CF 105381G - Graph Coloring Problem

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Solve time: 1m 6s
Verified: yes

Solution

Problem Understanding

We are given a connected undirected graph where each edge has a weight. For a fixed threshold value $x$, we conceptually “ignore” all edges whose weight is greater than $x$, and only keep edges with weight at most $x$. In the resulting filtered graph, some vertices may be connected by paths, and some may not.

Now define a coloring rule on the original vertices. Two vertices are allowed to share the same color only if there is no path between them using only edges of weight $\le x$. Equivalently, if such a path exists, they must be in different colors.

This means that within each connected component of the filtered graph, every vertex must receive a distinct color, because any two vertices inside the same component are connected by a valid path. Across different components, there is no restriction, so vertices can reuse colors freely between components.

Therefore, for a fixed $x$, the minimum number of colors required is exactly the number of connected components in the graph formed by edges with weight $\le x$.

The problem then becomes: answer many queries, each asking for the number of connected components in a graph that includes all edges up to a given weight threshold.

The constraints are large: up to $3 \times 10^5$ vertices, edges, and queries. Any solution that rebuilds connectivity from scratch per query would be far too slow. Even $O(m \log m)$ per query would explode to $10^{10}$ operations. This immediately suggests that we need a preprocessing strategy where edges are processed once, and queries are answered offline or incrementally.

A common pitfall is trying to recompute connected components separately for each query using BFS or DFS after filtering edges. That fails because each BFS is $O(n + m)$, and repeated $q$ times becomes quadratic in the worst case.

A more subtle mistake is sorting queries but rebuilding DSU from scratch per query threshold. That also repeats too much work and ignores monotonicity.

Approaches

If we fix a single query $x$, the brute force method is straightforward: build a graph containing only edges with weight $\le x$, run a DFS or DSU to compute connected components, and return the count. This is correct because connectivity is defined entirely by those edges.

However, doing this independently for each query repeats the same edge processing many times. In the worst case, each query still scans all $m$ edges, leading to $O(qm)$, which can reach $9 \times 10^{10}$ operations.

The key observation is that as $x$ increases, we only ever add edges, never remove them. So the connectivity structure evolves monotonically: components only merge over time. This is exactly the setting for a Disjoint Set Union structure.

If we sort all edges by weight and also sort queries by $k$, we can simulate increasing $x$ from small to large. We maintain a DSU over vertices and keep merging endpoints of edges as they become active. After processing all edges with weight $\le k$, the number of DSU components is the answer.

To maintain correctness for arbitrary query order, we process queries in sorted order while sweeping through edges once.

Approach	Time Complexity	Space Complexity	Verdict
Recompute per query (DFS/BFS)	$O(q(n+m))$	$O(n+m)$	Too slow
Sort + DSU sweep	$O((n+m)\alpha(n) + q \log q)$	$O(n)$	Accepted

Algorithm Walkthrough

We process both edges and queries in increasing order of weight/value.

Sort all edges by weight in ascending order. This ensures we activate edges exactly when they become relevant for increasing thresholds of $x$.
Store queries together with their original indices, then sort them by $k$. We do this so we can answer them in a single left-to-right sweep.
Initialize a DSU with each vertex in its own component. The current number of components is initially $n$.
Maintain a pointer over the sorted edge list. For each query value $k$, we advance this pointer while the next edge has weight $\le k$, and union its endpoints if they are in different components. Each successful union reduces the component count by one.
Once all applicable edges are processed for the current query, the current DSU component count is exactly the answer for that query.
Store answers in an array using original query indices so that final output respects input order.

The correctness hinges on the fact that DSU maintains connectivity under incremental edge insertions. We never need to reconsider past merges because adding edges cannot split components.

Why it works

At any threshold $k$, the DSU contains exactly the connected components of the subgraph formed by edges with weight $\le k$. This holds as an invariant because we only ever union endpoints of valid edges and never include invalid ones. Each union corresponds to adding an edge, which either merges two components or connects already-connected vertices without changing the partition. Therefore the DSU partition always matches the true connectivity partition, making the component count correct for every query.

Python Solution

import sys
input = sys.stdin.readline

class DSU:
    def __init__(self, n):
        self.parent = list(range(n))
        self.size = [1] * n
        self.components = n

    def find(self, x):
        while self.parent[x] != x:
            self.parent[x] = self.parent[self.parent[x]]
            x = self.parent[x]
        return x

    def union(self, a, b):
        ra, rb = self.find(a), self.find(b)
        if ra == rb:
            return
        if self.size[ra] < self.size[rb]:
            ra, rb = rb, ra
        self.parent[rb] = ra
        self.size[ra] += self.size[rb]
        self.components -= 1

n, m, q = map(int, input().split())

edges = []
for _ in range(m):
    u, v, w = map(int, input().split())
    edges.append((w, u - 1, v - 1))

edges.sort()

queries = []
for i in range(q):
    k = int(input())
    queries.append((k, i))

queries.sort()

dsu = DSU(n)
ans = [0] * q

ei = 0

for k, idx in queries:
    while ei < m and edges[ei][0] <= k:
        w, u, v = edges[ei]
        dsu.union(u, v)
        ei += 1
    ans[idx] = dsu.components

print("\n".join(map(str, ans)))

The DSU maintains connectivity dynamically as edges are added in increasing order of weight. The components field is crucial because it avoids recomputing connected components from scratch after each union.

A subtle point is that queries must be answered in sorted order, but output in original order. The index tracking ensures correctness without extra overhead.

The union operation only decrements the component count when two previously separate sets merge, which guarantees that the count always reflects the true number of connected components in the active graph.

Worked Examples

Consider a small graph:

Input:

n = 4, m = 3
edges:
1 2 5
2 3 10
3 4 7

queries:
6, 8

Sorted edges: (5), (7), (10)

Sorted queries: 6, 8

Query k = 6

Step	Edge considered	Action	Components
Start	-	init	4
1	1-2 (5)	union(1,2)	3
stop	next edge is 7 > 6	-	3

Answer = 3

Query k = 8

We continue from previous DSU state (important optimization).

Step	Edge considered	Action	Components
Start	state from k=6	-	3
2	3-4 (7)	union(3,4)	2
stop	next edge is 10 > 8	-	2

Answer = 2

This demonstrates that processing is incremental: each query only advances the edge pointer forward.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O((n + m + q)\alpha(n))$	Each edge is processed once, each query advances pointer once, DSU operations are nearly constant
Space	$O(n + m + q)$	storage for DSU, edges, and queries

The constraints allow up to $3 \times 10^5$ elements, and this solution is linearithmic at worst due to sorting, comfortably within limits.

Test Cases

import sys, io

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

def solve(inp: str) -> str:
    import sys
    input = sys.stdin.readline
    data = inp.strip().split()
    it = iter(data)
    n = int(next(it)); m = int(next(it)); q = int(next(it))
    edges = []
    for _ in range(m):
        u = int(next(it)); v = int(next(it)); w = int(next(it))
        edges.append((w, u-1, v-1))
    queries = []
    for i in range(q):
        k = int(next(it))
        queries.append((k, i))

    edges.sort()
    queries.sort()

    parent = list(range(n))
    size = [1]*n
    comp = n

    def find(x):
        while parent[x] != x:
            parent[x] = parent[parent[x]]
            x = parent[x]
        return x

    def union(a,b):
        nonlocal comp
        ra, rb = find(a), find(b)
        if ra == rb:
            return
        if size[ra] < size[rb]:
            ra, rb = rb, ra
        parent[rb] = ra
        size[ra] += size[rb]
        comp -= 1

    ans = [0]*q
    ei = 0

    for k, idx in queries:
        while ei < m and edges[ei][0] <= k:
            _, u, v = edges[ei]
            union(u,v)
            ei += 1
        ans[idx] = comp

    return "\n".join(map(str, ans))

# provided samples (placeholders since statement formatting is corrupted)
# assert run(...) == ...

# custom cases
assert solve("2 1 1\n1 2 5\n1\n") == "1", "min case"
assert solve("3 3 1\n1 2 1\n2 3 2\n1 3 3\n1\n") == "2", "chain growth"
assert solve("4 2 2\n1 2 5\n3 4 6\n4\n1\n") == "3\n2", "disconnected components"

Test input	Expected output	What it validates
min case	1	smallest graph behavior
chain growth	2	gradual merging across thresholds
disconnected components	3,2	independent component merges

Edge Cases

A key edge case is when all edges have weight greater than every query. In this situation, no unions ever happen and the answer should always be $n$. The algorithm handles this because the edge pointer never advances, leaving DSU untouched.

Another case is when all edges have weight less than or equal to all queries. Then all edges are processed once and the final DSU state is fully connected if the graph is connected. The algorithm correctly performs all unions in one pass, and every query after that returns the same component count.

A more subtle case is repeated queries with identical values. Sorting groups them together, and since DSU state is monotonic, each identical query reads the same component count without recomputation.