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: Accepted

Solution code:
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub

sys.setrecursionlimit(100000)
input = sys.stdin.readline


def read_int():
    return int(input())


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


def read_float():
    return float(input())


def read_float_n():
    return list(map(float, input().split()))


def read_str():
    return input().strip()


def read_str_n():
    return list(map(str, input().split()))


def error_print(*args):
    print(*args, file=sys.stderr)


def mt(f):
    import time

    def wrap(*args, **kwargs):
        s = time.time()
        ret = f(*args, **kwargs)
        e = time.time()

        error_print(e - s, 'sec')
        return ret

    return wrap


def BellmanFord(N, e, s, goal):
    dist = [sys.maxsize] * (N+1)
    dist[s] = 0
    for i in range(N+1):
        for u, v, c in e:
            if dist[v] > dist[u] + c:
                if i == N:
                    return 'inf'
                dist[v] = dist[u] + c
    return dist

@mt
def slv(N, M, P, ABC):
    g = [[] for _ in range(N+1)]
    for a, b, c in ABC:
        g[a].append(b)

    def r(s, e):
        q = deque()
        q.append(s)
        done = set([s])
        while q:
            u = q.popleft()
            for v in g[u]:
                if v not in done:
                    done.add(v)
                    q.append(v)
        return done

    d = set()
    for i in range(2, N):
        reachable = r(i, N)
        if N not in reachable:
            d.add(i)
    
    e = []
    sr = r(1, N)
    for a, b, c in ABC:
        if a not in d and b not in d and a in sr and b in sr:
            e.append((a, b, -(c - P)))

    d = BellmanFord(N, e, 1, N)
    if d == 'inf':
        return -1
    return max(0, -d[N])


def main():
    N, M, P = read_int_n()
    ABC = [read_int_n() for _ in range(M)]
    print(slv(N, M, P, ABC))

    # N = 2500
    # M = 5000
    # P = random.randint(0, 10**5)
    # P= 0
    # ABC = [
    #     [random.randint(1, N), random.randint(1, N), random.randint(1, 10**5)] for _ in range(M)
    # ]
    # print(slv(N, M, P, ABC))



if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
n,m,p = map(int,input().split())
graph = [[] for _ in range(n)]
INF = float("inf")
dist = [INF]*n
dist[0] = 0

for _ in range(m):
	f,t,c = map(int,input().split())
	graph[f-1].append((t-1,p-c))
#print(graph)

for _ in range(n):
	update = False
	for f,e in enumerate(graph):
		for t,c in e:
			if dist[f] != INF and dist[t] > dist[f] + c:
				dist[t] = dist[f] + c
				update = True
	if not update:
		break
else:
	prev = dist[-1]
	for _ in range(n):
		for f,e in enumerate(graph):
			for t,c in e:
				if dist[f] != INF and dist[t] > dist[f] + c:
					dist[t] = -INF
	if dist[-1] != prev:
		print(-1)
		exit()
#print(dist)

print(max(0,dist[-1]*(-1)))

Submission status: Accepted

Solution code:
def shortest_path(s,n,w,es):
    #s→iの最短距離
    # s:始点, n:頂点数, w:辺の数, es[i]: [辺の始点,辺の終点,辺のコスト]
    d = [float("inf")] * n
    #d[i] : s→iの最短距離
    d[s] = 0
    for _ in [0]*n:
        update = False
        for p,q,r in es:
            # e: 辺iについて [from,to,cost]
            if d[p] != float("inf") and d[q] > d[p] + r:
                d[q] = d[p] + r
                update = True
        if not update:
            break
    return d

def main():
    import sys
    input = sys.stdin.readline
    n,m,p = map(int,input().split())
    es = []
    edge = [[] for _ in range(n)]

    for _ in range(m):
        a,b,c = map(int,input().split())
        es.append([a-1,b-1,p-c])
        edge[b-1].append(a-1)
    
    ren = [False]*n
    tank = [n-1]
    while len(tank) != 0:
        new = []
        for elt in tank:
            for go in edge[elt]:
                if ren[go]:
                    continue
                ren[go] = True
                new.append(go)
        tank = new

    d = shortest_path(0,n,m,es)

    for p,q,r in es:
        # e: 辺iについて [from,to,cost]
        if d[p] != float("inf") and d[q] > d[p] + r:
            d[q] = d[p] + r
            if ren[q]:
                print(-1)
                exit()

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

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
import sys
stdin = sys.stdin

sys.setrecursionlimit(10 ** 7)

def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x) - 1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())


def dfs(graph: list, node: int, start: int) -> list:
    # 未探索のノードは距離INF
    INF = float("inf")
    dist = [INF] * node

    # 始点ノードの距離を0とし、dfsのためのstackを作成
    dist[start] = 0
    stack = [(0, start)]

    while stack:
        cost, cur_node = stack.pop()

        # 未探索のノードをstackに入れる
        for nex_cost, nex_node in graph[cur_node]:
            if dist[nex_node] != INF:
                continue
            else:
                dist[nex_node] = dist[cur_node] + nex_cost
                stack.append((dist[nex_node], nex_node))

    return dist


n, m, p = li()
edges = []
for _ in range(m):
    ai, bi, ci = li()
    ai -= 1
    bi -= 1
    ci -= p
    edges.append((ai, bi, ci))

# グラフ構築
one2n = [[] for _ in range(n)]
n2one = [[] for _ in range(n)]
for ai, bi, _ in edges:
    one2n[ai].append((ci, bi))
    n2one[bi].append((ci, ai))

# 1から到達可能な点を抽出
one2n_dist = dfs(one2n, n, 0)
one2n_reachable = set()
for i, di in enumerate(one2n_dist):
    if di != float('inf'):
        one2n_reachable.add(i)

# Nから到達可能な点を抽出
n2one_dist = dfs(n2one, n, n-1)
n2one_reachable = set()
for i, di in enumerate(n2one_dist):
    if di != float('inf'):
        n2one_reachable.add(i)

one2n_path = one2n_reachable & n2one_reachable


# ベルマンフォード
INF = float('inf')
dist = [-INF]*n
dist[0] = 0

for _ in range(n):
    for ai, bi, ci in edges:
        if ai in one2n_path and bi in one2n_path and dist[ai] + ci > dist[bi]:
            dist[bi] = dist[ai] + ci

aftern = dist[n-1]

modified = False

for ai, bi, ci in edges:
    if ai in one2n_path and bi in one2n_path and dist[ai] + ci > dist[bi]:
        modified = True

# N+1回で更新されていれば-1
if modified:
    print(-1)
else:
    print(max(0, aftern))

Submission status: Accepted

Solution code:
import sys
stdin = sys.stdin

sys.setrecursionlimit(10 ** 7)

def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x) - 1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())


def dfs(graph: list, node: int, start: int) -> list:
    # 未探索のノードは距離INF
    INF = float("inf")
    dist = [INF] * node

    # 始点ノードの距離を0とし、dfsのためのstackを作成
    dist[start] = 0
    stack = [(0, start)]

    while stack:
        cost, cur_node = stack.pop()

        # 未探索のノードをstackに入れる
        for nex_cost, nex_node in graph[cur_node]:
            if dist[nex_node] != INF:
                continue
            else:
                dist[nex_node] = dist[cur_node] + nex_cost
                stack.append((dist[nex_node], nex_node))

    return dist


n, m, p = li()
edges = []
for _ in range(m):
    ai, bi, ci = li()
    ai -= 1
    bi -= 1
    ci -= p
    edges.append((ai, bi, ci))

# グラフ構築
one2n = [[] for _ in range(n)]
n2one = [[] for _ in range(n)]
for ai, bi, _ in edges:
    one2n[ai].append((ci, bi))
    n2one[bi].append((ci, ai))

# 1から到達可能な点を抽出
one2n_dist = dfs(one2n, n, 0)
one2n_reachable = set()
for i, di in enumerate(one2n_dist):
    if di != float('inf'):
        one2n_reachable.add(i)

# Nから到達可能な点を抽出
n2one_dist = dfs(n2one, n, n-1)
n2one_reachable = set()
for i, di in enumerate(n2one_dist):
    if di != float('inf'):
        n2one_reachable.add(i)

one2n_path = one2n_reachable & n2one_reachable


# ベルマンフォード
INF = float('inf')
dist = [-INF]*n
dist[0] = 0

for _ in range(n):
    for ai, bi, ci in edges:
        if ai in one2n_path and bi in one2n_path and dist[ai] + ci > dist[bi]:
            dist[bi] = dist[ai] + ci

aftern = dist[n-1]

modified = False
for _ in range(n):
    for ai, bi, ci in edges:
        if ai in one2n_path and bi in one2n_path and dist[ai] + ci > dist[bi]:
            modified = True

# N回更新と2N回更新を比較
# 2N回で更新されていれば-1
if modified:
    print(-1)
else:
    print(max(0, aftern))

Submission status: Accepted

Solution code:
inf = float('inf')
V,E,P = 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,P-C])
    edge.append([A-1,B-1,P-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(-1)

else:
    ans = -d[-1]
    print(max(0,ans))

Submission status: Accepted

Solution code:
n,m,p = map(int, input().split())
es = []
g1 = [[]for _ in range(n)]
g2 = [[] for _ in range(n)]
li = []
for i in range(m):
  a,b,c = map(int, input().split())
  li.append((a - 1,b - 1,c))
  g1[a - 1].append((b - 1,(c-p)*-1))
  g2[b - 1].append((a - 1,(c-p)*-1))
done1 = [0] * n
done1[0] = 1
s = [0]
while s:
  d = s.pop()
  for node,_ in g1[d]:
    if not done1[node]:
      done1[node] = 1
      s.append(node)
done2 = [0] * n
done2[n - 1] = 1
s = [n - 1]
while s:
  d = s.pop()
  for node, _ in g2[d]:
    if not done2[node]:
      done2[node] = 1
      s.append(node)
done = [0] * n
for i in range(n):
  done[i] = done1[i] & done2[i]
for a,b,c in li:
  if done[a] & done[b]:
    es.append((a,b,(c-p)*-1))
m = len(es)
d = [0]*n
for i in range(n):
  for j in range(m):
    f, t, c = es[j]
    if d[t] > d[f] + c:
      d[t] = d[f] + c
      if i == n - 1:
        print(-1)
        exit(0)
d = [float('inf')] * n
d[0] = 0
while(True):
  update = False
  for i in range(m):
    f, t, c = es[i]
    if d[f] != float('inf') and d[t] > d[f] + c:
      d[t] = d[f] + c
      update = True
  if not update:
    break
print(max(0, -d[n - 1]))

Submission status: Accepted

Solution code:
from collections import deque
import sys
input = sys.stdin.readline


def bellman_ford(N, adj, start):
    ans = [float("inf")] * N
    q = deque()
    q.append((start, 0, 0))
    while len(q):
        i, cost, path = q.pop()
        if cost >= ans[i]:
            continue
        ans[i] = cost
        for j in adj[i]:
            ncost = cost + adj[i][j]
            if path > N:
                ncost = -float("inf")
            q.append((j, ncost, path+1))
    return ans


def main():
    N, M, P = map(int, input().split())
    adj = [{} for _ in range(N)]
    for _ in range(M):
        A, B, C = map(int, input().split())
        if B-1 in adj[A-1]:
            adj[A-1][B-1] = min(P-C, adj[A-1][B-1])
        else:
            adj[A-1][B-1] = P-C
    costs = bellman_ford(N, adj, 0)
    ans = -costs[N-1]
    if ans < 0:
        print(0)
    elif ans == float("inf"):
        print(-1)
    else:
        print(ans)


if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.readline

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])

    reachable_from = [False for _ in range(N)]
    reachable_from[0] = True
    reachable_to = [False for _ in range(N)]
    reachable_to[N-1] = True

    for i in range(N-1):
        for edge in edges:
            if reachable_from[edge[0]]:
                reachable_from[edge[1]] = True
            if reachable_to[edge[1]]:
                reachable_to[edge[0]] = True

    edges2 = []
    for edge in edges:
        if reachable_from[edge[0]] and reachable_to[edge[1]]:
            edges2.append(edge)

    inf = 10000*(10**5)
    dist = [inf for _ in range(N)]
    dist[0] = 0

    for i in range(N):
        for edge in edges2:
            if dist[edge[1]] > dist[edge[0]] + edge[2]:
                dist[edge[1]] = dist[edge[0]]+edge[2]
                if i==N-1:
                    print(-1)
                    break
        else:
            continue
        break
    else:
        print(max(0,-1*dist[N-1]))

if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
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))  # 辺の重みの正負反転（最長路→最短路にしたいので）

  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()

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, 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()