CH#17C 舞动的夜晚

原题链接

即求二分图的不可行边数量，因为不保证是完备匹配，所以需要通过网络流求出任意一组最大匹配，并建立新图判断。

建新图：对于跑完网络流的图上已经匹配的边，建立反边；对于没有匹配的边，建立正边（图只改变边的方向，别的结构不变）。

有结论：

1. 必须边的判定条件为：\((x,y)\)的流量为\(1\)，并且在残量网络上属于不同的强联通分量。
2. 可行边的判定条件为：\((x,y)\)的流量为\(1\)，或者在残量网络上属于同一个强联通分量。

所以我们在新图上跑\(tarjan\)，再逐边检验即可。

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 2e4 + 10;
const int M = 2e5 + 10;
struct dd {
	int x, y;
};
dd a[M];
int fi[N], di[M << 1], ne[M << 1], da[M << 1], cfi[N], cdi[M], cne[M], dfn[N], low[N], sta[N], bl[N], cu[N], de[N], q[M << 1], an[M], l = 1, tp, lc, SCC, st, ed, ti;
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, int z)
{
	di[++l] = y;
	da[l] = z;
	ne[l] = fi[x];
	fi[x] = l;
}
inline void add_c(int x, int y)
{
	cdi[++lc] = y;
	cne[lc] = cfi[x];
	cfi[x] = lc;
}
inline int minn(int x, int y)
{
	return x < y ? x : y;
}
bool bfs()
{
	int i, x, y, head = 0, tail = 1;
	memset(de, 0, sizeof(de));
	q[1] = st;
	de[st] = 1;
	while (head ^ tail)
	{
		x = q[++head];
		for (i = fi[x]; i; i = ne[i])
			if (!de[y = di[i]] && da[i] > 0)
			{
				de[y] = de[x] + 1;
				if (!(y ^ ed))
					return true;
				q[++tail] = y;
			}
	}
	return false;
}
int dfs(int x, int k)
{
	int y, mi;
	if (!(x ^ ed))
		return k;
	for (int &i = cu[x]; i; i = ne[i])
		if (!(de[y = di[i]] ^ (de[x] + 1)) && da[i] > 0)
		{
			mi = dfs(y, minn(k, da[i]));
			if (mi > 0)
			{
				da[i] -= mi;
				da[i ^ 1] += mi;
				return mi;
			}
		}
	return 0;
}
void tarjan(int x)
{
	int i, y;
	dfn[x] = low[x] = ++ti;
	sta[++tp] = x;
	v[x] = 1;
	for (i = cfi[x]; i; i = cne[i])
	{
		if (!dfn[y = cdi[i]])
		{
			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 = sta[tp--];
			bl[y] = SCC;
			v[y] = 0;
		} while (x ^ y);
	}
}
int main()
{
	int i, n, m, k, x, y, s = 0;
	n = re();
	m = re();
	k = re();
	st = n + m + 1;
	ed = st + 1;
	for (i = 1; i <= k; i++)
	{
		x = re();
		y = re() + n;
		a[i].x = x;
		a[i].y = y;
		add(x, y, 1);
		add(y, x, 0);
	}
	for (i = 1; i <= n; i++)
	{
		add(st, i, 1);
		add(i, st, 0);
	}
	for (i = 1; i <= m; i++)
	{
		add(i + n, ed, 1);
		add(ed, i + n, 0);
	}
	while (bfs())
	{
		for (i = 1; i <= ed; i++)
			cu[i] = fi[i];
		while (dfs(st, 1e9) > 0);
	}
	for (i = 2; i <= l; i++)
		if (da[i])
			add_c(di[i ^ 1], di[i]);
	for (i = 1; i <= ed; i++)
		if (!dfn[i])
			tarjan(i);
	for (i = 1; i <= k;i++)
		if (bl[a[i].x] ^ bl[a[i].y] && da[i << 1])
			an[++s] = i;
	printf("%d\n", s);
	if (!s)
	{
		printf("\n");
		return 0;
	}
	for (i = 1; i < s; i++)
		printf("%d ", an[i]);
	printf("%d", an[s]);
	return 0;
}