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;
}