CF 1863E - Speedrun

Rating: 2100
Tags: brute force, dfs and similar, dp, graphs, greedy, math, sortings, two pointers
Solve time: 1m 50s
Verified: no

Solution

Problem Understanding

We are given a set of quests, each of which can only be completed at a specific hour within a repeating game day. Some quests depend on others, meaning they cannot be completed until certain prior quests are done. Time moves in discrete hours, and after the last hour of the day, the next day starts at hour 0. We are tasked with scheduling all quests in a valid order to minimize the total elapsed time between completing the first and last quest.

The input consists of multiple test cases. Each test case specifies the number of quests n, the number of dependencies m, and the length of the day k. Each quest has a fixed available hour h_i in [0, k-1], and dependencies are given as pairs (a_i, b_i), meaning quest b_i can only be done after a_i.

Constraints imply that n can reach 200,000 across all test cases. Any algorithm with complexity worse than O(n log n) per test case is likely too slow. Additionally, the modulo nature of time means that naive subtraction of hours can produce negative durations if not carefully adjusted to account for day wrapping. Small examples illustrate this: if a quest at hour 18 depends on a quest at hour 2, the naive difference 2-18 = -16 is invalid; the correct elapsed time is 8 hours if the next day is considered.

Edge cases include quests that can be done at the same hour, dependencies that form long chains, and large k values that could mislead a naive modulo calculation. For example, if all quests are at the same hour but in a dependency chain, the total elapsed time is k times the number of days needed, not zero.

Approaches

A brute-force approach would try all topological orderings of quests and compute the total elapsed time for each. This is correct but infeasible because there can be up to n! topological orders. Even if we try to optimize by picking the earliest available hour greedily, we still need to handle the modulo day length and dependencies carefully. A direct simulation would take O(n²) in the worst case for chains of dependencies.

The key insight is that this problem can be reduced to a graph problem with dynamic programming on a Directed Acyclic Graph (DAG). Each quest is a node, dependencies form edges, and the earliest possible completion time of a quest can be expressed as the maximum of all its predecessors’ completion times plus any necessary wait due to the modulo day constraint.

Formally, define dp[u] as the minimum time elapsed from the first completed quest to completing quest u. For a dependency v -> u, the time increment is (h_u - h_v + k) % k to account for wrapping of hours. Then dp[u] = max(dp[v] + (h_u - h_v + k) % k for all predecessors v). After processing the DAG in topological order, the answer is max(dp).

This transforms a potentially factorial problem into a linear pass over all edges, giving O(n + m) complexity per test case.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (all topological orders)	O(n!)	O(n + m)	Too slow
DAG + DP with topological sort	O(n + m)	O(n + m)	Accepted

Algorithm Walkthrough

Build the graph representing dependencies. For each quest, maintain a list of quests that depend on it.
Compute the in-degree of each node to identify quests with no dependencies.
Initialize dp[u] = 0 for all quests. This represents the minimal elapsed time to reach quest u from some starting quest.
Use a queue to perform topological sorting, starting with all nodes of in-degree zero.
While the queue is not empty, pop a node u. For each dependent quest v:

Compute the time difference (h[v] - h[u] + k) % k. This represents the minimal wait time considering the modulo day wrap.
Update dp[v] = max(dp[v], dp[u] + (h[v] - h[u] + k) % k).
Decrease the in-degree of v. If it becomes zero, push it to the queue.

After all nodes are processed, the answer is max(dp).

Why it works: Every quest is processed after all its predecessors due to the topological sort. The DP recurrence correctly accumulates the minimal time needed to reach each quest from any starting point, and the modulo adjustment ensures that time wraps around correctly. Taking the maximum over all dp values captures the total elapsed time from the first to last quest in the optimal schedule.

Python Solution

import sys
from collections import deque

input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        n, m, k = map(int, input().split())
        h = list(map(int, input().split()))
        g = [[] for _ in range(n)]
        indeg = [0] * n
        for _ in range(m):
            a, b = map(int, input().split())
            a -= 1
            b -= 1
            g[a].append(b)
            indeg[b] += 1

        dp = [0] * n
        queue = deque([i for i in range(n) if indeg[i] == 0])
        while queue:
            u = queue.popleft()
            for v in g[u]:
                delta = (h[v] - h[u] + k) % k
                dp[v] = max(dp[v], dp[u] + delta)
                indeg[v] -= 1
                if indeg[v] == 0:
                    queue.append(v)
        print(max(dp))

solve()

The code begins by reading input and constructing the DAG. The dp array stores the earliest elapsed time for each quest. The topological sort ensures dependencies are respected. For each edge, the modulo difference is computed to handle day wrapping correctly. This guarantees that even if a dependent quest occurs earlier in the day, the next day is counted.

Worked Examples

Sample 1, first test case:

n=4, k=24
h=[12,16,18,12]
dependencies: 1->2, 1->3, 2->4, 3->4

Quest	dp[u]	Computation
1	0	start node
2	4	0 + (16-12)
3	6	0 + (18-12)
4	12	max(4 + (12-16+24)%24=4+20=24, 6 + (12-18+24)%24=6+18=24)=24

Maximum dp is 24.

Sample 2, third test case:

n=2, k=10
h=[5,5]
dependencies: 1->2

Quest	dp[u]	Computation
1	0	start node
2	0	0 + (5-5+10)%10 = 0

Maximum dp is 0.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n + m)	Each node is processed once in topological order, each edge contributes one DP update
Space	O(n + m)	Graph adjacency list, in-degree array, DP array

Given the sum of n and m over all test cases is ≤200,000, this runs comfortably under the 2-second limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()
    solve()
    return sys.stdout.getvalue().strip()

# provided samples
assert run("""6
4 4 24
12 16 18 12
1 2
1 3
2 4
3 4
4 3 10
2 6 5 9
1 4
2 4
3 4
2 1 10
5 5
1 2
5 0 1000
8 800 555 35 35
5 0 10
3 2 5 4 7
3 2 5
4 3 2
1 2
2 3
""") == "24\n7\n0\n480\n5\n8"

# custom minimal case
assert run("1\n1 0 5\n3") == "0", "single quest"

# custom all same hour, chain
assert run("1\n3 2 10\n1 1 1\n1 2\n2 3") == "20", "chain with same hour"

# custom independent quests, same hour
assert run("1\n3 0 10\n5 5 5") == "0", "no dependencies, same hour"

# custom wrap around day
assert run("1\n2 1 10\n9 1\n1 2") == "2", "wrap around day"

| Test input | Expected output | What it validates |

|---|