Problem:


Score : 500 points

Problem Statement

There is a directed graph with N vertices numbered 1 to N and M edges.
The i-th edge is directed from Vertex A_i to Vertex B_i, and there are C_i coins placed along that edge.
Additionally, there is a button on Vertex N.

We will play a game on this graph.
You start the game on Vertex 1 with zero coins, and head for Vertex N by traversing the edges while collecting coins.
It takes one minute to traverse an edge, and you can collect the coins placed along the edge each time you traverse it.
As usual in games, even if you traverse an edge once and collect the coins, the same number of coins will reappear next time you traverse that edge, which you can collect again.

When you reach Vertex N, you can end the game by pressing the button. (You can also choose to leave Vertex N without pressing the button and continue traveling.)
However, when you end the game, you will be asked to pay T \times P coins, where T is the number of minutes elapsed since the start of the game. If you have less than T \times P coins, you will have to pay all of your coins instead.

Your score will be the number of coins you have after this payment.
Determine if there exists a maximum value of the score that can be obtained. If the answer is yes, find that maximum value.

Constraints

2 \leq N \leq 2500

1 \leq M \leq 5000

1 \leq A_i, B_i \leq N

1 \leq C_i \leq 10^5

0 \leq P \leq 10^5

All values in input are integers.

Vertex N can be reached from Vertex 1.

Input

Input is given from Standard Input in the following format:

N M P
A_1 B_1 C_1
:
A_M B_M C_M

Output

If there exists a maximum value of the score that can be obtained, print that maximum value; otherwise, print -1.

Sample Input 1

3 3 10
1 2 20
2 3 30
1 3 45

Sample Output 1

35

There are two ways to travel from Vertex 1 to Vertex 3:

Vertex 1 \rightarrow 2 \rightarrow 3: You collect 20 + 30 = 50 coins on the way. After two minutes from the start of the game, you press the button, pay 2 \times 10 = 20 coins, and you have 50 - 20 = 30 coins left.

Vertex 1 \rightarrow 2: You collect 45 coins on the way. After one minute from the start of the game, you press the button, pay 1 \times 10 = 10 coins, and you have 45 - 10 = 35 coins left.

Thus, the maximum score that can be obtained is 35.

Sample Input 2

2 2 10
1 2 100
2 2 100

Sample Output 2

-1

The edge extending from Vertex 1 takes you to Vertex 2. If you then traverse the edge extending from Vertex 2 to itself t times and press the button, your score will be 90 + 90t. Thus, you can infinitely increase your score, which means there is no maximum value of the score that can be obtained.

Sample Input 3

4 5 10
1 2 1
1 4 1
3 4 1
2 2 100
3 3 100

Sample Output 3

0

There is no way to travel from Vertex 1 to Vertex 4 other than traversing the edge leading from Vertex 1 to Vertex 4 directly. You will pick up one coin along this edge, but after being asked to paying 10 coins, your score will be 0.

Note that you can collect an infinite number of coins if you traverse the edge leading from Vertex 1 to Vertex 2, but this is pointless since you can no longer reach Vertex 4 and end the game.



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


Sample output:
35

Submission status: Wrong Answer

Solution code:
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-10) for a, b, c in ABC]
    G = [[INF]*N for _ in range(N)]
    
    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

    dfs_from_start = []
    dfs_to_end = []
    reachable_to_direct = defaultdict(list)
    reachable_from_direct = defaultdict(list)

    for a, b, c in ABC:
        G[a][b] = c
        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)
    
    dist = [-INF for i in range(N)]
    dist[vertex_start] = 0
    for i in range(N):
        for a, b, c in ABC:
            if reachable_from_start[a] and reachable_to_end[a]:
                if dist[a] != INF and dist[a] + c > dist[b]:
                    dist[b] = dist[a] + c
                    if i == N-1:
                        print(-1)
                        return
    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()

Submission status: Wrong Answer

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

next = [0]*N
next[0] = 1

for i in range(N*2):
    nenext = [0]*N
    for j in range(N):
        if next[j] == 1:
            for b,c in edge[j]:
                if cost[b] < cost[j]+c:
                    cost[b] = cost[j]+c
                    A[b] = j
                    nenext[b] = 1
    next = nenext[:]

if A[0] != -1:
    print(-1)
    exit()

p = N-1
while 1:
    if p == -1 or p == N:
        break
    A[p],p = N,A[p]

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

Submission status: Wrong Answer

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

for i in range(N*2):
    for j in range(N):
        for b,c in edge[j]:
            if cost[b] < cost[j]+c:
                cost[b] = cost[j]+c
                A[b] = j

if A[0] != -1:
    print(-1)
    exit()

p = N-1
while 1:
    if p == -1 or p == N:
        break
    A[p],p = N,A[p]

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

Submission status: Wrong Answer

Solution code:
N, M, P = [int(_) for _ in input().split()]
ABC = [[int(_) for _ in input().split()] for _ in range(M)]
G = [{} for _ in range(N)]
for abc in ABC:
    a, b, c = abc
    a -= 1
    b -= 1
    G[a][b] = c - P


def bellman_ford(V, G, i):
    INF = float('inf')
    D = [-INF] * V
    D[i] = 0
    back = [None] * V
    for _ in range(V + 1):
        update_this = False
        update = [False] * V
        for fr in range(V):
            for to in G[fr]:
                if D[fr] != -INF and D[to] < D[fr] + G[fr][to]:
                    D[to] = D[fr] + G[fr][to]
                    update_this = update[to] = True
                    back[to] = fr
        if not update_this:
            return D[N - 1]
    to = N - 1
    while True:
        if update[to]:
            return False
        to = back[to]
        if to is None:
            return D[N - 1]


ans = bellman_ford(N, G, 0)
if ans is False:
    print(-1)
else:
    print(max(0, ans))

Submission status: Wrong Answer

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

for i in range(N-1):
    for j in range(N):
        for b,c in edge[j]:
            cost[b] = max(cost[b],cost[j]+c)
print(max(cost[-1],0))

Submission status: Wrong Answer

Solution code:
N, M, P = map(int, input().split())
ABC = []
for _ in range(M):
    a, b, c = map(int, input().split())
    ABC.append((a, b, c))
    
import sys
INT_MAX = sys.maxsize # 9223372036854775807

def bellman_ford(edges,num_v,source):

    dist=[INT_MAX] * num_v
    dist[source]=0

    #辺の緩和
    has_negative = False
    for i in range(num_v):
        for edge in edges:
            if edge[0] != INT_MAX and dist[edge[1]] > dist[edge[0]] + edge[2]:
                dist[edge[1]] = dist[edge[0]] + edge[2]
                if i==num_v-1:
                    has_negative = True
    
    if has_negative == False:
        return dist
    
    negative = [False] * num_v
    for i in range(num_v):
        for from_edge, to_edge, cost in edges:
            if dist[from_edge] == INT_MAX:
                continue
            if dist[to_edge] > dist[from_edge] + cost:
                dist[to_edge] = dist[from_edge] + cost
                negative[to_edge] = True
            
            if negative[from_edge] == INT_MAX:
                negative[to_edge] = True
                
    for i, is_negative in enumerate(negative):
        if is_negative:
            dist[i] = -1 * INT_MAX
        
    return dist

edges = ABC[:]
edges = [(a, b, -c+P) for a, b, c in edges]

shortest_path = bellman_ford(edges, N+1, 1)
ans = shortest_path[N]
if ans == -1 * INT_MAX:
    print(-1)
elif ans > 0:
    print(0)
else:
    print(-ans)

Submission status: Wrong Answer

Solution code:
# pdf
from collections import defaultdict

inf = 10 ** 16
fr = defaultdict(list)
n, m, p = map(int, input().split())
abc = [list(map(int, input().split())) for _ in range(m)]
dp = [inf] * n
dp[0] = 0
cnt=0
for _ in range(2*n):
    ndp = dp[:]
    for a, b, c in abc:
        a, b = a - 1, b - 1
        tmp = ndp[a] - c + p
        if tmp < dp[b]:
            dp[b] = tmp
    if dp[-1] == ndp[-1]:
        cnt+=1
    else:
        cnt=0
    if cnt==n:
        print(max(-dp[-1], 0))
        break
else:
    print(-1)

Submission status: Wrong Answer

Solution code:
N, M, P = map(int, input().split())
edges = []
for i in range(M):
    A, B, C = map(int, input().split())
    edges.append([A, B, P - C])


def BellmanFord(edges, num_v, source):
    minus_flag = 0
    inf_flag = 0
    # グラフの初期化
    inf = float("inf")
    dist = [inf for i in range(num_v)]
    dist[source - 1] = 0

    # 辺の緩和
    for i in range(num_v * 3):
        for edge in edges:
            if edge[0] != inf and dist[edge[1] - 1] > dist[edge[0] - 1] + edge[2]:
                dist[edge[1] - 1] = dist[edge[0] - 1] + edge[2]
                if i == num_v - 1:
                    minus_flag = 1
                if minus_flag == 1:
                    if edge[1] == N:
                        inf_flag = 1

    if inf_flag == 0:
        return dist
    else:
        dist.append(-1)
        return dist


ans = BellmanFord(edges, N, 1)
if len(ans) != N:
    print(-1)
else:
    print(max(0, ans[-1] * -1))

Submission status: Wrong Answer

Solution code:
# pdf
from collections import defaultdict

inf = 10 ** 16
fr = defaultdict(list)
n, m, p = map(int, input().split())
abc = [list(map(int, input().split())) for _ in range(m)]
dp = [inf] * n
dp[0] = 0
cnt=0
for _ in range(2*n):
    ndp = dp[:]
    for a, b, c in abc:
        a, b = a - 1, b - 1
        tmp = ndp[a] - c + p
        if tmp < dp[b]:
            dp[b] = tmp
    if dp[-1] == ndp[-1]:
        cnt+=1
    else:
        cnt=0
    if cnt==n-1:
        print(max(-dp[-1], 0))
        break
else:
    print(-1)

Submission status: Wrong Answer

Solution code:
# pdf
from collections import defaultdict

inf = 10 ** 9
fr = defaultdict(list)
n, m, p = map(int, input().split())
abc = [list(map(int, input().split())) for _ in range(m)]
dp = [inf] * n
dp[0] = 0
cnt=0
for _ in range(2*n):
    ndp = dp[:]
    for a, b, c in abc:
        a, b = a - 1, b - 1
        tmp = ndp[a] - c + p
        if tmp < dp[b]:
            dp[b] = tmp
    if dp[-1] == ndp[-1]:
        cnt+=1
    else:
        cnt=0
    if cnt==n-1:
        print(max(-dp[-1], 0))
        break
else:
    print(-1)

Submission status: Wrong Answer

Solution code:
# pdf
from collections import defaultdict

inf = 10 ** 9
fr = defaultdict(list)
n, m, p = map(int, input().split())
abc = [list(map(int, input().split())) for _ in range(m)]
dp = [inf] * n
dp[0] = 0
for _ in range(n):
    ndp = dp[:]
    stop = True
    for a, b, c in abc:
        a, b = a - 1, b - 1
        tmp = ndp[a] - c + p
        if tmp < dp[b]:
            dp[b] = tmp
            stop = False
    # print(dp)
    if stop:
        print(max(-dp[-1], 0))
        break
else:
    if dp[-1]==ndp[-1]:
        print(max(-dp[-1],0))
    else:
        print(-1)

Submission status: Wrong Answer

Solution code:
# pdf
from collections import defaultdict

inf = 10 ** 9
fr = defaultdict(list)
n, m, p = map(int, input().split())
for _ in range(m):
    a, b, c = map(int, input().split())
    a -= 1
    b -= 1
    fr[b].append([a, -(c - p)])
dp = [inf] * n
dp[0] = 0
for _ in range(n):
    ndp = dp[:]
    flag = False
    flagn = False
    for i in range(n):
        if fr[i]:
            tmp = min(dp[oi] + c for oi, c in fr[i])
            if tmp < ndp[i]:
                ndp[i] = tmp
                flag = True
                if i == n - 1:
                    flagn = True
#    print(dp)
    if not flag: break
    dp = ndp
if flagn:
    print(-1)
else:
    print(max(0,-dp[n - 1]))

Submission status: Wrong Answer

Solution code:
#入力
n,m,r=map(int,input().split())
abc=[list(map(int,input().split())) for _ in [0]*m]
#グラフ
g=[[] for _ in [0]*n]
[g[a-1].append([b-1,c]) for a,b,c in abc]
#逆辺
g2=[[] for _ in [0]*n]
[g2[b-1].append([a-1,c]) for a,b,c in abc]

#前処理用のDFS
def remove_edge(g,start):
    q=[start]
    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
    return {i for i,j in enumerate(memo) if j==True}

#前処理
for i in remove_edge(g,0)|remove_edge(g2,n-1):
    g[i]=[]

#ベルマンフォード
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]
        
#1からnの間に負の閉路があるときは-1
if d1==d2:
    print(max(d1[-1],0))
else:
    print(-1)

Submission status: Wrong Answer

Solution code:
#単一始点最短経路問題
#負の辺があっても動作する

#最短経路を求める
#計算量O(N * M)

INF = 10 ** 15
#入力
# N = int(input()) #頂点の数
# M = int(input()) #辺の数
N, M, P = map(int, input().split())

# 入力は1-index
# 内部で0-indexにして処理
G = []
for _ in range(M): #M個の辺の情報を受け取る
    l, r, s = map(int, input().split()) #lからrへ重みsの辺が存在
    G += [[l-1, r-1, -(s-P)]] #有向グラフのときはここだけ
    # G += [[r-1, l-1, s]] #無向グラフのときはこの行も必要


#メイン文
d = [0] * N #最短距離を記録しておくリスト
# prev = [-1] * N
def shortest_path(s): #S番目の頂点から各頂点への最短距離を求める
    s -= 1
    for i in range(N):
        d[i] = INF
    d[s] = 0
    count = 0
    while True:
        count += 1
        update = False
        for i in range(M): #無向グラフのときは辺は実質的には2倍
            e = G[i]
            if d[e[0]] != INF and d[e[1]] > d[e[0]] + e[2]:
                d[e[1]] = d[e[0]] + e[2]
                update = True
                # prev[e[1]] = e[0]
        if not update: #更新が終わったとき
            return True
        if count == N - 1: #永遠に更新が起きているとき
            tmp = d[N -1]
        if count == 2 * N:
            if d[N - 1] == tmp:
                return True #N-1の値が更新されていないとき
            else:
                return False #N + 2回目でもN - 1の値が小さくなっている-->得点が無限に大きくなる

#呼び出しスタート地点を入れる
if not shortest_path(1):
    print (-1)
    # print (d)
    exit()
# print (d)

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

Submission status: Wrong Answer

Solution code:
import sys
inf = 10000000000

def BellmanFord(edges,points,start):#edgeは1-indexedの起点/終点/距離
  #グラフの初期化
  dist=[inf for i in range(points)]
  dist[start-1]=0
  
  #辺の緩和
  for i in range(points):
    for edge in edges:
      if dist[edge[0]-1]!=inf and  dist[edge[1]-1] > dist[edge[0]-1] + edge[2]:
        dist[edge[1]-1] = dist[edge[0]-1] + edge[2]
    dist1 = dist[N-1]
  for i in range(points):
    for edge in edges:
      if dist[edge[0]-1]!=inf and  dist[edge[1]-1] > dist[edge[0]-1] + edge[2]:
        dist[edge[1]-1] = dist[edge[0]-1] + edge[2]
    dist2 = dist[N-1]
  if dist1 != dist2:return "Loop"
  return (dist)

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

Submission status: Wrong Answer

Solution code:
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]

for a,b,c in ABC:
    Ef[a-1][b-1] = P-c
    Eb[b-1][a-1] = P-c
    
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)

Submission status: Wrong Answer

Solution code:
from collections import deque
N,M,P = map(int,input().split())
side = [[]for i in range(N)]
for i in range(M):
    a,b,c = map(int,input().split())
    side[a-1].append([b-1,c-P])
node = [-float('inf')]*N
node[0] = 0
A = [-1]*N
A[0] = 0
queue = deque([0,-1])
i = 0
while i < N*2:
    p = queue.popleft()
    if p != -1:
        for b,v in side[p]:
            if node[b] < node[p]+v:
                node[b] = node[p]+v
                A[b] = p
                queue.append(b)
    else:
        queue.append(-1)
        i += 1
if A[0] == 0:
    p = N-1
    for i in range(N):
        p = A[p]
    if p == 0:
        print(max(node[-1],0))
    else:
        print(-1)
else:
    print(-1)

Submission status: Wrong Answer

Solution code:
from collections import deque
N,M,P = map(int,input().split())
side = [[]for i in range(N)]
for i in range(M):
    a,b,c = map(int,input().split())
    side[a-1].append([b-1,c-P])
node = [-float('inf')]*N
node[0] = 0
A = [-1]*N
A[0] = 0
queue = deque([0,-1])
i = 0
while i < N+5:
    p = queue.popleft()
    if p != -1:
        for b,v in side[p]:
            if node[b] < node[p]+v:
                node[b] = node[p]+v
                A[b] = p
                queue.append(b)
    else:
        queue.append(-1)
        i += 1
if A[0] == 0:
    p = N-1
    for i in range(N):
        p = A[p]
    if p == 0:
        print(max(node[-1],0))
    else:
        print(-1)
else:
    print(-1)

Submission status: Wrong Answer

Solution code:
from collections import defaultdict

inf = 10 ** 9
to = defaultdict(list)
fr = defaultdict(list)
n, m, p = map(int, input().split())
for _ in range(m):
    a, b, c = map(int, input().split())
    a -= 1
    b -= 1
    to[a].append([b, c])
    fr[b].append([a, c])
dp = [[-inf] * n for _ in range(n + 2)]
dp[0][0] = 0
for i in range(1, n + 2):
    for node in range(n):
        if fr[node]:
            dp[i][node] = max(dp[i - 1][node], max(c + dp[i - 1][k] for k, c in fr[node]))
#print(dp)
if dp[-1][-1] - dp[-2][-1] > p:
    print(-1)
else:
    print(max(0, max(k[-1] - p * i for i, k in enumerate(dp))))