結果

問題 No.2536 同値性と充足可能性
ユーザー k1suxuk1suxu
提出日時 2023-11-10 23:48:58
言語 C++23
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 5,438 bytes
コンパイル時間 4,130 ms
コンパイル使用メモリ 264,024 KB
実行使用メモリ 23,656 KB
最終ジャッジ日時 2023-11-10 23:49:07
合計ジャッジ時間 8,660 ms
ジャッジサーバーID
(参考情報)
judge15 / judge11
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 WA -
testcase_01 AC 2 ms
6,676 KB
testcase_02 WA -
testcase_03 AC 2 ms
6,676 KB
testcase_04 WA -
testcase_05 WA -
testcase_06 WA -
testcase_07 WA -
testcase_08 WA -
testcase_09 WA -
testcase_10 WA -
testcase_11 WA -
testcase_12 WA -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 WA -
testcase_17 WA -
testcase_18 WA -
testcase_19 AC 2 ms
6,676 KB
testcase_20 AC 5 ms
6,676 KB
testcase_21 AC 6 ms
6,676 KB
testcase_22 AC 6 ms
6,676 KB
testcase_23 AC 43 ms
16,232 KB
testcase_24 AC 42 ms
15,336 KB
testcase_25 WA -
testcase_26 WA -
testcase_27 WA -
testcase_28 WA -
testcase_29 WA -
testcase_30 WA -
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ソースコード

diff #

// #pragma GCC target("avx")
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i = 0; i < (int)n; i++)
#define FOR(n) for(int i = 0; i < (int)n; i++)
#define repi(i,a,b) for(int i = (int)a; i < (int)b; i++)
#define all(x) x.begin(),x.end()
//#define mp make_pair
#define vi vector<int>
#define vvi vector<vi>
#define vvvi vector<vvi>
#define vvvvi vector<vvvi>
#define pii pair<int,int>
#define vpii vector<pair<int,int>>

template<typename T>
bool chmax(T &a, const T b) {if(a<b) {a=b; return true;} else {return false;}}
template<typename T>
bool chmin(T &a, const T b) {if(a>b) {a=b; return true;} else {return false;}}

using ll = long long;
using ld = long double;
using ull = unsigned long long;

const ll INF = numeric_limits<long long>::max() / 2;
const ld pi = 3.1415926535897932384626433832795028;
const ll mod = 998244353;
int dx[] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};

#define int long long

namespace internal {
    template <class E> struct csr {
    std::vector<int> start;
    std::vector<E> elist;
    explicit csr(int n, const std::vector<std::pair<int, E>>& edges)
        : start(n + 1), elist(edges.size()) {
        for (auto e : edges) {
            start[e.first + 1]++;
        }
        for (int i = 1; i <= n; i++) {
            start[i] += start[i - 1];
        }
        auto counter = start;
        for (auto e : edges) {
            elist[counter[e.first]++] = e.second;
        }
    }
};

// Reference:
// R. Tarjan,
// Depth-First Search and Linear Graph Algorithms
struct scc_graph {
  public:
    explicit scc_graph(int n) : _n(n) {}

    int num_vertices() { return _n; }

    void add_edge(int from, int to) { edges.push_back({from, {to}}); }

    // @return pair of (# of scc, scc id)
    std::pair<int, std::vector<int>> scc_ids() {
        auto g = csr<edge>(_n, edges);
        int now_ord = 0, group_num = 0;
        std::vector<int> visited, low(_n), ord(_n, -1), ids(_n);
        visited.reserve(_n);
        auto dfs = [&](auto self, int v) -> void {
            low[v] = ord[v] = now_ord++;
            visited.push_back(v);
            for (int i = g.start[v]; i < g.start[v + 1]; i++) {
                auto to = g.elist[i].to;
                if (ord[to] == -1) {
                    self(self, to);
                    low[v] = std::min(low[v], low[to]);
                } else {
                    low[v] = std::min(low[v], ord[to]);
                }
            }
            if (low[v] == ord[v]) {
                while (true) {
                    int u = visited.back();
                    visited.pop_back();
                    ord[u] = _n;
                    ids[u] = group_num;
                    if (u == v) break;
                }
                group_num++;
            }
        };
        for (int i = 0; i < _n; i++) {
            if (ord[i] == -1) dfs(dfs, i);
        }
        for (auto& x : ids) {
            x = group_num - 1 - x;
        }
        return {group_num, ids};
    }

    std::vector<std::vector<int>> scc() {
        auto ids = scc_ids();
        int group_num = ids.first;
        std::vector<int> counts(group_num);
        for (auto x : ids.second) counts[x]++;
        std::vector<std::vector<int>> groups(ids.first);
        for (int i = 0; i < group_num; i++) {
            groups[i].reserve(counts[i]);
        }
        for (int i = 0; i < _n; i++) {
            groups[ids.second[i]].push_back(i);
        }
        return groups;
    }

  private:
    int _n;
    struct edge {
        int to;
    };
    std::vector<std::pair<int, edge>> edges;
};

}  // namespace internal

struct two_sat {
  public:
    two_sat() : _n(0), scc(0) {}
    explicit two_sat(int n) : _n(n), _answer(n), scc(2 * n) {}

    void add_clause(int i, bool f, int j, bool g) {
        assert(0 <= i && i < _n);
        assert(0 <= j && j < _n);
        scc.add_edge(2 * i + (f ? 0 : 1), 2 * j + (g ? 1 : 0));
        scc.add_edge(2 * j + (g ? 0 : 1), 2 * i + (f ? 1 : 0));
    }
    bool satisfiable() {
        auto id = scc.scc_ids().second;
        for (int i = 0; i < _n; i++) {
            if (id[2 * i] == id[2 * i + 1]) return false;
            _answer[i] = id[2 * i] < id[2 * i + 1];
        }
        return true;
    }
    std::vector<bool> answer() { return _answer; }

  private:
    int _n;
    std::vector<bool> _answer;
    internal::scc_graph scc;
};

void solve() {
    int n, m;
    cin >> n >> m;

    two_sat sat(n);
    FOR(m) {
        int u, v;
        string s;
        cin >> u >> s >> v;
        --u;
        --v;

        if(s == "<==>") {
            sat.add_clause(u, true, v, false);
            sat.add_clause(u, false, v, true);
        }else {
            sat.add_clause(u, true, v, true);
            sat.add_clause(u, false, v, false);
        }
    }

    bool satisfy = sat.satisfiable();
    vector<bool> ans = sat.answer();

    if(!satisfy) {
        cout << "No" << endl;
        return;
    }
    
    cout << "Yes" << endl;

    int sz = (int)count(all(ans), true);

    if(2*sz >= n) {
        cout << sz << endl;
        FOR(n) if(ans[i]) cout << i+1 << " "; cout << "\n";
    }else {
        cout << n-sz << endl;
        FOR(n) if(!ans[i]) cout << i+1 << " "; cout << "\n";
    }
}

signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    solve();
    return 0;
}
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