2-SAT
(sat/twosat.hpp)

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• Last update: 2024-01-08 13:32:33+09:00
• Include: #include "sat/twosat.hpp"

Description

2-SAT は，節内のリテラル数が高々2つであるような乗法標準形の論理式に対する充足可能性問題 (SAT) である．

節 $(u \lor v)$ が $(\lnot u \rightarrow v) \land (\lnot v \rightarrow u)$ と同値であることを利用すると，2-SAT を強連結成分分解を用いて解くことができる．

Operations

• vector<bool> two_sat(int n, vector<tuple<int, bool, int, bool>> clauses)
  • $n$ リテラルを含む節のリストが与えられた時，すべての節を充足するリテラルの真偽値の組み合わせを一つ返す．節は {i, f, j, g} の形で与え，$((x_i = f) \lor (x_j = g))$ を追加する．問題が充足可能でない場合，空リストを返す．
  • 時間計算量: $O(n)$

Depends on

• Strongly Connected Components (graph/scc.hpp)

Verified with

• test/yosupo/two_sat.test.cpp

Code

#pragma once
#include <vector>

#include "../graph/scc.hpp"

std::vector<bool> two_sat(
    int n, const std::vector<std::tuple<int, bool, int, bool>>& clauses) {
    std::vector<std::vector<int>> G(2 * n);
    std::vector<bool> val(n);

    for (auto& [i, f, j, g] : clauses) {
        G[n * f + i].push_back(n * (!g) + j);
        G[n * g + j].push_back(n * (!f) + i);
    }

    auto comp = scc(G);
    for (int i = 0; i < n; ++i) {
        if (comp[i] == comp[n + i]) {
            // not satisfiable
            return {};
        }
        val[i] = comp[i] > comp[n + i];
    }
    return val;
}

#line 2 "sat/twosat.hpp"
#include <vector>

#line 2 "graph/scc.hpp"
#include <algorithm>
#include <ranges>
#line 5 "graph/scc.hpp"

std::vector<int> scc(const std::vector<std::vector<int>>& G) {
    const int n = G.size();
    std::vector<std::vector<int>> G_rev(n);
    for (int u = 0; u < n; ++u) {
        for (int v : G[u]) G_rev[v].push_back(u);
    }
    std::vector<int> comp(n, -1), order(n);
    std::vector<bool> visited(n);

    auto dfs = [&](const auto& dfs, int u) -> void {
        if (visited[u]) return;
        visited[u] = true;
        for (int v : G[u]) dfs(dfs, v);
        order.push_back(u);
    };

    for (int v = 0; v < n; ++v) dfs(dfs, v);
    int c = 0;

    auto rdfs = [&](const auto& rdfs, int u, int c) -> void {
        if (comp[u] != -1) return;
        comp[u] = c;
        for (int v : G_rev[u]) rdfs(rdfs, v, c);
    };

    for (int v : order | std::views::reverse) {
        if (comp[v] == -1) rdfs(rdfs, v, c++);
    }
    return comp;
}

std::vector<std::vector<int>> contract(const std::vector<std::vector<int>>& G,
                                       const std::vector<int>& comp) {
    const int n = *std::ranges::max_element(comp) + 1;
    std::vector<std::vector<int>> G2(n);
    for (int i = 0; i < (int)G.size(); ++i) {
        for (int j : G[i]) {
            if (comp[i] != comp[j]) {
                G2[comp[i]].push_back(comp[j]);
            }
        }
    }
    for (int i = 0; i < n; ++i) {
        std::ranges::sort(G2[i]);
        G2[i].erase(std::ranges::unique(G2[i]).begin(), G2[i].end());
    }
    return G2;
}
#line 5 "sat/twosat.hpp"

std::vector<bool> two_sat(
    int n, const std::vector<std::tuple<int, bool, int, bool>>& clauses) {
    std::vector<std::vector<int>> G(2 * n);
    std::vector<bool> val(n);

    for (auto& [i, f, j, g] : clauses) {
        G[n * f + i].push_back(n * (!g) + j);
        G[n * g + j].push_back(n * (!f) + i);
    }

    auto comp = scc(G);
    for (int i = 0; i < n; ++i) {
        if (comp[i] == comp[n + i]) {
            // not satisfiable
            return {};
        }
        val[i] = comp[i] > comp[n + i];
    }
    return val;
}

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