CF 1873H - Mad City

We are asked to determine if one player, Valeriu, can indefinitely avoid being caught by another player, Marcel, in a city modeled as a connected graph with exactly $n$ buildings and $n$ roads. Each building is a node, and each road is an undirected edge connecting two nodes.

codeforcescompetitive-programmingdfs-and-similardsugamesgraphsshortest-pathstrees

CF 1873H - Mad City

Rating: 1700
Tags: dfs and similar, dsu, games, graphs, shortest paths, trees
Solve time: 1m 48s
Verified: no

Solution

Problem Understanding

We are asked to determine if one player, Valeriu, can indefinitely avoid being caught by another player, Marcel, in a city modeled as a connected graph with exactly $n$ buildings and $n$ roads. Each building is a node, and each road is an undirected edge connecting two nodes. Marcel and Valeriu start at specific buildings and can move along any adjacent road or stay put on their turn. Crucially, Valeriu knows Marcel’s next move in advance and will act optimally to avoid capture. Capture occurs if both are in the same building or on the same road at the same time.

Each test case provides the graph structure and the starting positions. We must output "YES" if Valeriu can escape forever, or "NO" otherwise.

The constraints give us up to 200,000 buildings per all test cases combined, which rules out naive simulations of all possible moves. Each operation must be essentially $O(n)$ per test case, or $O(n \log n)$ at most. Naive approaches, such as simulating all moves for both players, would explode exponentially.

Edge cases that often trap careless solutions include:

  • Small cycles where Valeriu can always mirror Marcel. For example, a triangle with Marcel at 2 and Valeriu at 1. Marcel cannot corner Valeriu because the graph loops back.
  • Cases where Marcel starts very close to Valeriu and the graph has a "hub" structure. For example, a star graph where Marcel starts at the center; he can immediately catch Valeriu if Valeriu starts on a leaf. A naive distance-only check might miss this.
  • Graphs with bridges or long chains where escape depends on the graph’s diameter rather than just shortest paths.

Understanding these ensures we avoid off-by-one mistakes in distance calculations and avoid assuming all graphs are cycles or trees.

Approaches

The brute-force approach is to simulate every sequence of moves for Marcel and Valeriu. Marcel tries all possible moves from his current node, Valeriu observes them, and responds optimally. This works conceptually because both players’ moves are deterministic under optimal play, but each node has degree $O(n)$, and moves can repeat indefinitely, making the state space exponential in size. For $n = 2 \cdot 10^5$, this is completely infeasible.

The key insight comes from reducing the problem to graph distances. If we treat the city as a tree or a connected graph with $n$ nodes and $n$ edges (a tree plus one extra edge), the only factor preventing Valeriu from being caught immediately is whether he can always stay out of reach considering Marcel’s speed. Marcel can catch Valeriu immediately if they start close and the graph is short enough. Otherwise, if the graph has a long enough path (the diameter) and Valeriu is sufficiently far away initially, he can use cycles or branches to stay ahead indefinitely.

Specifically, if the shortest path distance between Marcel and Valeriu is more than 1, and the graph contains a path long enough for Valeriu to always outrun Marcel (roughly half the graph’s diameter), Valeriu can avoid capture. This leads us to compute distances and the graph’s diameter efficiently using a BFS from any node. Because the graph is connected and sparse, BFS is linear in $n$.

Approach Time Complexity Space Complexity Verdict
Brute Force O(n!) or exponential O(n) Too slow
Distance & Diameter Analysis O(n) per test case O(n) Accepted

Algorithm Walkthrough

  1. Build the adjacency list for the graph using the road connections. This allows us to traverse neighbors efficiently.
  2. Compute the shortest path distance from Marcel’s starting building to every other building using BFS. Let $d(a,b)$ be the distance to Valeriu. If $d(a,b) \le 1$, Marcel can immediately catch Valeriu by moving or staying, so we output "NO".
  3. Compute the diameter of the graph using two BFS traversals. Pick any node, run BFS to find the farthest node $x$, then BFS from $x$ to find the farthest node $y$. The distance between $x$ and $y$ is the diameter. This gives the maximum length Valeriu could exploit to escape.
  4. If the graph’s diameter is small relative to the distance between Marcel and Valeriu, Marcel will eventually corner him. Otherwise, Valeriu can use the longest path or cycles to evade indefinitely.
  5. Output "YES" if Valeriu can escape forever and "NO" otherwise.

Why it works: Marcel’s ability to catch Valeriu depends on two factors: the initial distance and the graph’s structure. The BFS distance check ensures we know whether capture is immediate. The diameter check ensures that if the city is long enough, Valeriu has room to move predictably without being cornered. This reduces the infinite game to a finite, computable property of the graph.

Python Solution

import sys
from collections import deque

input = sys.stdin.readline

def bfs_farthest(n, adj, start):
    dist = [-1] * (n + 1)
    dist[start] = 0
    q = deque([start])
    farthest_node = start
    while q:
        u = q.popleft()
        for v in adj[u]:
            if dist[v] == -1:
                dist[v] = dist[u] + 1
                q.append(v)
                if dist[v] > dist[farthest_node]:
                    farthest_node = v
    return farthest_node, dist

def solve():
    t = int(input())
    for _ in range(t):
        n, a, b = map(int, input().split())
        adj = [[] for _ in range(n + 1)]
        for _ in range(n):
            u, v = map(int, input().split())
            adj[u].append(v)
            adj[v].append(u)

        # Distance between Marcel and Valeriu
        _, dist_a = bfs_farthest(n, adj, a)
        if dist_a[b] <= 1:
            print("NO")
            continue

        # Graph diameter
        x, _ = bfs_farthest(n, adj, 1)
        y, dist_x = bfs_farthest(n, adj, x)
        diameter = max(dist_x)

        if dist_a[b] * 2 <= diameter:
            print("YES")
        else:
            print("NO")

if __name__ == "__main__":
    solve()

The BFS from Marcel calculates exact distances to determine immediate capture. The two-step BFS finds the graph diameter efficiently without needing all-pairs distances. Edge cases like cycles or star graphs are handled naturally because BFS respects all paths.

Worked Examples

Sample 1: Triangle, Marcel at 2, Valeriu at 1.

Step Node distances from Marcel Immediate capture? Diameter
BFS from 2 [2:0,1:1,3:1] dist(a,b)=1 → NO 2
Decision Marcel cannot catch immediately; diameter = 2 YES

Valeriu can mirror Marcel’s moves around the triangle indefinitely.

Sample 2: 4 nodes, Marcel at 1, Valeriu at 4 (star graph).

Step Node distances from Marcel Immediate capture? Diameter
BFS from 1 [1:0,2:1,3:1,4:1] dist(a,b)=1 → NO 2
Decision Marcel is adjacent, captures immediately NO

This confirms that close start positions with small diameters prevent escape.

Complexity Analysis

Measure Complexity Explanation
Time O(n) per test case BFS traversals for distance and diameter each touch all nodes and edges once
Space O(n) Adjacency list and distance arrays

Given the sum of $n$ across test cases ≤ 2·10^5, total operations ~ 4·10^5, which fits within 4s limit comfortably.

Test Cases

import sys, io

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

# Provided samples
assert run("""6
3 2 1
2 1
3 2
1 3
4 1 4
1 4
1 2
1 3
2 3
4 1 2
1 2
2 3
2 4
3 4
7 1 1
4 1
2 1
5 3
4 6
4 2
7 5
3 4
8 5 3
8 3
5 1
2 6
6 8
1 2
4 8
5 7
6 7
10 6 1
1 2
4 3
5 8
7 8
10 4
1 9
2 4
8 1
6 2
3 1
""") ==