洛谷4782 【模板】2-SAT 问题

原题链接

\(2-SAT\)模板

#include<cstdio>
using namespace std;
const int N = 2e6 + 10;
int fi[N], di[N], ne[N], dfn[N], low[N], st[N], bl[N], l, tp, ti, SCC;
bool v[N];
inline int re()
{
	int x = 0;
	char c = getchar();
	bool p = 0;
	for (; c < '0' || c > '9'; c = getchar())
		p |= c == '-';
	for (; c >= '0' && c <= '9'; c = getchar())
		x = x * 10 + c - '0';
	return p ? -x : x;
}
inline void add(int x, int y)
{
	di[++l] = y;
	ne[l] = fi[x];
	fi[x] = l;
}
inline int minn(int x, int y)
{
	return x < y ? x : y;
}
void tarjan(int x)
{
	int i, y;
	dfn[x] = low[x] = ++ti;
	st[++tp] = x;
	v[x] = 1;
	for (i = fi[x]; i; i = ne[i])
	{
		y = di[i];
		if (!dfn[y])
		{
			tarjan(y);
			low[x] = minn(low[x], low[y]);
		}
		else
			if (v[y])
				low[x] = minn(low[x], dfn[y]);
	}
	if (!(dfn[x] ^ low[x]))
	{
		++SCC;
		do
		{
			y = st[tp--];
			bl[y] = SCC;
			v[y] = 0;
		} while (x ^ y);
	}
}
int main()
{
	int i, n, m, x, y, a, b;
	n = re();
	m = re();
	for (i = 1; i <= m; i++)
	{
		x = re();
		a = re();
		y = re();
		b = re();
		add(x + (a ^ 1) * n, y + b * n);
		add(y + (b ^ 1) * n, x + a * n);
	}
	for (i = 1; i <= (n << 1); i++)
		if (!dfn[i])
			tarjan(i);
	for (i = 1; i <= n; i++)
		if (!(bl[i] ^ bl[i + n]))
		{
			printf("IMPOSSIBLE");
			return 0;
		}
	printf("POSSIBLE\n");
	for (i = 1; i <= n; i++)
		printf("%d ", bl[i] > bl[i + n]);
	return 0;
}