Loj 2276 新型城市化

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二分图最大匹配的必须边.

首先可以在反图上考虑这个问题(即将边集替换为其补集).

问题变为给出一张二分图,询问删掉哪些边后最大匹配会减少,即哪些边是最大匹配的必须边.

首先用最大流跑出一个最大匹配,仅保留未满流的边做 tarjan,将每个点所属的 SCC 求出.

边 $(u,v)$ 是最大匹配的可行边,当且仅当它满流,或 $u,v$ 在相同的 SCC 中.

证明:若其满流,显然是可行边,若在相同的 SCC 中,则可以替换掉原来的一条匹配边.

边 $(u,v)$ 是最大匹配的必须边,当且仅当它满流,且 $u,v$ 在不同的 SCC 中.

证明:若其不满流,显然不是必须边,若在相同的 SCC 中,则可以被其他边替换掉.

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//%std
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read()
{
	int out = 0, fh = 1;
	char jp = getchar();
	while ((jp > '9' || jp < '0') && jp != '-')
		jp = getchar();
	if (jp == '-')
		fh = -1, jp = getchar();
	while (jp >= '0' && jp <= '9')
		out = out * 10 + jp - '0', jp = getchar();
	return out * fh;
}
void print(int x)
{
	if (x >= 10)
		print(x / 10);
	putchar('0' + x % 10);
}
void write(int x, char c)
{
	if (x < 0)
		putchar('-'), x = -x;
	print(x);
	putchar(c);
}
const int N = 1e4 + 10, M = 3e5 + 10, inf = 1e9;
int S, T, n, m, ecnt = 1, head[N], cur[N], dep[N];
struct Edge
{
	int nx, to, flow;
} E[M];
void addedge(int u, int v, int w)
{
	E[++ecnt] = (Edge){head[u], v, w};
	head[u] = ecnt;
}
void ins(int u, int v, int w)
{
	addedge(u, v, w);
	addedge(v, u, 0);
}
queue<int> q;
bool bfs()
{
	for (int i = 1; i <= T; ++i)
		dep[i] = -1, cur[i] = head[i];
	dep[S] = 0, q.push(S);
	while (!q.empty())
	{
		int u = q.front();
		q.pop();
		for (int i = head[u]; i; i = E[i].nx)
		{
			int v = E[i].to;
			if (E[i].flow && dep[v] == -1)
				dep[v] = dep[u] + 1, q.push(v);
		}
	}
	return dep[T] != -1;
}
int dfs(int u, int limit)
{
	if (u == T || !limit)
		return limit;
	int flow = 0, f;
	for (int &i = cur[u]; i; i = E[i].nx)
	{
		int v = E[i].to;
		if (dep[v] == dep[u] + 1 && E[i].flow && (f = dfs(v, min(limit, E[i].flow))))
		{
			E[i ^ 1].flow += f;
			E[i].flow -= f;
			flow += f;
			limit -= f;
			if (!limit)
				break;
		}
	}
	return flow;
}
vector<int> G[N];
int col[N];
void paint(int u)
{
	for (int v : G[u])
	{
		if (!col[v])
		{
			col[v] = 3 - col[u];
			paint(v);
		}
	}
}
void Dinic()
{
	while (bfs())
		dfs(S, inf);
}
int dfn[N], low[N], idx = 0, in[N], stk[N], tp = 0, cnt = 0, scc[N];
void tarjan(int u)
{
	dfn[u] = low[u] = ++idx;
	in[u] = 1, stk[++tp] = u;
	for (int i = head[u]; i; i = E[i].nx)
	{
		if (E[i].flow == 0)
			continue;
		int v = E[i].to;
		if (!dfn[v])
		{
			tarjan(v);
			low[u] = min(low[u], low[v]);
		}
		else if (in[v])
			low[u] = min(low[u], dfn[v]);
	}
	if (dfn[u] == low[u])
	{
		++cnt;
		int x = 0;
		while (x != u)
		{
			x = stk[tp--];
			scc[x] = cnt;
			in[x] = 0;
		}
	}
}
vector<pair<int, int> > ans;
int main()
{
	n = read(), m = read();
	S = n + 1, T = n + 2;
	for (int i = 1; i <= m; ++i)
	{
		int u = read(), v = read();
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for (int i = 1; i <= n; ++i)
		if (!col[i])
		{
			col[i] = 1;
			paint(i);
		}
	for (int i = 1; i <= n; ++i)
	{
		if (col[i] == 1)
			ins(S, i, 1);
		else
			ins(i, T, 1);
		if (col[i] == 1)
			for (int j : G[i])
				ins(i, j, 1);
	}
	Dinic();
	for (int i = 1; i <= T; ++i)
		if (!dfn[i])
			tarjan(i);
	for (int u = 1; u <= n; ++u) if (col[u] == 1)
		for (int i = head[u]; i; i = E[i].nx)
		{
			int v = E[i].to;
			if (v <= n && E[i].flow == 0 && scc[u] != scc[v])
				ans.push_back(make_pair(min(u, v), max(u, v)));
		}
	sort(ans.begin(), ans.end());
	printf("%llu\n", ans.size());
	for (auto p : ans)
		write(p.first, ' '), write(p.second, '\n');
	return 0;
}