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):
N,M,P = map(int,input().split())
edge = []
cost = [-float('inf')]*N
cost[0] = 0
for i in range(M):
    a,b,c = map(int,input().split())
    edge.append([a-1,b-1,c-P])

for i in range(N-1):
    for a,b,c in edge:
        cost[b] = max(cost[b],cost[a]+c)


for a,b,c in edge:
    if cost[b] < cost[a]+c:
        cost[b] = float('inf')


for i in range(N-1):
    for a,b,c in edge:
        cost[b] = max(cost[b],cost[a]+c)

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

print(max(-cost[N], 0))

Solution (Python):
import sys
input = sys.stdin.readline
sys.setrecursionlimit(100000)
 
def getN():
    return int(input())
def getList():
    return list(map(int, input().split()))
import math
import queue
 
NINF = -1000000000
 
def reachable(n, vtxs):
    visited = [0 for i in range(n)]
    visited[0] = 1
 
    qu = queue.Queue()
    qu.put(0)
    while(not qu.empty()):
        curr = qu.get()
        for next in vtxs[curr]:
            if visited[next] == 0:
                visited[next] = 1
                qu.put(next)
 
    return visited
 
def reachable_rev(n, revs):
    visited = [0 for i in range(n)]
    visited[n-1] = 1
    qu = queue.Queue()
    qu.put(n - 1)
    while(not qu.empty()):
        curr = qu.get()
        for next in revs[curr]:
            if visited[next] == 0:
                visited[next] = 1
                qu.put(next)
 
    return visited
 
 
 
def bellman(n, m, edges, vtxs, revs):
    cost = [NINF for i in range(n)]
    cost[0] = 0
    tgts = [0]
    loop = 0
    final = False
    reach = reachable(n, vtxs)
    revreach = reachable_rev(n, revs)
    while(True):
        flag = False
        if loop > n:
            final = True
            break
        for e in edges:
            
            if cost[e[1]] < cost[e[0]] + e[2]:
                point = e[1]
                ppoint = e[0]
                
                if reach[point] and revreach[point]:
                    if reach[ppoint] and revreach[ppoint]:
                        cost[e[1]] = cost[e[0]] + e[2]
                        flag = True
                        
        if not flag:
            break
        loop += 1
        
 
 
    
    if final:
        for e in edges:
            
            if cost[e[1]] < cost[e[0]] + e[2]:
                point = e[1]
                ppoint = e[0]
                
                if reach[point] and revreach[point]:
                    if reach[ppoint] and revreach[ppoint]:
                        return -1
 
    if reach[-1] == 0:
        return -1
 
    return max(0, cost[-1])
 
n, m , p= getList()
vtxs = [[] for i in range(n)]
revs = [[] for i in range(n)]
edges = []
 
for i in range(m):
    a, b, c = getList()
    vtxs[a-1].append(b-1)
    revs[b-1].append(a-1)
    edges.append((a-1, b-1, c - p))
 

print(bellman(n, m, edges, vtxs, revs))

Solution (Python):
import sys
sys.setrecursionlimit(10**9)

def BF(edges,n,s):
    inf=float("inf")
    d=[inf for i in range(n+1)]
    d[s]=0
    update=True
    for i in range(n+1):
      update=False
      for v,u,cost in edges:
        if cand[v]!=1 or cand[u]!=1:
          continue
        if d[v]!=inf and d[u]>d[v]+cost:
          d[u]=d[v]+cost
          update=True
      if update==False:
        break
      if i==n:
        return inf
    return -d[-1]

def DFS_1toN(v):
  for u in to[v]:
    if toflag[u]==0:
      toflag[u]=1
      DFS_1toN(u)
      
def DFS_Nto1(v):
  for u in ot[v]:
    if otflag[u]==0:
      otflag[u]=1
      DFS_Nto1(u)
    
n,m,p=map(int,input().split())
g=[]
to=[[] for _ in range(n+1)]
ot=[[] for _ in range(n+1)]
for _ in range(m):
  a,b,c=map(int,input().split())
  g.append([a,b,-(c-p)])
  to[a].append(b)
  ot[b].append(a)
toflag=[0]*(n+1)
toflag[1]=1
DFS_1toN(1)
otflag=[0]*(n+1)
otflag[n]=1
DFS_Nto1(n)
cand=[0]*(n+1)
for i in range(1,n+1):
  cand[i]=toflag[i]*otflag[i]
ans=BF(g,n,1)
if ans==float('inf'):
  print(-1)
else:
  print(max(ans,0))

Solution (Python):
import sys


def solve(n, links, parents, costs):
    q = [n - 1]
    reachable = set()
    while q:
        v = q.pop()
        if v in reachable:
            continue
        reachable.add(v)
        q.extend(u for u in parents[v] if u not in reachable)

    INF = 10 ** 12
    dist = [INF] * n
    dist[0] = 0
    for _ in range(len(reachable)):
        updated = False
        for v in reachable:
            d = dist[v]
            if d == INF:
                continue
            cv = costs[v]
            for u in links[v]:
                if dist[u] == INF:
                    links[u] &= reachable
                if dist[u] > d + cv[u]:
                    dist[u] = d + cv[u]
                    updated = True
        if not updated:
            break
    else:
        return -1

    return max(0, -dist[n - 1])


n, m, p = list(map(int, input().split()))
INF = 10 ** 12
links = [set() for _ in range(n)]
parents = [set() for _ in range(n)]
costs = [[INF] * n for _ in range(n)]
for line in sys.stdin:
    a, b, c = map(int, line.split())
    a -= 1
    b -= 1
    links[a].add(b)
    parents[b].add(a)
    costs[a][b] = min(costs[a][b], p - c)

print(solve(n, links, parents, costs))

Solution (Python):
import sys


def reverse_search(n, parents):
    q = [n - 1]
    visited = set()
    while q:
        v = q.pop()
        if v in visited:
            continue
        visited.add(v)
        q.extend(u for u in parents[v] if u not in visited)
    return visited


n, m, p = list(map(int, input().split()))
INF = 10 ** 12
links = [set() for _ in range(n)]
parents = [set() for _ in range(n)]
costs = [[INF] * n for _ in range(n)]
for line in sys.stdin:
    a, b, c = map(int, line.split())
    a -= 1
    b -= 1
    links[a].add(b)
    parents[b].add(a)
    costs[a][b] = min(costs[a][b], p - c)
reachable = reverse_search(n, parents)
for v in reachable:
    links[v].intersection_update(reachable)

dist = [INF] * n
dist[0] = 0
for _ in range(len(reachable)):
    updated = False
    for v in reachable:
        d = dist[v]
        if d == INF:
            continue
        cv = costs[v]
        for u in links[v]:
            if dist[u] > d + cv[u]:
                dist[u] = d + cv[u]
                updated = True
    if not updated:
        break
else:
    print(-1)
    exit()

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

Solution (Python):
import sys
from collections import defaultdict


def reverse_search(n, parents):
    q = [n - 1]
    visited = set()
    while q:
        v = q.pop()
        if v in visited:
            continue
        visited.add(v)
        q.extend(u for u in parents[v] if u not in visited)
    return visited


n, m, p = list(map(int, input().split()))
INF = 10 ** 12
links = [set() for _ in range(n)]
parents = [set() for _ in range(n)]
costs = [[INF] * n for _ in range(n)]
for line in sys.stdin:
    a, b, c = map(int, line.split())
    a -= 1
    b -= 1
    links[a].add(b)
    parents[b].add(a)
    costs[a][b] = min(costs[a][b], p - c)
reachable = reverse_search(n, parents)
links = list(map(reachable.intersection, links))

dist = [INF] * n
dist[0] = 0
for _ in range(len(reachable)):
    updated = False
    for v in reachable:
        d = dist[v]
        if d == INF:
            continue
        cv = costs[v]
        for u in links[v]:
            if dist[u] > d + cv[u]:
                dist[u] = d + cv[u]
                updated = True
    if not updated:
        break
else:
    print(-1)
    exit()

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

Solution (Python):
import sys
from collections import defaultdict


def reverse_search(n, parents):
    q = [n - 1]
    visited = set()
    while q:
        v = q.pop()
        if v in visited:
            continue
        visited.add(v)
        q.extend(u for u in parents[v] if u not in visited)
    return visited


n, m, p = list(map(int, input().split()))
INF = 10 ** 12
links = [set() for _ in range(n)]
parents = [set() for _ in range(n)]
costs = [[INF] * n for _ in range(n)]
for line in sys.stdin:
    a, b, c = map(int, line.split())
    a -= 1
    b -= 1
    links[a].add(b)
    parents[b].add(a)
    costs[a][b] = min(costs[a][b], p - c)
reachable = reverse_search(n, parents)

dist = [INF] * n
dist[0] = 0
for _ in range(len(reachable)):
    updated = False
    for v in reachable:
        d = dist[v]
        if d == INF:
            continue
        cv = costs[v]
        for u in links[v]:
            if dist[u] > d + cv[u]:
                dist[u] = d + cv[u]
                updated = True
    if not updated:
        break
else:
    print(-1)
    exit()

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

Solution (Python):
import sys
from collections import defaultdict


def reverse_search(n, parents):
    q = [n - 1]
    visited = set()
    while q:
        v = q.pop()
        if v in visited:
            continue
        visited.add(v)
        q.extend(u for u in parents[v] if u not in visited)
    return visited


n, m, p = list(map(int, input().split()))
INF = 10 ** 12
links = [set() for _ in range(n)]
parents = [set() for _ in range(n)]
costs = [defaultdict(lambda: INF) for _ in range(n)]
for line in sys.stdin:
    a, b, c = map(int, line.split())
    a -= 1
    b -= 1
    links[a].add(b)
    parents[b].add(a)
    costs[a][b] = min(costs[a][b], p - c)
reachable = reverse_search(n, parents)

dist = [INF] * n
dist[0] = 0
for _ in range(len(reachable)):
    updated = False
    for v in reachable:
        d = dist[v]
        if d == INF:
            continue
        cv = costs[v]
        for u in links[v]:
            if dist[u] > d + cv[u]:
                dist[u] = d + cv[u]
                updated = True
    if not updated:
        break
else:
    print(-1)
    exit()

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

Solution (Python):
import sys


def reverse_search(n, parents):
    q = [n - 1]
    visited = set()
    while q:
        v = q.pop()
        if v in visited:
            continue
        visited.add(v)
        q.extend(u for u in parents[v] if u not in visited)
    return visited


n, m, p = list(map(int, input().split()))
INF = 10 ** 12
links = [set() for _ in range(n)]
parents = [set() for _ in range(n)]
for line in sys.stdin:
    a, b, c = map(int, line.split())
    a -= 1
    b -= 1
    links[a].add((b, p - c))
    parents[b].add(a)
reachable = reverse_search(n, parents)

dist = [INF] * n
dist[0] = 0
for _ in range(len(reachable)):
    updated = False
    for v in reachable:
        d = dist[v]
        if d == INF:
            continue
        for u, c in links[v]:
            if dist[u] > d + c:
                dist[u] = d + c
                updated = True
    if not updated:
        break
else:
    print(-1)
    exit()

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

Solution (Python):
def connect_dfs(N,E,start):
    connect = {start}
    q = [start]
    while q:
        i = q.pop()
        for j in E[i]:
            if j in connect:continue
            connect.add(j)
            q.append(j)
    return connect

def remove_Edge(N,Ef,Eb,start,goal):
    v = connect_dfs(N,Ef,start) & connect_dfs(N,Eb,goal)
    for i in range(N):
        if i in v:continue
        for j in Eb[i]:
            Ef[j].pop(i)

def Bellman_Ford(N,E,start,INFTY=10**18):
    def update(dist,N,E,start):
        done = [False]*N
        done[start] = True
        q = [start]
        d = dist[:]
        while q:
            i = q.pop()
            di = d[i]
            for j,c in E[i].items():
                d[j] = min(d[j],d[i]+c)
                if not done[j]:
                    q.append(j)
                    done[j] = True
        return d
    
    d = [INFTY]*N
    d[start] = 0
    for _ in range(N-1):
        d = update(d,N,E,start)
    if d!=update(d,N,E,start): return None
    else: return d

N,M,P = map(int,input().split())
ABC = [list(map(int,input().split())) for _ in [0]*M]

Ef = [{} for _ in [0]*N]
Eb = [{} for _ in [0]*N]
INFTY = 10**18
for a,b,c in ABC:
    Ef[a-1][b-1] = min(P-c,Ef[a-1].get(b-1,INFTY))
    Eb[b-1][a-1] = min(P-c,Eb[b-1].get(a-1,INFTY))
    
remove_Edge(N,Ef,Eb,0,N-1)
d = Bellman_Ford(N,Ef,0)

if d==None:ans=-1
else:ans = max(0,-d[-1])
print(ans)

Solution (Python):

n,m,r=map(int,input().split())
abc=[list(map(int,input().split())) for _ in [0]*m]

g=[[] for _ in [0]*n]
[g[a-1].append([b-1,c]) for a,b,c in abc]

g2=[[] for _ in [0]*n]
[g2[b-1].append([a-1,c]) for a,b,c in abc]


dist=[-10**15 for _ in [0]*n]
dist[0]=0
a,b=-10**15,-10**15

for k in range(n):
    for p in range(n):
        for i,j in g[p]:
            dist[i]=max(dist[i],dist[p]+j-r)
    if k==n-2:
        d1=[t for t in dist]
    if k==n-1:
        d2=[t for t in dist]
    

loop=set()
for i in range(n):
    if d1[i]!=d2[i]:
        loop.add(i)    
ans=max(d2[n-1],0)


q=[0]
memo=[True for _ in range(n)]
memo[0]=False
while q:
    qq=q.pop()
    for i,j in g[qq]:
        if memo[i]:
            q.append(i)
            memo[i]=False
            

q=[n-1]
memo2=[True for _ in range(n)]
memo2[n-1]=False
while q:
    qq=q.pop()
    for i,j in g2[qq]:
        if memo2[i]:
            q.append(i)
            memo2[i]=False
            

for i in loop:
    if memo[i]==False and memo2[i]==False:
        ans=-1
        
print(ans)

Solution (Python):
N, M, P = map(int, input().split())
edges = []
edges1 = [[] for _ in range(N)]
for i in range(M):
    a, b, c = map(int, input().split())
    edges1[b-1].append(a-1)
    edges.append((a-1, b-1, -c+P))
from collections import deque
que = deque([N-1])
flag = [True]*N
flag[-1] = False
while que:
    v = que.pop()
    for e in edges1[v]:
        if flag[e]:
            que.append(e)
            flag[e] = False
def f(n,w,es):
    
    
    d = [float("inf")] * n
    d[0] = 0 
    
    for i in range(n):
        for j in range(w):
            e = es[j]
            if flag[e[1]]:
                continue
            if d[e[1]] > d[e[0]] + e[2]:
                d[e[1]] = d[e[0]] + e[2]
                if i == n-1:
                    return False
    return d
d = f(N, M, edges)
if d:
    print(max(-d[-1], 0))
else:
    print(-1)

Solution (Python):
N,M,P = map(int,input().split())
ABC = [tuple(map(int,input().split())) for i in range(M)]

g = [[] for i in range(N)]
revg = [[] for i in range(N)]

for a,b,_ in ABC:
    a,b = a-1, b-1
    g[a].append(b)
    revg[b].append(a)

st = set([0])
stack = [0]
while stack:
    v = stack.pop()
    for to in g[v]:
        if to in st: continue
        st.add(to)
        stack.append(to)

revst = set([N-1])
stack = [N-1]
while stack:
    v = stack.pop()
    for to in revg[v]:
        if to in revst: continue
        revst.add(to)
        stack.append(to)

use = st & revst
es = []
for a,b,c in ABC:
    a,b = a-1, b-1
    if a in use and b in use:
        es.append((a,b,c))

INF = float('inf')
d = [INF] * N
d[0] = 0
for i in range(N):
    for fr,to,c in es:
        cost = -(c-P)
        if d[to] > d[fr]+cost:
            d[to] = d[fr] + cost
            if i == N-1:
                print(-1)
                exit()

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

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

print(max(-cost[N], 0))

Solution (Python):
import sys
N, M, P = [int(x) for x in input().split()]
forward = {}
backward = {}
for _ in range(M):
  A, B, C = [int(x) for x in input().split()]
  forward[A] = forward.get(A, {})
  if B in forward[A]:
    forward[A][B] = max(forward[A][B], C-P)
  else:
    forward[A][B] = C-P
  backward[B] = backward.get(B, {})
  if A in backward[B]:
    backward[B][A] = max(backward[B][A], C-P)
  else:
    backward[B][A] = C-P


reachable = set([N])
to_track = set([N])
while to_track:
  x = to_track.pop()
  if x not in backward:
    continue
  for y in backward[x]:
    if y not in reachable:
      to_track.add(y)
      reachable.add(y)

if 1 not in reachable:
  print(-1)
  sys.exit()

forward = {k: {k2: v2 for k2, v2 in v.items() if k2 in reachable}
           for k, v in forward.items() if k in reachable}





dist = {k: -float('inf') for k in reachable} 
dist[1] = 0



for _ in range(len(reachable)):
  renewed = False
  for x, y_dict in forward.items():
    for y in y_dict:
 
      if dist[y] < dist[x] + forward[x][y]:
        renewed = True
        dist[y] = dist[x] + forward[x][y]
if renewed:
  
  print(-1)
else:
  print(max(0,dist[N]))

Solution (Python):
import sys, math, itertools, bisect, copy, re
from collections import Counter, deque, defaultdict

from itertools import accumulate, permutations, combinations, takewhile, compress, cycle




import heapq

INF = float('inf')
MOD = 10 ** 9 + 7
EPS = 10 ** -7
sys.setrecursionlimit(1000000)









def dfs(n):
    if n in to:
        return
    else:
        to.add(n)
        for a, b, c in edges:
            if a == n:
                dfs(b)


def dfs2(n):
    if n in ot:
        return
    else:
        ot.add(n)
        for a, b, c in edges:
            if b == n:
                dfs2(a)


N, M, P = [int(x) for x in input().split()]
edges = []

to = set()

ot = set()
for _ in range(M):
    a, b, c = [int(x) for x in input().split()]
    a, b, c = a - 1, b - 1, c - P
    edges.append((a, b, -c))

dfs(0)
dfs2(N - 1)

W = [INF] * N
W[0] = 0
for cnt in range(N + 1):
    isShusoku = True
    for a, b, c in edges:
        if b in to and b in ot:
            if W[b] > W[a] + c:
                W[b] = W[a] + c
                isShusoku = False

if isShusoku == True:
    print(max(0,-W[N - 1]))
else:
    print(-1)
pass

Solution (Python):

import sys
from collections import deque
INF = float("inf")



class Graph(object):
    def __init__(self, N):
        self.N = N
        self.V = list(range(N))
        self.E = [[] for _ in range(N)]

    def add_edge(self, edge):
        """辺を加える。edgeは(始点, 終点、重み)からなるリスト
        重みがなければ、重み1とする。
        """
        if len(edge) == 2:
            edge.append(1)
        elif len(edge) != 3:
            print("error in add_edge")
            pass

        s, t, w = edge
        self.E[s].append([t, w])

        pass


def BellmanFord(g: Graph, s):
    """ベルマンフォード法による最短経路
    """
    N = g.N
    dist = [INF]*N
    dist[s] = 0
    prev = [None]*N

    
    for _ in range(N):
        for from_ in range(N):      
            for to_, weight in g.E[from_]:
                if dist[to_] > dist[from_] + weight:
                    dist[to_] = dist[from_] + weight
                    prev[to_] = from_
    
    for from_ in range(N):
        for to_, weight in g.E[from_]:
            if dist[from_]+weight < dist[to_]:
                return None, None

    return dist, prev


def solve(N: int, M: int, P: int, A: "List[int]", B: "List[int]", C: "List[int]"):
    for i in range(M):
        C[i] = P - C[i]

    rev_g = Graph(N)
    for a, b, c in zip(A, B, C):
        rev_g.add_edge([b-1, a-1, c])

    
    visitable = [False]*N
    visitable[N-1] = True
    q = deque()
    q.append(N-1)
    while len(q) > 0:
        curr = q.popleft()
        for t_, w in rev_g.E[curr]:
            if not visitable[t_]:
                visitable[t_] = True
                q.append(t_)

    g = Graph(N)
    for a, b, c in zip(A, B, C):
        if visitable[a-1] and visitable[b-1]:
            g.add_edge([a-1, b-1, c])

    dist, prev = BellmanFord(g, 0)
    if dist is None:
        print(-1)
    else:
        print(max(-dist[-1], 0))

    
    return


def main():

    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  
    M = int(next(tokens))  
    P = int(next(tokens))  
    A = [int()] * (M)  
    B = [int()] * (M)  
    C = [int()] * (M)  
    for i in range(M):
        A[i] = int(next(tokens))
        B[i] = int(next(tokens))
        C[i] = int(next(tokens))
    solve(N, M, P, A, B, C)


if __name__ == '__main__':
    main()

Solution (Python):
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))


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

Solution (Python):
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))


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):
		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
	if dist[-1] != prev:
		print(-1)
		exit()


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

Solution (Python):
from collections import deque

n, m, p = list(map(int, input().split()))


edges = [list(map(int, input().split())) for i in range(m)]
for edge in edges:
    edge[0] -= 1
    edge[1] -= 1
    edge[2] -= p

start = 0
goal = n - 1


dist = [-float("inf")] * n
dist[start] = 0


reachable = [False] * n
reachable[goal] = True
d = deque()
d.append(goal)
reachable[goal] = True
while len(d) > 0:
    x = d.popleft()
    for edge in edges:
        if edge[1] == x and reachable[edge[0]] is False:
            d.append(edge[0])
            reachable[edge[0]] = True

cant_reach = []
for i in range(n):
    if reachable[i] is False:
        cant_reach.append(i)

for edge in edges:
    if edge[0] in cant_reach or edge[1] in cant_reach:
        edges.remove(edge)




negative_cycle = False
for i in range(n):
    
    for edge in edges:
        if dist[edge[0]] != -float("inf") and dist[edge[1]] < dist[edge[0]] + edge[2]:
            dist[edge[1]] = dist[edge[0]] + edge[2]
            if i == n - 1:
                negative_cycle = True

print(-1 if negative_cycle else max(dist[n - 1], 0))

Solution (Python):
inf = float('inf')
V,E,P = map(int,input().split())
edge = []
for i in range(E):
    A,B,C = map(int,input().split())
    c = P-C
    if A == B:
        c *= 10**6
    edge.append([A-1,B-1,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

ans = d[V-1]

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
loop = ans != d[V-1]

if loop:
    print(-1)

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

Solution (Python):
n, m, p = map(int,input().split())
A = [list(map(int,input().split())) for i in range(m)]

for i in range(m):
    A[i][2] = - (A[i][2]-p)







S = [[float('inf'), 0, 0] for i in range(n+1)]

S[1][0]=0
S[1][1]=1
S[n][2]=1

for i in range(m):
    if A[i][0] == 1:
        S[A[i][1]][0] = A[i][2]

for j in range(n-1):
    for i in range(m):
        S[A[i][1]][0] = min(S[A[i][1]][0], S[A[i][0]][0] + A[i][2])
        S[A[i][1]][1] = max(S[A[i][1]][1], S[A[i][0]][1])
        S[A[i][0]][2] = max(S[A[i][1]][2], S[A[i][0]][2])

ANS0 = [S[i][0] for i in range(len(S))]

for j in range(1):
    for i in range(m):
        S[A[i][1]][0] = min(S[A[i][1]][0], S[A[i][0]][0] + A[i][2])
        
        




imp = 0

for i in range(len(S)):
    if S[i][1] == 1 and S[i][2] == 1:
        if ANS0[i] != S[i][0]:
            imp = 1

if imp == 1:
    print(-1)
else:
    print(max(0, -ANS0[n]))

Solution (Python):


import sys

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7

def bellman_ford(N: int, edges: list, src: int) -> list:
    """ ベルマンフォード(頂点数, 辺集合(0-indexed), 始点) """

    
    res = [INF] * N
    res[src] = 0
    
    for i in range(N-1):
        for src, dest, cost in edges:
            if res[dest] > res[src] + cost:
                res[dest] = res[src] + cost
    
    for i in range(N):
        for src, dest, cost in edges:
            if res[dest] > res[src] + cost:
                res[dest] = -INF
                if dest == N-1:
                    
                    return []
    
    return res

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

res = bellman_ford(N, edges, 0)
if not len(res):
    print(-1)
else:
    
    ans = -res[N-1]
    
    print(max(0, ans))

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 can_go_from_0():
    
    ret = [False] * N

    stack = [0]
    while stack:
        v = stack.pop()
        ret[v] = True
        for u, _ in graph[v]:
            if not ret[u]:
                ret[u] = True
                stack.append(u)
    return ret


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

    
    ret = [False] * N
    stack = [N - 1]
    while stack:
        v = stack.pop()
        ret[v] = True
        for u in revs[v]:
            if not ret[u]:
                ret[u] = True
                stack.append(u)
    return ret


vertices = []
for i, (v1, v2) in enumerate(zip(can_go_from_0(), can_go_to_n())):
    if v1 & v2:
        vertices.append(i)


dist = [INF] * N
dist[0] = 0
for i in range(N):
    for v in vertices:
        for u, d in graph[v]:
            dist[u] = min(dist[v] + d, dist[u])


prev = dist[:]
for v in vertices:
    for u, d in graph[v]:
        dist[u] = min(dist[v] + d, dist[u])
updated = False
for v in vertices:
    updated |= prev[v] != dist[v]

if updated or dist[N - 1] == INF:
    print(-1)
else:
    print(max(-dist[N - 1], 0))