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[New Problem] Graph Isomorphism (180) and Reduced Binary Quadratic Assignment (181) (#68)
* Fix: Support unbounded ratio in scoring formula for p36, p119, p120 * Fix: Support unbounded ratio in scoring formula for p125 * Add Problem 174, 175 * Fix: problem 174, 175 larger public dataset * Fix: Problem 175 Data Range * Add Problem 176, 177 * Add Problem 178, 179 * Add Problem 180, 181 * Revert "Add Problem 180, 181" This reverts commit 94de9f3. * Add Problem 180, 181
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‎algorithmic/problems/180/chk.cc‎

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#include "testlib.h"
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#include <bits/stdc++.h>
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using namespace std;
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int main(int argc, char* argv[]) {
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registerTestlibCmd(argc, argv);
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int n = inf.readInt(2, 2000);
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int m = inf.readInt(1, n * (n - 1) / 2);
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// Read G1 edges
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set<pair<int,int>> E1;
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for (int i = 0; i < m; i++) {
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int u = inf.readInt(1, n);
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int v = inf.readInt(1, n);
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if (u == v) quitf(_wa, "self-loop in G1");
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if (u > v) swap(u, v);
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E1.insert({u, v});
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}
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// Read G2 edges
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vector<pair<int,int>> E2(m);
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for (int i = 0; i < m; i++) {
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int u = inf.readInt(1, n);
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int v = inf.readInt(1, n);
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if (u == v) quitf(_wa, "self-loop in G2");
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if (u > v) swap(u, v);
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E2[i] = {u, v};
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}
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// Read permutation p (G2 -> G1)
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vector<int> p(n + 1), used(n + 1, 0);
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for (int i = 1; i <= n; i++) {
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p[i] = ouf.readInt(1, n);
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if (used[p[i]]) {
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quitf(_wa, "not a permutation");
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}
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used[p[i]] = 1;
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}
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// Count matched edges
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int matched = 0;
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for (auto &e : E2) {
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int u = p[e.first];
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int v = p[e.second];
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if (u > v) swap(u, v);
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if (E1.count({u, v})) {
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matched++;
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}
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}
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double score = 0.0;
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if (m > 0) score = (double)matched / m;
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score = max(0.0, min(1.0, score));
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// EXACT same output format as old checker
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quitp(score, "Ratio: %.6f", score);
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}

‎algorithmic/problems/180/config.yaml‎

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type: default
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time: 1s
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memory: 1024m
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subtasks:
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- score: 100
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n_cases: 3
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checker: chk.cc
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checker_type: testlib
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filename: std.cc

‎algorithmic/problems/180/statement.txt‎

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# Graph Isomorphism
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## Problem
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You are given two undirected graphs with the same number of vertices.
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Your task is to find a permutation of vertices of the second graph that makes it as isomorphic as possible to the first graph.
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This is the Graph Isomorphism problem, evaluated with a soft scoring rule.
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## Input Format
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- Line 1: two integers n and m
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(2 ≤ n ≤ 2000, 1 ≤ m ≤ n*(n-1)/2)
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- Next m lines: two integers u and v
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(1 ≤ u, v ≤ n, u ≠ v)
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These describe edges of the first graph G₁. Note that there are no duplicate edges.
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- Next m lines: two integers u and v
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(1 ≤ u, v ≤ n, u ≠ v)
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These describe edges of the second graph G₂. Note that there are no duplicate edges.
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## Output Format
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- Output exactly one line:
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- n integers p₁, p₂, ..., pₙ
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- This is a permutation of {1, 2, ..., n}
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- pᵢ = j means vertex i of G₂ is mapped to vertex j of G₁
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## Scoring
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Let:
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- E₁ = set of edges in G₁
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- E₂ = set of edges in G₂
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Define:
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matched_edges = |{(u,v) : (p(u), p(v)) ∈ E₁ and (u,v) ∈ E₂}|
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total_edges = m
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Your score is:
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score = matched_edges / total_edges
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Notes:
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- The score is always in the range [0, 1]
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- Perfect isomorphism yields score = 1
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- Any valid permutation is accepted and scored

‎algorithmic/problems/180/testdata/1.ans‎

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‎algorithmic/problems/180/testdata/1.in‎

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‎algorithmic/problems/180/testdata/2.ans‎

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‎algorithmic/problems/180/testdata/2.in‎

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‎algorithmic/problems/180/testdata/3.ans‎

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