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:
import collections


def bellman_ford(es, n, start):
    infinity = float("inf")
    m = len(es)

    d = [infinity] * (n + 1)
    d[start] = 0

    for _ in range(m):
        updated = False

        for u, v, w in es:
            if d[u] == infinity: continue

            dd = d[u] + w

            if dd < d[v]:
                d[v] = dd
                updated = True

        if not updated: break

    for u, v, w in es:
        if d[u] < infinity and d[u] + w < d[v]:
            return None, True

    return d, False


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

Es = [None] * M

for i in range(M):
    a, b, c = ([int(e) for e in input().split()])
    Es[i] = (a - 1, b - 1, P - c)

q = collections.deque()
q.append(N - 1)

valid = [False] * N
valid[N - 1] = True

while 0 < len(q):
    cur = q.popleft()

    for u, v, _ in Es:
        if cur == v and not valid[u]:
            valid[u] = True
            q.append(u)

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

Es = es_tmp

distances, le = bellman_ford(Es, N - 1, 0)

if le:
    print(-1)
else:
    print(max(-distances[-1], 0))

Submission status: Accepted

Solution code:
n, m, p = map(int, input().split())

abc = []
# edge = [[] for _ in range(n)]
rev_edge = [[] for _ in range(n)]
for _ in range(m):
    a, b, c = map(int, input().split())
    abc.append((a - 1, b - 1, p - c))
    # edge[a - 1].append(b - 1)
    rev_edge[b - 1].append(a - 1)


def dfs(e, s):
    v_set = {s}
    stack = [s]
    while stack:
        v = stack.pop()
        for next_v in e[v]:
            if next_v in v_set:
                continue
            v_set.add(next_v)
            stack.append(next_v)
    return v_set


def bellman_ford(sub_abc, n):
    inf = float("inf")
    dist = [inf] * n
    dist[0] = 0

    for _ in range(n):
        updated = False
        for u, v, d in sub_abc:
            if dist[u] + d < dist[v]:
                dist[v] = dist[u] + d
                updated = True
        if not updated:
            break
    else:
        return -1

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


use = dfs(rev_edge, n - 1)
print(bellman_ford([(a, b, c)
                    for (a, b, c) in abc if a in use and b in use], n))

Submission status: Accepted

Solution code:
n, m, p = map(int, input().split())

abc = []
edge = [[] for _ in range(n)]
rev_edge = [[] for _ in range(n)]
for _ in range(m):
    a, b, c = map(int, input().split())
    abc.append((a - 1, b - 1, p - c))
    edge[a - 1].append(b - 1)
    rev_edge[b - 1].append(a - 1)


def dfs(e, s):
    v_set = {s}
    stack = [s]
    while stack:
        v = stack.pop()
        for next_v in e[v]:
            if next_v in v_set:
                continue
            v_set.add(next_v)
            stack.append(next_v)
    return v_set


def bellman_ford(sub_abc, n):
    inf = float("inf")
    dist = [inf] * n
    dist[0] = 0

    for _ in range(n):
        updated = False
        for u, v, d in sub_abc:
            if dist[u] + d < dist[v]:
                dist[v] = dist[u] + d
                updated = True
        if not updated:
            break
    else:
        return -1

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


use = dfs(edge, 0) & dfs(rev_edge, n - 1)
print(bellman_ford([(a, b, c)
                    for (a, b, c) in abc if a in use and b in use], n))

Submission status: Accepted

Solution code:
N,M,P = map(int,input().split())
abc = []
edge = [[] for _ in range(N)]
for _ in range(M):
    a,b,c = map(int,input().split())
    #向かってくる矢印
    edge[b-1].append(a-1)
    abc.append((a-1,b-1,P-c))

def dfs(edge,goal):
    #goal = n-1
    used = {goal}
    q = [goal]
    while q:
        v = q.pop()
        for w in edge[v]:
            if w in used:continue
            used.add(w)
            q.append(w)
    
    return used

def bellman_ford(v,s,g,e):

    '''
    v: vertex(頂点)
    s: start.  in this case,0
    g: goal. in this case,n-1
    e: (edge_start,egde_goal,score)
    今回はp-cの最小値を求めていく.
    '''
    inf = 10**10
    # cost : d[V] 
    d = [inf] * v
    #開始点のコストを0とする
    d[s] = 0
    for _ in range(v):
        flag = False
        for a,b,c in e:
            # a -> b の向きを考える
            if d[a] == inf:continue
            cost = d[a] + c
            if cost < d[b]:
                d[b] = cost
                flag = True
        if not flag: break
    if flag:
        return -1
    return max(-d[g],0)

vertex = dfs(edge,N-1)
abc_ = [(a,b,c) for a,b,c in abc if a in vertex and b in vertex]
print(bellman_ford(N,0,N-1,abc_))

Submission status: Accepted

Solution code:
N, M, P = map(int, input().split())
E = [None] * M
for i in range(M):
    a, b, c = map(int, input().split())
    E[i] = (a - 1, b - 1, P - c)
INF = float("inf")
d = [INF for _ in range(N)]
d[0] = 0
for i in range(2*N):
    for f, t, w in E:
        if d[t] > d[f] + w:
            d[t] = d[f] + w
            if i >= N:
                d[t] = -INF
    if i == N-1:
        dN = d[-1]
if dN != d[-1]:
    print(-1)
elif dN > 0:
    print(0)
else:
    print(-dN)

Submission status: Accepted

Solution code:
import collections
import sys
import time
sys.setrecursionlimit(10 ** 5)
N, M, P = map(int, input().split())
e = collections.defaultdict(list)
st = time.time()
for _ in range(M):
  u, v, c = map(int, input().split())
  e[u].append((v, c))
def isConnected(n):
  if n == N:
    return True
  vis = [0] * (N + 1)
  que = collections.deque([])
  que.append(n)
  while que:
    nc = que.popleft()
    vis[nc] = 1
    for o in e[nc]:
      if o[0] == N:
        return True
      if vis[o[0]] == 0:
        que.append(o[0])
  return False
inf = 10 ** 10
visited = [0] * (N + 1)
earn = [-inf] * (N + 1)
earn[1] = 0
def dfs(now):
  global visited
  global earn
  nt = time.time() - st
  if nt > 1.95:
    print(max(earn[N], 0))
    exit(0)
  visited[now] = 1
  for n in e[now]:
    if visited[n[0]] == 0:
      if n[1] + earn[now] - P > earn[n[0]]:
        earn[n[0]] = n[1] + earn[now] - P
        dfs(n[0])
        visited[n[0]] = 0
    else:
      if earn[n[0]] < n[1] + earn[now] - P:
        if isConnected(n[0]) == True:
          print(-1)
          exit(0)
dfs(1)
print(max(earn[N], 0))

Submission status: Accepted

Solution code:
from collections import defaultdict,deque
import sys,heapq,bisect,math,itertools,string,queue,copy,time
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_str(): return list(sys.stdin.readline().split())

N,M,P = inpl()
lines = defaultdict(set)

for _ in range(M):
    a,b,c = inpl()
    a,b = a-1,b-1
    lines[a].add((b,-c+P))

#ベルマンフォード(始点と終点が決まってる時)
def BellmanFord(Start,Goal):
    Costs=[INF]*N
    Costs[Start] = 0
    #upd8s = [True]*N
    for i in range(2*N): #2N回ループ(負閉路の検出までみる)
        #will_upd8s = [False]*N
        for s in range(N):
            #if not upd8s[s]: continue    #前回更新してないので見ない
            for t,c in lines[s]:
                if c + Costs[s] < Costs[t]:
                    Costs[t] = Costs[s]+c
                    #will_upd8s[t] = True #更新した点だけ次に見る
                    if i >= N:
                        Costs[t] = -INF

        if i == N-1: #Nループ目のGoalのCostを記録
            tmp = Costs[Goal]

        #upd8s = will_upd8s[:]

    if tmp != Costs[Goal]:
        return -INF
    else:
        return Costs[Goal]

ans = -BellmanFord(0,N-1)
if ans == INF:
    print(-1)
else:
    print(max(0,ans))

Submission status: Accepted

Solution code:
from collections import defaultdict,deque
import sys,heapq,bisect,math,itertools,string,queue,copy,time
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_str(): return list(sys.stdin.readline().split())

N,M,P = inpl()
lines = defaultdict(set)

for _ in range(M):
    a,b,c = inpl()
    a,b = a-1,b-1
    lines[a].add((b,-c+P))

#ベルマンフォード(始点と終点が決まってる時)
def BellmanFord(Start,Goal):
    Costs=[INF]*N
    Costs[Start] = 0
    upd8s = [True]*N
    for i in range(2*N): #2N回ループ(負閉路の検出までみる)
        will_upd8s = [False]*N
        for s in range(N):
            if not upd8s[s]: continue    #前回更新してないので見ない
            for t,c in lines[s]:
                if c + Costs[s] < Costs[t]:
                    Costs[t] = Costs[s]+c
                    will_upd8s[t] = True #更新した点だけ次に見る
                    if i >= N:
                        Costs[t] = -INF

        if i == N-1: #Nループ目のGoalのCostを記録
            tmp = Costs[Goal]

        upd8s = will_upd8s[:]

    if tmp != Costs[Goal]:
        return -INF
    else:
        return Costs[Goal]

ans = -BellmanFord(0,N-1)
if ans == INF:
    print(-1)
else:
    print(max(0,ans))

Submission status: Accepted

Solution code:
INF = float('inf')

def main():
    n, m, p = map(int,input().split())
    es = []
    for _ in range(m):
        e = list(map(int,input().split()))
        es.append([e[0]-1,e[1]-1,- e[2] + p])

    v_cost = [INF]*n
    v_cost[0] = 0
    for v in range(n*2):
        for e in es:
            frm = e[0]
            to = e[1]
            if v_cost[to] > v_cost[frm]+e[2]:
                v_cost[to] = v_cost[frm]+e[2]
                if v >= n:
                    v_cost[to] = -INF
        if v == n-1:
            prev = v_cost[-1]

    if prev != v_cost[-1]:
        return -1

    return max(0,-v_cost[-1])
   

if __name__ =="__main__":
    print (main())

Submission status: Accepted

Solution code:
import sys
sys.setrecursionlimit(10**9)
INF = float('inf')

def main():
    n, m, p = map(int,input().split())
    es = []
    for _ in range(m):
        e = list(map(int,input().split()))
        es.append([e[0]-1,e[1]-1,- e[2] + p])

    v_cost = [INF]*n
    v_cost[0] = 0
    for v in range(n*2):
        for e in es:
            frm = e[0]
            to = e[1]
            if v_cost[to] > v_cost[frm]+e[2]:
                v_cost[to] = v_cost[frm]+e[2]
                if v >= n:
                    v_cost[to] = -INF
        if v == n-1:
            prev = v_cost[-1]

    if prev != v_cost[-1]:
        return -1

    return max(0,-v_cost[-1])
   

if __name__ =="__main__":
    print (main())

Submission status: Accepted

Solution code:
import sys
sys.setrecursionlimit(10**9)
INF = float('inf')

def main():
    n, m, p = map(int,input().split())
    es = []
    for _ in range(m):
        e = list(map(int,input().split()))
        es.append([e[0]-1,e[1]-1,- e[2] + p])

    v_cost = [INF]*n
    v_cost[0] = 0
    for v in range(n*2):
        for e in es:
            frm = e[0]
            to = e[1]
            if v_cost[to] > v_cost[frm]+e[2]:
                v_cost[to] = v_cost[frm]+e[2]
                if v >= n:
                    v_cost[to] = -INF
            if v == n-1:
                prev = v_cost[-1]
                
    if prev != v_cost[-1]:
        return -1

    return max(0,-v_cost[-1])
   

if __name__ =="__main__":
    print (main())

Submission status: Accepted

Solution code:
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline

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, -c+P])


def bellnabFord(edges, src, N):
    inf = float('inf')
    dist = [inf for i in range(N)]
    dist[src] = 0

    for i in range(2*N):
        for s, d, c in edges:
            if s != inf and dist[d] > dist[s]+c:
                dist[d] = dist[s]+c
                if i >= N:
                    dist[d] = -float('inf')

            if i == N-1:
                prev = dist[-1]
    return (prev, dist[-1])


prev, dist = bellnabFord(edges, 0, N)
if prev != dist:
    ans = -1
else:
    ans = max(0, -dist)
print(ans)

Submission status: Accepted

Solution code:
def validate(paths, start):
    """
    有効な経路をDFS(深さ優先探索)
    経路をsetで返す
    """

    valid_paths = {start}
    stack = [start]

    while stack:
        v = stack.pop()
        if v not in paths:
            continue
        for w in paths[v]:
            if w in valid_paths:
                continue
            valid_paths.add(w)
            stack.append(w)
    return valid_paths


def bellman_ford(edges, num_v, start, end):
    """
    有向グラフ(負閉路有り)の最短距離

    num_v: 頂点の数(number_of_vertex)
    edges: 頂点(from, to, cost)
    """

    # グラフの初期化
    INF = 10**18
    dist = [INF] * num_v
    # 始点は0にする
    dist[start] = 0

    # 辺の緩和
    # 負の閉路が無ければ、num_v-1回までで収束する
    for _ in range(num_v):
        updated = False
        for f, t, c in edges:
            if dist[f] == INF:
                continue

            # コストが以前のものより小さければ更新
            cost = dist[f] + c
            if cost < dist[t]:
                dist[t] = cost
                updated = True
        if not updated:
            break
    else:
        # num_v回の時に更新があるなら負閉路が存在する
        return -1

    return max(-dist[end], 0)


def main():
    from collections import defaultdict

    n, m, p, *tmp = map(int, open(0).read().split())

    paths_rev = defaultdict(list)
    edges = []
    for a, b, c in zip(*[iter(tmp)]*3):
        # 頂点nからの経路をdfsするため、辺の向きを逆にする
        paths_rev[b-1].append(a-1)
        edges.append((a-1, b-1, p-c))

    valid_paths = validate(paths_rev, n-1)
    edges = [
        (a, b, c)
        for a, b, c in edges
        if a in valid_paths and b in valid_paths
    ]
    res = bellman_ford(edges, n, 0, n-1)
    print(res)

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
from collections import deque, defaultdict

N, M, P = map(int, input().split())

edge_dic = defaultdict(lambda: [])
reverse_edge_dic = defaultdict(lambda: [])

for _ in range(M):
    A, B, C = map(int, input().split())
    edge_dic[A-1].append([B-1, C-P])
    reverse_edge_dic[B-1].append([A-1, C-P])

forward_set = set()
node_deque = deque()
node_deque.append(0)
forward_set.add(0)
while len(node_deque):
    edge = node_deque.pop()
    for v in edge_dic[edge]:
        goal, cost = v
        if goal not in forward_set:
            forward_set.add(goal)
            node_deque.append(goal)

backward_set = set()
node_deque = deque()
node_deque.append(N-1)
backward_set.add(N-1)
while len(node_deque):
    edge = node_deque.pop()
    for v in reverse_edge_dic[edge]:
        goal, cost = v
        if goal not in backward_set:
            backward_set.add(goal)
            node_deque.append(goal)
can_use = forward_set & backward_set
# print(forward_set, backward_set, can_use)

new_edge_dic = {i: edge_dic[i] for i in can_use}

INF = 10 ** 10

ans_array = [-INF] * N
ans_array[0] = 0


for i in range(len(can_use)+1):
    change_flag = 0
    for key, value in new_edge_dic.items():
        for v in value:
            goal, cost = v
            if ans_array[goal] < ans_array[key] + cost:
                ans_array[goal] = ans_array[key] + cost
                change_flag = 1
    if change_flag == 0:
        break
else:
    print(-1)
    exit()

ans = max(0, ans_array[N-1])
print(ans)

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
#ABC137 E

import sys
import math
import bisect
sys.setrecursionlimit(1000000000)
from heapq import heappush, heappop
from collections import defaultdict
from itertools import accumulate
from collections import Counter
from collections import deque
from operator import itemgetter
from itertools import permutations
mod = 10**9 + 7
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))

def Bellman_Ford(edges,num_v, start):
    #start → iの最短距離
    #start:始点, num_v:頂点数, edges:[辺の始点, 辺の終点, 辺のコスト]
    dist = [-float('inf') for _ in range(num_v)]
    #dist[i] := s→iの最短距離
    dist[start] = 0
    flg = 0
    for i in range(num_v):
        update = False
        for edge in edges:
            e_from, e_to, e_cost = edge
            if e_from != -float('inf') and visited[e_to] and dist[e_to] < dist[e_from] + e_cost:
                dist[e_to] = dist[e_from] + e_cost
                update = True
        if not update:
            flg = 1
            break
    return dist, flg

n,m,p = LI()
edge = []
f = defaultdict(dict)
for _ in range(m):
    a,b,c = LI()
    edge.append([a-1,b-1,c-p])
    f[b-1][a-1] = 1
visited = [False]*n
def dfs(x):
    visited[x] = True
    for t in f[x].keys():
        if not visited[t]:
            dfs(t)
dfs(n-1)
d,f = Bellman_Ford(edge,n,0)
if not f:
    print(-1)
else:
    print(max(0,d[-1]))

Submission status: Accepted

Solution code:
import sys
sys.setrecursionlimit(10000000)
def input():
    return sys.stdin.readline()[:-1]

f_inf = float("inf") # sys.maxsize

N, M, P = map(int, input().split())
ABC_d = {}
for m in range(M):
    A, B, C = map(int, input().split())
    if A in ABC_d:
        ABC_d[A].append((B, C - P))
    else:
        ABC_d[A] = [(B, C - P)]

cost = [-f_inf] * N
cost[0] = 0

max_num_sum = (N * (N + 1)) // 2

def solve(pos, cnt, num_sum):
    global cost

    if not(pos in ABC_d): return
    if cnt > N or num_sum > max_num_sum:
        cost[pos-1] = f_inf;

    for B, C in ABC_d[pos]:
        if cost[B - 1] == f_inf or cost[B - 1] >= cost[pos - 1] + C:
            pass
        elif cost[B - 1] < cost[pos - 1] + C:
            cost[B - 1] = cost[pos - 1] + C
            solve(B, cnt + 1, num_sum + B)

solve(1, 1, 1)

if (cost[-1] == f_inf):
    print (-1)
else:
    print (max(0, int(cost[-1])))

Submission status: Accepted

Solution code:
words = lambda t : list(map(t, input().split()))
n,m,p = words(int)

w_edges = []
f_edges = [[] for _ in range(n)]
b_edges = [[] for _ in range(n)]

for i in range(m):
    a,b,c = words(int)
    w_edges.append((a-1,b-1,c))
    f_edges[a-1].append((b-1))
    b_edges[b-1].append((a-1))

from queue import Queue
def bfs(edges,stt):
    visited = [False] * n
    q = Queue()
    q.put(stt)
    visited[stt] = True
    while not q.empty():
        v = q.get()
        for c in edges[v]:
            if not visited[c]:
                visited[c] = True
                q.put(c)
    return visited

f_visited = bfs(f_edges, 0)
b_visited = bfs(b_edges, n-1)

edges = []
for a,b,c in w_edges:
    if f_visited[a] and b_visited[a] and f_visited[b] and b_visited[b]:
        edges.append((a,b,p-c))

def bellman_ford(edges, num_v, stt):
    inf = float('inf')
    dist = [inf]*num_v
    dist[stt] = 0
    for i in range(num_v+1):
        for a,b,c in edges:
            prev = dist[b]
            dist[b] = min(dist[b], dist[a]+c)
            if i == num_v and prev != dist[b]:
                return -1
    return dist

d = bellman_ford(edges, n, 0)
if d == -1:
    print(-1)
else:
    print(max(0,-d[n-1]))

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.readline
from collections import deque
from copy import deepcopy
def main():
    N, M, P = map( int, input().split())
    E = []
    rE = [[] for _ in range(N)]
    
    for _ in range(M):
        a, b, c = map( int, input().split())
        a, b = a-1, b-1
        E.append((a,b,c-P))
        rE[b].append(a)
    # 頂点Nに到達する頂点
    G = [False]*N
    G[N-1] = True
    q = deque([N-1])
    while q:
        v = q.popleft()
        for w in rE[v]:
            if G[w] == False:
                G[w] = True
                q.append(w)
    # Bellman-Ford
    V = [None]*N
    V[0] = 0
    I = sum([1 for i in range(N) if G[i] == True])
    for _ in range(I-1):
        for s, t, cost in E:
            if V[s] == None or G[t] == False:
                continue
            if V[t] == None:
                V[t] = V[s] + cost
                continue
            if V[s] + cost > V[t]:
                V[t] = V[s] + cost
    W = deepcopy(V) #[ V[i] for i in range(N)]
    for _ in range(I-1):
        for s, t, cost in E:
            if V[s] == None or G[t] == False:
                continue
            if V[t] == None:
                V[t] = V[s] + cost
                continue
            if V[s] + cost > V[t]:
                V[t] = V[s] + cost
    for i in range(N):
        if W[i] == None or G[i] == False:
            continue
        if V[i] > W[i]: # 頂点iは、頂点Nに到達するので、負（正？）閉路は伝播する
            print(-1)
            return

    print( max(W[-1], 0))
if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
import sys
def main():
    N, M, P = map( int, input().split())
    INF = 1e18
    d = list()
    for _ in range(M):
        a, b, c = map( int, input().split())
        a, b = a-1, b-1
        d.append((a,b,P-c))
    V = [INF]*N
    V[0] = 0
    for _ in range(N):
        for (s,t,cost) in d:
            if V[s] != INF:
                V[t] = min(V[t], V[s] + cost)

    for _ in range(N):
        for (s,t,cost) in d:
            if V[s] != INF:
                if V[t] > V[s] + cost:
                    V[t]=-INF

    if V[-1] == -INF:
        print(-1)
    else:
        print( max(-V[-1], 0))
if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
def e():
    n, m, p = map(int, input().split())

    edges = []
    for i in range(m):
        edges.append(tuple(map(int, input().split())))

    edges.sort(key=lambda x: x[1])
    dist = [float('inf') for i in range(n)]
    dist[0] = 0

    for i in range(n):
        for e in edges:
            if dist[e[1]-1] > dist[e[0]-1] - (e[2] - p):
                dist[e[1]-1] = dist[e[0]-1] - (e[2] - p)

    for i in range(n):
        for e in edges:
            if dist[e[1] - 1] > dist[e[0] - 1] - (e[2] - p):
                dist[e[1]-1] = float('-inf')

    if dist[n-1] == float('-inf'):
        print(-1)
    else:
        print(-dist[n-1] if -dist[n-1] >= 0 else 0)


if __name__ == '__main__':
    e()

Submission status: Accepted

Solution code:
#### import ####
import sys
import math
from collections import defaultdict

#### 設定 ####
sys.setrecursionlimit(10**7)

#### 定数 ####
mod = 10**9 + 7

#### 読み込み ####
def I(): return int(input())
def II(): return map(int, input().split())
def III(): return list(map(int, input().split()))
def Line(N):
    read_all = [tuple(map(int, input().split())) for _ in range(N)]
    return map(list,zip(*read_all))

#################

def BF(E,s,n):
    inf=float("inf")
    d=[inf for _ in range(n)]
    d[s]=0
    for i in range(2*n):
        for e in E:
            if d[e[0]]!=inf and d[e[1]]>d[e[0]]+e[2]:
                d[e[1]] = d[e[0]] + e[2]
                if i>=n:
                    d[e[1]] = -float('inf')
    return d

N,M,P = II()
A,B,C = Line(M)
E = []
for i in range(M):
    E.append([A[i]-1, B[i]-1, -C[i]+P])

d = BF(E,0,N)
if d[N-1]==-float('inf'):
    print(-1)
else:
    print(max(0,-d[N-1]))

Submission status: Accepted

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

N,M,P = map(int,input().split())
edges = []
to = [[] for i in range(N)]
rto = [[] for i in range(N)]
for i in range(M) :
    u,v,cost = map(int,input().split())
    u -= 1
    v -= 1
    cost -= P
    cost *= -1
    edges.append([u,v,cost])
    to[u].append(v)
    rto[v].append(u)

canVisitFromS = [False]*N
canVisitFromG = [False]*N

def dfs(u) :
    if canVisitFromS[u] : return
    canVisitFromS[u] = True
    for v in to[u] :
        dfs(v)

def rdfs(v) :
    if canVisitFromG[v] : return
    canVisitFromG[v] = True
    for u in rto[v] :
        rdfs(u)

dfs(0)
rdfs(N-1)

OK = [canVisitFromS[i]*canVisitFromG[i] for i in range(N)]

dist = [float("inf")]*N
dist[0] = 0

ans = 0
step = 0
update = True
while update :
    update = False
    for i in range(M) :
        u,v,cost = edges[i]
        if OK[u]*OK[v] == 0 : continue
        if dist[v] > dist[u] + cost :
            dist[v] = dist[u] + cost
            update = True
    step += 1
    if step > N :
        ans = -1
        break

if ans != -1 :
    ans = max(0, dist[N-1]*(-1))

print(ans)

Submission status: Accepted

Solution code:
from collections import deque
NegInf = -float('inf')


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


def bellman_ford(vst, graph, node_num, NegInf):
    costs = [NegInf] * node_num
    costs[0] = 0
    v_nows = set([vst])
    for _ in range(n):
        v_changeds = set()
        for v_from in v_nows:
            for v_to, cost in graph[v_from]:
                c2 = costs[v_from] + cost
                if c2 > costs[v_to]:
                    costs[v_to] = c2
                    v_changeds.add(v_to)
        v_nows = v_changeds



    if not v_changeds:
        return costs

    for i in v_nows:
        costs[i] = NegInf

    que = deque(v_nows)
    while que:
        v_from = que.popleft()
        for v_to, cost in graph[v_from]:
            if costs[v_to] != NegInf:
                costs[v_to] = NegInf
                que.append(v_to)
    return costs


costs = bellman_ford(0, graph, n, -float('inf'))
if costs[n-1] == NegInf:
    print(-1)
else:
    print(max(0, costs[n-1]))

Submission status: Accepted

Solution code:
import sys
N, M, P = map(int, input().split())
Edge = []
G = [[] for i in range(N)]
G_rev = [[] for i in range(N)]
for i in range(M):
    a, b, c = map(int, sys.stdin.readline().split())
    a, b = a - 1, b - 1
    c = -(c - P)
    Edge.append([a, b, c])
    G[a].append(b)
    G_rev[b].append(a)


# 0とNの両方から辿れない頂点を削除
def dfs(g, stack):
    can_reach = set()
    while stack:
        n = stack.pop()
        can_reach.add(n)
        for to in g[n]:
            if to in can_reach:
                continue
            stack.append(to)
    return can_reach


can_reach_from_0 = dfs(G, [0])
can_reach_from_N = dfs(G_rev, [N - 1])

can_reach_from_0_and_N = can_reach_from_0 & can_reach_from_N
EdgeShifted = []
for a, b, c in Edge:
    if (a in can_reach_from_0_and_N) and (b in can_reach_from_0_and_N):
        EdgeShifted.append([a, b, c])


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

for n in range(len(can_reach_from_0_and_N)):
    for fr, to, cost in EdgeShifted:
        if dist[to] > dist[fr] + cost:
            dist[to] = dist[fr] + cost
            # 負閉路検出
            if n == len(can_reach_from_0_and_N) - 1:
                print(-1)
                exit()

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

Submission status: Accepted

Solution code:
def bellman_ford(v, G):
    import copy
    INF = 10 ** 10
    ret = [INF] * len(G)
    ret[v] = 0

    for _ in range(len(G) - 1):
        for u, g in enumerate(G):
            for w, l in g:
                if not ret[u] == INF and ret[w] > ret[u] + l:
                    ret[w] = ret[u] + l

    for _ in range(len(G) - 1):
        for u, g in enumerate(G):
            for w, l in g:
                if not ret[w] == INF and ret[w] > ret[u] + l:
                    ret[w] = -INF

    return ret


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

for i in range(M):
    A, B, C = map(int, input().split())
    G[A - 1].append((B - 1, - C + P))

ans = bellman_ford(0, G)[-1]
print("-1" if ans == -10 ** 10 else max(-ans, 0))

Submission status: Accepted

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

N,M,P = map(int,input().split())
edges = []
L = [[] for i in range(N)]
utov = [[] for i in range(N)]
vtou = [[] for i in range(N)]

for i in range(M) :
    u,v,cost = map(int,input().split())
    u -= 1
    v -= 1
    cost -= P
    cost *= -1
    edges.append([u,v,cost])
    utov[u].append(v)
    vtou[v].append(u)

reachableFromS = [False]*N
reachableFromG = [False]*N

def dfs(u) :
    if reachableFromS[u] :
        return
    reachableFromS[u] = True
    for v in utov[u] :
        dfs(v)

def dfs2(v) :
    if reachableFromG[v] :
        return
    reachableFromG[v] = True
    for u in vtou[v] :
        dfs2(u)

dfs(0)
dfs2(N-1)

OK = [reachableFromS[i] * reachableFromG[i] for i in range(N)]

dist = [float("inf")]*N
dist[0] = 0

ans = 0
step = 0
update = True

while update :
    update = False
    for i in range(M) :
        u,v,cost = edges[i]
        if OK[u] * OK[v] == 0 :
            continue
        if dist[v] > dist[u] + cost :
            dist[v] = dist[u] + cost
            update = True
    
    step += 1
    if step > N :
        ans = -1
        break

if ans != -1 :
    ans = max(0, dist[N-1] * (-1))
print(ans)