P3469 [POI2008]BLO-Blockade 割点，强联通分量

 

 

 

 

// 题意：对于每一个点，求删去这个点的所有边会形成多少个点对满足两点之间不互通
// 思路：思路很简单，分为这个点是否是割点，但写法上就有点讲究，详情见博客
//

/*#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, m;
vector<int> e[N];
int dfn[N], low[N], idx, cut[N], ans[N], id, sz[N];
void dfs(int u, int f) {
    dfn[u] = low[u] = ++idx;
    int ch = 0;
    sz[id]++;
    for (auto v : e[u]) {
        if (!dfn[v]) {
            dfs(v, u);
            ch++;
            low[u] = min(low[u], low[v]);
            if (low[v] >= dfn[u]) {
                ans[u] = 2 * (n - sz[id]) * sz[id] + 2 * (n - sz[id] - 1);
                cut[u] = 1;
                ++id;
            }
        }
        else if (v != f) {
            low[u] = min(low[u], dfn[v]);
            sz[id]++;
        }
    }
    if (u == 1 && ch <= 1) cut[u] = 0;
    if (cut[u] == 0)ans[u] = 2 * (n - 1);
}
int main() {
    cin >> n >> m;
    id = 1;
    for (int i = 1; i <= m; i++) {
        int a, b; cin >> a >> b;
        e[a].push_back(b);
        e[b].push_back(a);
    }
    dfs(1, -1);
    for (int i = 1; i <= n; i++) cout << ans[i] << endl;
    return 0;
}*/


#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, m;
vector<int> e[N];
int dfn[N], low[N], idx, sz[N];
long long ans[N];
void dfs(int u, int f) {
    dfn[u] = low[u] = ++idx;
    sz[u] = 1;
    ans[u] = n - 1;
    int cut = n - 1;
    for (auto v : e[u]) {
        if (!dfn[v]) {
            dfs(v, u);
            sz[u] += sz[v];
            low[u] = min(low[u], low[v]);
            if (low[v] >= dfn[u]) {
                ans[u] += (long long)sz[v] * (n - sz[v]);
                cut -= sz[v];
            }
        }
        else if (v != f) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    ans[u] += (long long)cut * (n - cut);
}
int main() {
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
        int a, b; cin >> a >> b;
        e[a].push_back(b);
        e[b].push_back(a);
    }
    dfs(1, -1);
    for (int i = 1; i <= n; i++) cout << ans[i] << endl;
    return 0;
}