2-SAT

2-SAT

例题：洛谷P4782

struct SCC {
    int n;
    std::vector<std::vector<int>> edges; // 边
    std::vector<int> stk; // 栈
    std::vector<int> dfn, low, bel; // 公共祖先，强连通分量
    int cur, cnt; // cur: dfn序，cnt: 强连通分量数

    SCC() {}
    SCC(int n) {
        init(n);
    }

    void init(int n) {
        this -> n = n;
        edges.assign(n, {});
        stk.clear();
        dfn.assign(n, -1);
        low.resize(n);
        bel.assign(n, -1);
        cur = cnt = 0;
    }

    void addEdge(int u, int v) {
        edges[u].push_back(v);
    }

    void dfs(int x) {
        dfn[x] = low[x] = cur ++ ;
        stk.push_back(x);

        for (auto y : edges[x]) {
            if (dfn[y] == -1) {
                dfs(y);
                low[x] = std::min(low[x], low[y]);
            } else if (bel[y] == -1) {
                low[x] = std::min(low[x], dfn[y]);
            }
        }

        if (dfn[x] == low[x]) {
            int y;
            do {
                y = stk.back();
                bel[y] = cnt;
                stk.pop_back();
            } while (y != x);
            cnt ++ ;
        }
    }

    std::vector<int> work() {
        for (int i = 0; i < n; i++) {
            if (dfn[i] == -1) {
                dfs(i);
            }
        }
        return bel;
    }

};

void solve() {
    int n, m;
    std::cin >> n >> m;
    
    SCC g(2 * n);
    for (int i = 0; i < m; i++) {
        int u, c_u, v, c_v;
        std::cin >> u >> c_u >> v >> c_v;
        u -- , v -- ;
        // !u 为 n + u, !v 为 n + v
        if (c_u && c_v) { // u和v都为真 { (!u) -> v, (!v) -> u }
        	g.addEdge(n + u, v);
        	g.addEdge(n + v, u);
        } else if (!c_u && c_v) { // u为假, v为真 { u -> v, (!v) -> (!u) }
        	g.addEdge(u, v);
        	g.addEdge(n + v, n + u);
        } else if (c_u && !c_v) { // u为真, v为假 { (!u) -> (!v), v -> u }
        	g.addEdge(n + u, n + v);
        	g.addEdge(v, u);
        } else { // u和v都为假 { u -> (!v), v -> (!u) }
        	g.addEdge(u, n + v);
        	g.addEdge(v, n + u);
        }
    }

    g.work();

    for (int i = 0; i < n; i++) {
    	if (g.bel[i] == g.bel[n + i]) { // 在同一强联通分量内, 无解
    		std::cout << "IMPOSSIBLE\n";
    		return ;
    	}
    }

    std::cout << "POSSIBLE\n";
    for (int i = 0; i < n; i++) {
    	if (g.bel[i] < g.bel[n + i]) {
    		std::cout << "1 ";
    	} else {
    		std::cout << "0 ";
    	}
    }

    std::cout << "\n";

}