洛谷2417 课程

原题链接

对于每一个课堂，向能够来这堂课的学生连边，然后跑二分图最大匹配，判断是否是完全匹配即可。
这里我是用的匈牙利算法。

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1e5 + 10;
int fi[N], di[N], ne[N], mtc[N], l;
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;
}
bool dfs(int x)
{
	int i, y;
	for (i = fi[x]; i; i = ne[i])
		if (!v[y = di[i]])
		{
			v[y] = 1;
			if (!mtc[y] || dfs(mtc[y]))
			{
				mtc[y] = x;
				return true;
			}
		}
	return false;
}
int main()
{
	int i, j, x, n, m, t, s;
	t = re();
	while (t--)
	{
		n = re();
		re();
		memset(mtc, 0, sizeof(mtc));
		memset(fi, 0, sizeof(fi));
		l = 0;
		for (i = 1; i <= n; i++)
		{
			m = re();
			for (j = 1; j <= m; j++)
			{
				x = re();
				add(i, x + n);
			}
		}
		s = 0;
		for (i = 1; i <= n; i++)
		{
			memset(v, 0, sizeof(v));
			if (dfs(i))
				s++;
		}
		s ^ n ? printf("NO\n") : printf("YES\n");
	}
	return 0;
}