题解 洛谷P3469

题目每个割点去掉后会导致多少对点不能连通

考虑跑Tarjan的时候记录每个儿子的size，那么去掉这个割点后其他的点都不能和这个儿子连通

注意每个点去掉后它本身就不能与其他所有点连通

还有就是题目里求的是有序点对，所以应将总方案数(×2)

#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <iostream>
#include <queue>
#include <vector>
#define il inline
#define re register
#define gc getchar
#define LL long long
#define int LL
#define D() 
// cerr << __LINE__ << endl
using namespace std;
template <typename T>
void read(T &s)
{
	s = 0;
	char ch;
	while (ch = gc(), !isdigit(ch))
		;
	while (s = s * 10 + ch - '0', ch = gc(), isdigit(ch))
		;
}
const int MAXN = 600000;
int cnt = 0;
int vis[MAXN];
int tot = 0;
int del[MAXN];
int dfn[MAXN], low[MAXN];
int sta[MAXN], top;
int ans[MAXN];
int sze[MAXN];
vector<int> edge[MAXN];
vector<int> edge2[MAXN];
il void insert(int u, int v)
{
	edge[u].push_back(v);
	edge[v].push_back(u);
}
il void insert2(int u, int v)
{
	edge2[u].push_back(v);
}
int n;
void tarjan(int u, int fa)
{
	int t = 0;
	int child = 0;
	vis[u] = 1;
	dfn[u] = low[u] = ++tot;
	sta[++top] = u;
	sze[u] = 1;
	for (auto v : edge[u])
		if (!dfn[v])
		{
			tarjan(v, u);
			sze[u] += sze[v];
			++child;
			if ((fa == -1 && child > 1) || (fa != -1 && low[v] >= dfn[u]))
			{
				ans[u] += sze[v] * t;
				t += sze[v];
			}
			low[u] = min(low[u], low[v]);
		}
		else if (vis[v])
			low[u] = min(low[u], dfn[v]);
	ans[u] += t * (n - t - 1);
	if (low[u] == dfn[u])
	{
		while (sta[top] != u)
		{
			int y = sta[top--];
			vis[y] = 0;
		}
		top--;
		vis[u] = 0;
	}
}
signed main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int m, a, b;
	cin >> n >> m;
	for (int i = 1; i <= m; ++i)
	{
		cin >> a >> b;
		insert(a, b);
	}
	for (int i = 1; i <= n; ++i)
		if (!dfn[i])
			tarjan(i, -1);
	for (int i = 1; i <= n; ++i)
		cout << (((n - 1 + ans[i]) << 1)) << endl;
}