POJ3683 Priest John's Busiest Day

原题链接

(2-SAT)模板题。
对于有时间重叠的婚礼转换成(2-SAT)的命题形式连边，用(tarjan)找强连通分量并判断，确定方案即可。
然而一道模板题，我因为数组开小了调了一晚上。。。

#include<cstdio>
using namespace std;
const int N = 2010;
const int M = 1e6 + 10;
struct dd {
	int x[2], y[2], z, S[2];
};
dd a[N >> 1];
int fi[N], di[M << 2], ne[M << 2], dfn[N], low[N], st[N], bl[N], l, ti, tp, 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);
	}
}
void pr(int p, int i)
{
	printf("%02d:%02d ", a[i].x[p], a[i].y[p]);
	a[i].y[p] += a[i].z;
	a[i].x[p] += a[i].y[p] / 60;
	a[i].y[p] %= 60;
	printf("%02d:%02d
", a[i].x[p], a[i].y[p]);
}
int main()
{
	int i, n, j, x, y, z;
	n = re();
	for (i = 1; i <= n; i++)
	{
		a[i].x[0] = re();
		a[i].y[0] = re();
		a[i].x[1] = re();
		a[i].y[1] = re();
		a[i].z = re();
		a[i].S[0] = 60 * a[i].x[0] + a[i].y[0];
		a[i].S[1] = 60 * a[i].x[1] + a[i].y[1] - a[i].z;
		while (a[i].y[1] < a[i].z)
		{
			a[i].x[1]--;
			a[i].y[1] += 60;
		}
		a[i].y[1] -= a[i].z;
	}
	for (i = 1; i < n; i++)
		for (j = i + 1; j <= n; j++)
			for (x = 0; x < 2; x++)
				for (y = 0; y < 2; y++)
					if (!(a[i].S[x] + a[i].z <= a[j].S[y] || a[i].S[x] >= a[j].S[y] + a[j].z))
					{
						add(i + x * n, j + (y ^ 1) * n);
						add(j + y * n, i + (x ^ 1) * 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("NO");
			return 0;
		}
	printf("YES
");
	for (i = 1; i <= n; i++)
	{
		if (bl[i] > bl[i + n])
			pr(1, i);
		else
			pr(0, i);
	}
	return 0;
}