
        divisors.append(n//i)
  return divisors

#####MakePrimes######
def make_primes(N):
  max = int(math.sqrt(N))
  seachList = [i for i in range(2,N+1)]
  primeNum = []
  while seachList[0] <= max:
    primeNum.append(seachList[0])
    tmp = seachList[0]
    seachList = [i for i in seachList if i % tmp != 0]
  primeNum.extend(seachList)
  return primeNum

#####GCD#####
def gcd(a, b):
    while b: a, b = b, a % b
    return a

#####LCM#####
def lcm(a, b):
    return a * b // gcd (a, b)

#####BitCount#####
def count_bit(n):
  count = 0
  while n:
    n &= n-1
    count += 1
  return count

#####ChangeBase#####
def base_10_to_n(X, n):
  if X//n: return base_10_to_n(X//n, n)+[X%n]
  return [X%n]

def base_n_to_10(X, n):
  return sum(int(str(X)[-i-1])*n**i for i in range(len(str(X))))

def base_10_to_n_without_0(X, n):
  X -= 1
  if X//n: return base_10_to_n_without_0(X//n, n)+[X%n]
  return [X%n]

#####IntLog#####
def int_log(n, a):
  count = 0
  while n>=a:
    n //= a
    count += 1
  return count

#############
# Main Code #
#############

N,M,P = IL()

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

s = 0
d = [INF] * N
d[s] = 0
update = True
for _ in range(N):
  update = [0 for i in range(N)]
  for v, e in enumerate(graph):
    for t, cost in e:
      if d[v] != INF and d[v] + cost < d[t]:
        d[t] = min(0,d[v] + cost)
        update[t] = 1
  if not max(update):
    break
else:
  if update[-1]:
    print(-1)
    exit()
print(-1)
#print(max(-d[-1],0))