CF 104741L - 兄弟校问题

We are given a collection of schools. Each school has a name and a city. We also have a list of keyword strings. A school is considered directly related to a keyword if that keyword appears as a whole token inside the school name, where tokens are the parts separated by…

codeforcescompetitive-programming

CF 104741L - \u5144\u5f1f\u6821\u95ee\u9898

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Solve time: 52s
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Solution

Problem Understanding

We are given a collection of schools. Each school has a name and a city. We also have a list of keyword strings. A school is considered directly related to a keyword if that keyword appears as a whole token inside the school name, where tokens are the parts separated by underscores. Keyword matching is case insensitive, while school names may contain uppercase letters.

There is also a second relation defined on schools: two schools are directly related if they are in the same city or if they share at least one keyword appearing in their names. This relation is then expanded transitively, meaning if A is related to B and B is related to C, then A is related to C as well, even if A and C share no direct property.

The task is, for every school, to compute how many schools belong to its full connected component under this relation.

The constraints n ≤ 1000 and m ≤ 1000 indicate that an O(n²) or O(n² α(n)) solution is acceptable, while anything like O(n³) may already be borderline but still potentially acceptable with tight constant factors. String parsing can be linear in total input size since total name length is at most about 10⁶.

A subtle point is case normalization. Keywords are lowercase, while school names may include uppercase letters, so comparisons must be normalized consistently. Another subtlety is that keywords match only whole underscore-separated tokens, not substrings. For example, keyword "tech" does not match "biotech".

A second subtle issue is transitivity: connectivity is not just “same city or share keyword”, but the transitive closure of that graph. A naive approach that counts only direct neighbors will undercount in chains such as A shares city with B, B shares keyword with C, so A must include C.

Approaches

A direct way to view the problem is as a graph problem. Each school is a node. We connect two nodes with an edge if they are in the same city or if their name token sets intersect with the keyword list in a way that both share at least one keyword. Once edges are built, each answer is simply the size of the connected component containing that node.

The brute-force construction checks every pair of schools. For each pair, we compare cities and also compare keyword intersections by scanning tokens. If two schools share a city or a keyword, we union them. This is correct because it explicitly encodes all direct edges. However, checking all pairs costs O(n²). With up to 1000 schools, this is about 10⁶ pairs, and inside each pair we may scan up to 1000 characters or multiple tokens, leading to roughly 10⁹ operations in the worst case, which is too slow for 1 second.

The key observation is that we do not need to compare every pair directly. Instead, we can group schools by shared attributes. Schools in the same city form a natural group. Keywords also form groups: each keyword connects all schools containing it. This suggests building connectivity incrementally using a union-find structure. Instead of comparing all pairs, we union schools that share a city and union schools that share a keyword via a representative mechanism.

To avoid connecting all schools in a keyword group pairwise, we pick one representative school per city and per keyword bucket, and union all members into that representative. This reduces work to near linear in total occurrences.

Approach Time Complexity Space Complexity Verdict
Brute Force Pairwise Comparison O(n² · L) O(n) Too slow
Union-Find via grouping (city + keyword) O(n α(n) + total tokens) O(n + m) Accepted

Algorithm Walkthrough

We model the problem as building connected components incrementally using disjoint set union.

  1. Normalize all keywords to lowercase and store them in a hash set for O(1) lookup. This ensures case-insensitive matching can be handled consistently.
  2. For each school, parse its name by splitting on underscores to extract tokens. Convert each token to lowercase and check whether it is a keyword. Collect all matching keywords for that school.
  3. Create a disjoint set union structure over n schools, initially each school is its own component.
  4. Maintain a dictionary mapping each city to the first school index seen in that city. When processing a school, if the city has been seen before, union the current school with the stored representative. Otherwise store it as representative.
  5. Maintain a dictionary mapping each keyword to the first school index that contains it. For each keyword found in a school, if it already has a representative school, union the current school with that representative. Otherwise assign it.
  6. After processing all schools, compute component sizes by counting final DSU roots and output the size for each school’s root.

The reason we only store a single representative per city or keyword works because union-find guarantees transitivity. Once all schools sharing a property are connected through a chain of unions, they form one connected component regardless of order.

Why it works

The algorithm builds a graph implicitly where edges are introduced only when two schools share a city or a keyword. Every such edge is represented by a union operation. Since union-find maintains transitive closure, any path formed by these unions merges all reachable nodes into one set. Every valid relation path in the problem corresponds to a sequence of shared attributes, and each step in that path is captured by a union operation at the moment the shared attribute is processed. Therefore, every connected component in the true graph is exactly one DSU set.

Python Solution

import sys
input = sys.stdin.readline

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

def solve():
    n, m = map(int, input().split())
    keywords = set()
    for _ in range(m):
        keywords.add(input().strip().lower())

    dsu = DSU(n)

    city_rep = {}
    word_rep = {}

    schools = []

    for i in range(n):
        line = input().strip().split()
        name = line[0]
        city = line[1]

        tokens = name.split('_')
        kw_list = []

        for t in tokens:
            t_low = t.lower()
            if t_low in keywords:
                kw_list.append(t_low)

        schools.append((city, kw_list))

        if city in city_rep:
            dsu.union(i, city_rep[city])
        else:
            city_rep[city] = i

        for w in kw_list:
            if w in word_rep:
                dsu.union(i, word_rep[w])
            else:
                word_rep[w] = i

    ans = [0] * n
    for i in range(n):
        ans[dsu.find(i)] += 1

    for i in range(n):
        print(ans[dsu.find(i)])

if __name__ == "__main__":
    solve()

The DSU implementation uses path compression via path halving in the iterative find function and union by size. This ensures near constant amortized time per operation. City and keyword maps guarantee that we only connect each school once per shared attribute instead of enumerating all pairs.

Token extraction uses underscore splitting because the problem defines underscores as word separators. Lowercasing both keywords and tokens ensures consistent matching.

Finally, we compute component sizes by counting root frequencies, then output the size for each node's root.

Worked Examples

Example 1

Input:

4 1
jimei_University Xiamen
xiamen_University Xiamen
genshin_University Mihoyo
genshin_Impact Mihoyo
genshin

We have keyword "genshin". Only schools 3 and 4 contain it.

Step School City Keywords found DSU action
1 0 Xiamen [] city_rep[Xiamen]=0
2 1 Xiamen [] union(1,0)
3 2 Mihoyo [] city_rep[Mihoyo]=2
4 3 Mihoyo genshin union(3,2), word_rep[genshin]=3

After unions, component A = {0,1}, component B = {2,3}. Output becomes:

2
2
2
2

This shows that city-based connectivity merges the first two schools even without keywords.

Example 2

Input:

3 2
a_b City1
c_d City2
a_x City2
a
c

Tokens: "a" is keyword, "c" is keyword.

Step School City Keywords Action
1 0 City1 a city_rep[City1]=0, word_rep[a]=0
2 1 City2 c city_rep[City2]=1, word_rep[c]=1
3 2 City2 a union(2,1), union(2,0 via word a)

All nodes become connected due to chain: 0 shares keyword a with 2, 2 shares city with 1. Final output:

3
3
3

This demonstrates transitive closure through mixed relations.

Complexity Analysis

Measure Complexity Explanation
Time O((n + total tokens) α(n)) Each union/find is almost constant, and each token/city processed once
Space O(n + m) DSU arrays plus hash maps for cities and keywords

The constraints n ≤ 1000 and total string length ≤ 10⁶ ensure the algorithm runs comfortably within limits. DSU operations dominate but remain efficient due to path compression.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    from collections import defaultdict

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

    n, m = map(int, input().split())
    keywords = set(input().strip() for _ in range(m))

    dsu = DSU(n)
    city_rep = {}
    word_rep = {}

    for i in range(n):
        name, city = input().split()
        tokens = name.split('_')
        kws = [t.lower() for t in tokens if t.lower() in keywords]

        if city in city_rep:
            dsu.union(i, city_rep[city])
        else:
            city_rep[city] = i

        for w in kws:
            if w in word_rep:
                dsu.union(i, word_rep[w])
            else:
                word_rep[w] = i

    res = [0]*n
    for i in range(n):
        res[dsu.find(i)] += 1

    return "\n".join(str(res[dsu.find(i)]) for i in range(n)) + "\n"

# provided samples (placeholders)
# assert run("...") == "..."

# custom cases
assert run("1 0\nA B\n") == "1\n", "single node"
assert run("2 0\nA B\nC B\n") == "2\n2\n", "same city"
assert run("2 1\nA_x B\nC_x D\nx\n") == "2\n2\n", "keyword merge"
assert run("3 2\na_b C\nc_d D\na C\nd\n") == "3\n3\n3\n", "transitive merge"
Test input Expected output What it validates
1 node 1 trivial base
same city 2,2 city union correctness
keyword merge 2,2 keyword connectivity
transitive 3,3,3 transitive closure

Edge Cases

A corner case arises when a school belongs to both a city group and a keyword group that do not overlap directly. For example, one chain may connect through city and another through keyword.

Input:

3 1
a_b X
c_d Y
a_x Y
a

School 0 connects to keyword a, school 2 also connects to keyword a, and schools 1 and 2 share a city. The algorithm unions 0-2 and 2-1, producing a full component {0,1,2}. The DSU ensures that the order of processing does not matter, since union operations accumulate connectivity regardless of direction.

Another edge case is case sensitivity. Without lowercasing tokens, a keyword match would fail even when logically present. The algorithm explicitly normalizes both sides before comparison, ensuring consistent matching.