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:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    # 

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array, use [[0] * N for _ in range(N)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    R = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main

    N, M, P = R()
    ABC = [R() for _ in range(M)]
    ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
    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
    reachable_from_both_sides = [False] * N

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

    for a, b, c in ABC:
        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)
    
    for i in range(N):
        reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
    
    dist = [-INF for i in range(N)]
    dist[vertex_start] = 0
    for i in range(N):
        for a, b, c in ABC:
            if reachable_from_both_sides[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: Accepted

Solution code:
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))

Submission status: Accepted

Solution code:
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))

Submission status: Accepted

Solution code:
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:
            # print(e, cost)
            if cost[e[1]] < cost[e[0]] + e[2]:
                point = e[1]
                ppoint = e[0]
                # print("point", point)
                if reach[point] and revreach[point]:
                    if reach[ppoint] and revreach[ppoint]:
                        cost[e[1]] = cost[e[0]] + e[2]
                        flag = True
                        # print("update")
        if not flag:
            break
        loop += 1
        # print("loop", loop, cost)
 
 
    # print("judge", reach, revreach)
    if final:
        for e in edges:
            # print(e, cost)
            if cost[e[1]] < cost[e[0]] + e[2]:
                point = e[1]
                ppoint = e[0]
                # print("point", point)
                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(edges)
print(bellman(n, m, edges, vtxs, revs))

Submission status: Accepted

Solution code:
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))

Submission status: Accepted

Solution code:
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]))

Submission status: Accepted

Solution code:
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]))

Submission status: Accepted

Solution code:
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]))

Submission status: Accepted

Solution code:
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]))

Submission status: Accepted

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]
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)

Submission status: Accepted

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]

#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]
    
#負の経路上の頂点の一覧
loop=set()
for i in range(n):
    if d1[i]!=d2[i]:
        loop.add(i)    
ans=max(d2[n-1],0)

#1からの到達可能地点
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
            
#nからの逆辺による到達可能地点
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
            
#1からでもnからでも行ける点でかつ負の閉路に入っているものがあるときは-1
for i in loop:
    if memo[i]==False and memo2[i]==False:
        ans=-1
        
print(ans)

Submission status: Accepted

Solution code:
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):
    #負の経路の検出
    #n:頂点数, w:辺の数, es[i]: [辺の始点,辺の終点,辺のコスト]
    d = [float("inf")] * n
    d[0] = 0 
    #この始点はどこでもよい
    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)

Submission status: Accepted

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

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

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

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

Submission status: Accepted

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

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

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

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