[洛谷P4171][JSOI2010]满汉全席

题目大意：有$n$个点，每个点可以选或不选，有$m$组约束，形如$a,u,b,v$，表示$u=a,v=b$中至少要满足一个条件，问是否存在一组解，多组询问

题解：$2-SAT$，感觉是板子题呀，最后判断一下每一个点选与不选是否在同一个强连通分量内即可

卡点：无

C++ Code：

#include <algorithm>
#include <cstdio>
#include <iostream>
#define maxn 210
#define maxm 2010

int head[maxn], cnt;
struct Edge {
	int to, nxt;
} e[maxm];
inline void addedge(int a, int b) {
	e[++cnt] = (Edge) { b, head[a] }; head[a] = cnt;
}

int Tim, n, m, nn;
inline int getpos(int a, int b) { return a * n + b; }
inline void addedge(bool a, int b, bool c, int d) {
	addedge(getpos(!a, b), getpos(c, d));
	addedge(getpos(!c, d), getpos(a, b));
}

int DFN[maxn], low[maxn], idx;
int S[maxn], top, bel[maxn], scc;
bool inS[maxn];
void tarjan(int u) {
	DFN[u] = low[u] = ++idx;
	inS[S[++top] = u] = true;
	int v;
	for (int i = head[u]; i; i = e[i].nxt) {
		v = e[i].to;
		if (!DFN[v]) {
			tarjan(v);
			low[u] = std::min(low[u], low[v]);
		} else if (inS[v]) low[u] = std::min(low[u], DFN[v]);
	}
	if (DFN[u] == low[u]) {
		++scc;
		do {
			inS[v = S[top--]] = false;
			bel[v] = scc;
		} while (v != u);
	}
}

int main() {
	std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
	std::cin >> Tim;
	while (Tim --> 0) {
		std::cin >> n >> m; nn = n << 1;
		for (int i = 0; i < m; ++i) {
			static int b, d;
			static char a, c;
			std::cin >> a >> b >> c >> d;
//			std::cout << a << b << ' ' << c << d << std::endl;
			addedge(a == 'h', b, c == 'h', d);
		}
		for (int i = 1; i <= nn; ++i) if (!DFN[i]) tarjan(i);
		bool solution = true;
		for (int i = 1; i <= n; ++i) if (bel[i] == bel[i + n]) {
			solution = false;
			break;
		}
//		for (int i = 1; i <= n; ++i) printf("%d: %d %d\n", i, bel[i], bel[n + i]);
		std::cout << (solution ? "GOOD" : "BAD") << '\n';
		if (Tim) {
			__builtin_memset(head, 0, sizeof head), cnt = 0;
			__builtin_memset(DFN, 0, sizeof DFN), idx = 0;
			scc = 0;
		}
	}
	return 0;
}