Luogu P3731 [HAOI2017]新型城市化

题目显然可以转化为求每一条边对二分图最大独立集的贡献，二分图最大独立集\(=\)点数\(-\)最大匹配数，我们就有了\(50pts\)做法。

正解的做法是在原图上跑\(Tarjan\)，最开始我想复杂了，后来才意识到，只要存在这样一个强连通分量，那么断掉分量内的任意一条边都不会破坏其连通性，即不管删掉哪个连边都一定会有新的匹配补充。只要让两个点不在同一个分量里面，而且原来是满流的（匹配可行边），那么它就是一个可用边（匹配必须边）。

#include <bits/stdc++.h>
using namespace std;

const int N = 400010;
const int M = 800010;
const int INF = 0x3f3f3f3f;

struct Graph {
	int cnt, head[N];
	
	struct edge {int nxt, to, f;}e[M];
	
	Graph () {
		cnt = -1;
		memset (head, -1, sizeof (head));
	}
	
	void add_edge (int u, int v, int f) {
		e[++cnt] = (edge) {head[u], v, f}; head[u] = cnt;
	}
	
	void add_len (int u, int v, int f) {
		add_edge (u, v, f);
		add_edge (v, u, 0);
	}
	
	queue <int> q;
	int cur[N], deep[N];
	
	bool bfs (int s, int t) {
		memcpy (cur, head, sizeof (head));
		memset (deep, 0x3f, sizeof (deep));
		deep[s] = 0; q.push (s);
		while (!q.empty ()) {
			int u = q.front (); q.pop ();
			for (int i = head[u]; ~i; i = e[i].nxt) {
				int v = e[i].to;
				if (deep[v] == INF && e[i].f) {
					deep[v] = deep[u] + 1;
					q.push (v);
				}
			}
		}
		return deep[t] != INF;
	}
	
	int dfs (int u, int t, int lim) {
		if (u == t || !lim) {
			return lim;
		}
		int tmp = 0, flow = 0;
		for (int &i = cur[u]; ~i; i = e[i].nxt) {
			int v = e[i].to;
			if (deep[v] == deep[u] + 1) {
				tmp = dfs (v, t, min (lim, e[i].f));
				lim -= tmp;
				flow += tmp;
				e[i ^ 0].f -= tmp;
				e[i ^ 1].f += tmp;
				if (!lim) break;
			}
		}
		return flow;
	}
	
	int Dinic (int s, int t) {
		int max_flow = 0;
		while (bfs (s, t)) {
			max_flow += dfs (s, t, INF);
		}
		return max_flow;
	}
}G;

int n, m, _ans, id[N];

struct Query {
	int u, v;
	
	bool operator < (Query rhs) const {
		return u == rhs.u ? v < rhs.v : u < rhs.u;
	}	
	
	bool operator == (Query rhs) const {
		return u == rhs.u && v == rhs.v;
	}
}q[N], ans[N];

int A (int x) {return n * 0 + x;}
int B (int x) {return n * 1 + x;}

int read () {
	int s = 0, w = 1, ch = getchar ();
	while ('9' < ch || ch < '0') {
		if (ch == '-') w = -1;
		ch = getchar ();
	}
	while ('0' <= ch && ch <= '9') {
		s = s * 10 + ch - '0';
		ch = getchar ();
	}
	return s * w;
} 

stack <int> sta;
int dfn[N], low[N], col[N], vis[N]; 

void Tarjan (int u) {
	sta.push (u);
	vis[u] = true;
	dfn[u] = low[u] = ++dfn[0];	
	for (int i = G.head[u]; ~i; i = G.e[i].nxt) {
		int v = G.e[i].to;
		if (!G.e[i].f) continue;
		if (!dfn[v]) {
			Tarjan (v);
			low[u] = min (low[u], low[v]);
		} else if (vis[v]) {
			low[u] = min (low[u], dfn[v]);
		}
	}
	if (dfn[u] == low[u]) {
		int tmp; ++col[0];
		do {
			tmp = sta.top ();
			vis[tmp] = false;
			col[tmp] = col[0];
			sta.pop ();
		}while (tmp != u);
	}
}


int main () {
	cin >> n >> m;
	int s = n * 2 + 1;
	int t = n * 2 + 2;
	for (int i = 1; i <= m; ++i) {
		q[i].u = read ();
		q[i].v = read ();
		if (q[i].u > q[i].v) {
			swap (q[i].u, q[i].v);
		}
	}
	sort (q + 1, q + 1 + m);
	for (int i = 1; i <= m; ++i) {
		id[i] = G.cnt + 1;
		G.add_len (A (q[i].u), B (q[i].v), 1);
		G.add_len (A (q[i].v), B (q[i].u), 1);
	}
	for (int i = 1; i <= n; ++i) {
		G.add_len (s, A (i), 1);
		G.add_len (B (i), t, 1);
	}
	G.Dinic (s, t);
	for (int i = 1; i <= t; ++i) {
		if (!dfn[i]) {
			Tarjan (i);
		}
	}
	for (int i = 1; i <= m; ++i) {
		if (col[A (q[i].u)] != col[B (q[i].v)] && !G.e[id[i]].f) {
			ans[++_ans] = q[i];
		}
	}
	cout << _ans << endl;
	for (int i = 1; i <= _ans; ++i) {
		printf ("%d %d\n", ans[i].u, ans[i].v);
	}
}