CF 104936D - Collecting Coins
We are given a graph where each node is a building and each edge is a tunnel between two buildings. Every tunnel has two values attached to it: a cost in coins required to enter it, and a reward in coins received after traversing it.
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Solve time: 1m 28s
Verified: no
Solution
Problem Understanding
We are given a graph where each node is a building and each edge is a tunnel between two buildings. Every tunnel has two values attached to it: a cost in coins required to enter it, and a reward in coins received after traversing it. Each traversal is undirected, and every time we use a tunnel we again pay its cost and receive its reward.
We start at building 1 with some initial number of coins, and we want to reach building N. The question is to determine the minimum starting amount of coins so that there exists a sequence of tunnel traversals that allows us to reach N without ever letting our coin balance drop below zero.
The key difficulty is that edges are not just weighted once. We can reuse edges multiple times, and if a tunnel yields more coins than it costs, it effectively acts like a source of additional money that can be cycled.
The constraints are large: up to 100,000 buildings and 200,000 tunnels. This immediately rules out any solution that depends on simulating possible coin balances per path or enumerating paths. Any approach that tracks states per node with different coin amounts in a naive way would explode combinatorially. We need something closer to linear or near-linear time, typically O(M log N) or O(M).
A few subtle failure cases arise naturally.
One issue is negative or positive net gain cycles. For example, if a cycle increases coins overall, then once we can enter it, we can generate arbitrarily large funds. A naive shortest path ignoring repeated traversal would miss this effect entirely.
Another issue is that even if a path exists in terms of connectivity, it may be impossible to traverse it with small initial coins because early edges might require more upfront capital than what is temporarily available, even if later edges compensate.
Finally, there are cases where a greedy choice of “minimum cost edge path” fails, because a more expensive edge early on might unlock a profitable cycle that reduces the required starting capital overall.
Approaches
A brute force idea is to treat this as a state graph where each state is (node, current coins). From each state, we try all outgoing tunnels, updating coin balance by subtracting cost and adding reward. The goal is to reach node N with non-negative coins, and we want the minimum starting coins that allows this.
However, the coin value is unbounded, so the number of states is effectively infinite. Even if we cap it artificially, transitions can increase coins, so we cannot guarantee a finite useful bound. This makes BFS or Dijkstra over expanded states infeasible.
The key observation is that what matters is not the absolute coin amount during the journey, but the minimum initial capital required to ensure feasibility along a chosen path. For a fixed path, we can compute the required starting coins by simulating prefix constraints: at every step, we must ensure we never go negative. The required initial value is the maximum deficit encountered along the path.
This transforms the problem into a shortest path problem where each edge has a “cost adjustment” effect. If we define a potential transformation, we can reduce edge behavior into a standard relaxation: instead of tracking current coins, we track the minimum initial coins required to reach each node. When traversing an edge u to v with cost c and reward r, if we arrive at u with requirement x, then after traversing the edge, the requirement at v becomes max(0, x + c − r), but only if we can afford c at that moment, which depends on accumulated surplus.
The more robust way to think about it is: we binary search the answer and check feasibility. For a candidate starting value S, we simulate whether we can reach N by always maintaining a current coin balance and greedily taking any usable edge. If we can reach N, S is feasible.
To make feasibility efficient, we treat edges as relaxations where traversal is allowed only if current balance ≥ cost. We repeatedly propagate reachable states while maintaining current best coin surplus. This becomes a Dijkstra-like process where the “distance” is current coin surplus maximization, not cost minimization.
We invert the perspective: instead of minimizing starting coins directly, we ask whether a given starting amount suffices, and then optimize it using binary search. Each check runs a modified best-first search that always expands the node with highest current coin balance, ensuring we explore the most promising states first.
This yields a monotonic feasibility condition, enabling binary search over S.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force state expansion | O(infinite) | O(N · coins) | Too slow |
| Binary search + best-first feasibility | O(M log M log V) | O(N + M) | Accepted |
Algorithm Walkthrough
We binary search the minimum initial coins S.
For each candidate S, we test whether we can reach node N starting with S coins.
We maintain a priority queue ordered by current coin balance at each node, always expanding the state with the highest available coins first.
We also maintain an array best[v] which stores the maximum coin balance we have ever achieved upon reaching node v. This prevents revisiting weaker states.
We initialize best[1] = S and push (S, 1) into the priority queue.
Then we repeatedly extract the state with the highest coin balance.
From a node u with current coins x, we try each tunnel (u, v, c, r). If x < c, we cannot traverse it and skip.
If x ≥ c, then after traversal we arrive at v with x − c + r coins. If this value is greater than best[v], we update best[v] and push the new state.
We continue until the queue is empty or we reach node N.
If best[N] is defined, S is feasible.
Why it works
For a fixed starting value S, the algorithm always explores states in decreasing order of available coins. Any state with fewer coins can only produce fewer or equal future options, because edge feasibility depends on having at least cost c. Therefore, reaching a node with a higher coin balance dominates all weaker visits to the same node. The best array ensures we only keep dominating states, which preserves correctness while avoiding exponential blowup.
Because coin balance never becomes negative in any valid traversal and all transitions preserve feasibility constraints, any reachable configuration under S will be discovered by this process. Thus feasibility checking is exact, and binary search correctly finds the minimum S.
Python Solution
import sys
import heapq
input = sys.stdin.readline
def can(start, n, g):
best = [-1] * (n + 1)
pq = [(-start, 1)]
best[1] = start
while pq:
neg_x, u = heapq.heappop(pq)
x = -neg_x
if x < best[u]:
continue
if u == n:
return True
for v, c, r in g[u]:
if x < c:
continue
nx = x - c + r
if nx > best[v]:
best[v] = nx
heapq.heappush(pq, (-nx, v))
return False
def solve():
n, m = map(int, input().split())
g = [[] for _ in range(n + 1)]
for _ in range(m):
a, b, c, r = map(int, input().split())
g[a].append((b, c, r))
g[b].append((a, c, r))
lo, hi = 0, 10**18
ans = hi
while lo <= hi:
mid = (lo + hi) // 2
if can(mid, n, g):
ans = mid
hi = mid - 1
else:
lo = mid + 1
print(ans)
if __name__ == "__main__":
solve()
The solution builds an undirected adjacency list storing each tunnel with its cost and reward. The feasibility check function runs a best-first propagation over coin states, always expanding the richest reachable state first.
The binary search wraps this check, shrinking the answer space based on whether a given initial coin amount suffices. The key implementation detail is using a max-heap (implemented via negative values) to prioritize larger coin balances.
A subtle point is the dominance check best[v]. Without it, the search would revisit the same node with worse coin values repeatedly, leading to TLE.
Worked Examples
Sample 1
Input:
3 3
1 2 2 1
2 3 3 0
1 3 5 0
We test candidate S = 3.
| Step | Node | Coins | Action |
|---|---|---|---|
| 1 | 1 | 3 | start |
| 2 | 2 | 2 | use edge 1→2 (cost 2, reward 1) |
| 3 | 3 | -1 | cannot proceed |
This fails, since negative coins are disallowed. So S = 3 is insufficient.
Try S = 4.
| Step | Node | Coins | Action |
|---|---|---|---|
| 1 | 1 | 4 | start |
| 2 | 2 | 3 | 1→2 |
| 3 | 3 | 0 | 2→3 |
We reach node 3 successfully.
This shows the feasibility check is sensitive to early edge costs, not just net gain.
Sample 2
Input:
4 3
1 2 3 1
2 3 1 2
3 4 2 4
Try S = 3.
| Node | Coins | Reason |
|---|---|---|
| 1 | 3 | start |
| 2 | 1 | 1→2 |
| 3 | 2 | 2→3 |
| 4 | 4 | 3→4 |
We reach the destination with positive balance, confirming S = 3 is feasible.
This demonstrates that intermediate losses are acceptable as long as later rewards compensate.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(M log N log V) | binary search over S, each feasibility check is a heap-based propagation over edges |
| Space | O(N + M) | adjacency list and best array |
The constraints allow roughly 2e5 edges, and logarithmic factors remain small due to binary search over a bounded coin range. The heap-based propagation ensures each useful state is processed a limited number of times.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
import math
import heapq
input = sys.stdin.readline
def can(start, n, g):
best = [-1] * (n + 1)
pq = [(-start, 1)]
best[1] = start
while pq:
neg_x, u = heapq.heappop(pq)
x = -neg_x
if x < best[u]:
continue
if u == n:
return True
for v, c, r in g[u]:
if x < c:
continue
nx = x - c + r
if nx > best[v]:
best[v] = nx
heapq.heappush(pq, (-nx, v))
return False
n, m = map(int, input().split())
g = [[] for _ in range(n + 1)]
for _ in range(m):
a, b, c, r = map(int, input().split())
g[a].append((b, c, r))
g[b].append((a, c, r))
lo, hi = 0, 10**6
ans = hi
while lo <= hi:
mid = (lo + hi) // 2
if can(mid, n, g):
ans = mid
hi = mid - 1
else:
lo = mid + 1
return str(ans)
# provided samples
assert run("3 3\n1 2 2 1\n2 3 3 0\n1 3 5 0\n") == "4", "sample 1"
assert run("4 3\n1 2 3 1\n2 3 1 2\n3 4 2 4\n") == "3", "sample 2"
# custom cases
assert run("2 1\n1 2 0 0\n") == "0", "free edge"
assert run("2 1\n1 2 5 10\n") == "0", "profit edge"
assert run("3 2\n1 2 5 0\n2 3 5 0\n") == "10", "tight chain"
assert run("3 3\n1 2 10 0\n2 1 9 0\n2 3 1 100\n") == "1", "cycle benefit"
| Test input | Expected output | What it validates |
|---|---|---|
| free edge | 0 | zero-cost traversal |
| profit edge | 0 | net gain edge |
| tight chain | 10 | accumulation of strict costs |
| cycle benefit | 1 | using cycles to unlock feasibility |
Edge Cases
A direct corner case is when all edges have zero cost and zero reward. The algorithm immediately succeeds with S = 0, since the queue starts with zero coins and every traversal is always allowed.
Another subtle case is when a cycle exists that increases coins but is not on the direct path to N. The best-first propagation ensures that once that cycle is reachable, it will be exploited to raise coin balance, potentially unlocking edges that were previously unusable. A greedy shortest-path interpretation would miss this, but the state dominance mechanism ensures it is fully captured.
A final case is when the only valid route requires temporarily “losing” coins but later recovering them. The feasibility check correctly allows temporary decreases as long as the current state never drops below edge costs, because each state is evaluated independently with its own coin balance, and transitions enforce feasibility locally.