CF1986F Non-academic Problem

传送门

Description

一张连通无向图，你可以剪断任意一条边。求剪断后存在路径的点对数的最小值。

Solution

Tarjan 求出无向图中所有的桥，并在 Tarjan 过程中记录下每个点 $u$ 在 dfs 树中的子树规模 $siz_u$。若剪断桥 $(u,v)$（不妨设 $dfn_u<dfn_v$），设

$x = siz_v,y=n-x$

则答案为

$\dfrac{x(x-1)}{2}+\dfrac{y(y-1)}{2}$

枚举每一条桥，取结果最小值即可。

时间复杂度 $\Theta(n+m)$

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 1e5 + 10;

ll T, n, m, head[N], edge_cnt, fa[N], low[N], dfn[N], tar_cnt, isbri[N], siz[N], ans;
struct edge {
	ll nxt, v;
} e[N << 1];

void add_edge(ll u, ll v)
{
	e[++edge_cnt].nxt = head[u];
	head[u] = edge_cnt;
	e[edge_cnt].v = v;
}

void Tarjan(ll u, ll pre) 
{
	fa[u] = pre;
	siz[u] = 1;
	low[u] = dfn[u] = ++tar_cnt;
	for (int i = head[u]; i; i = e[i].nxt) {
		ll v = e[i].v;
		if (!dfn[v]) {
			Tarjan(v, u);
			siz[u] += siz[v];
			low[u] = min(low[u], low[v]);
			if (low[v] > dfn[u])
				isbri[v] = 1;
		} else if (dfn[v] < dfn[u] && v != pre)
			low[u] = min(low[u], dfn[v]);
	}
}

void WORK()
{
	scanf("%lld%lld", &n, &m);
	memset(dfn, 0, sizeof(ll) * (n + 10));
	memset(low, 0, sizeof(ll) * (n + 10));
	tar_cnt = edge_cnt = 0;
	memset(fa, 0, sizeof(ll) * (n + 10));
	memset(isbri, 0, sizeof(ll) * (n + 10));
	memset(siz, 0, sizeof(ll) * (n + 10));
	memset(head, 0, sizeof(ll) * (n + 10));
	for (int i = 1; i <= m; i++) {
		ll u, v;
		scanf("%lld%lld", &u, &v);
		add_edge(u, v);
		add_edge(v, u);
	}
	Tarjan(1, 0);
	ans = n;
	for (int i = 1; i <= n; i++)
		if (isbri[i])
			ans = min(ans, max(siz[i], n - siz[i]));
	printf("%lld\n", ans * (ans - 1) / 2 + (n - ans) * (n - ans - 1) / 2);
}

int main()
{
	scanf("%lld", &T);
	while (T--) WORK();
	return 0;
}