Sol
就是求割点，把贡献算一下就好。。。直接tarjan
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e5 + 10);
IL ll Read(){
RG ll x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, fst[_], nxt[_ << 2], to[_ << 2], cnt, m, dfn[_], size[_], Index, low[_];
ll ans[_];
IL void Add(RG int u, RG int v){ to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++; }
IL void Dfs(RG int u, RG int ff){
dfn[u] = low[u] = ++Index; size[u] = 1; RG int sum = 0;
for(RG int e = fst[u]; e != -1; e = nxt[e]){
if(to[e] == ff) continue;
if(!dfn[to[e]]){
Dfs(to[e], u);
size[u] += size[to[e]];
low[u] = min(low[u], low[to[e]]);
if(low[to[e]] >= dfn[u]) ans[u] += 1LL * sum * size[to[e]] * 2, sum += size[to[e]];
}
else low[u] = min(low[u], dfn[to[e]]);
}
ans[u] += 1LL * sum * (n - sum - 1) * 2;
}
int main(RG int argc, RG char* argv[]){
n = Read(); m = Read();
for(RG int i = 1; i <= n; ++i) ans[i] = n + n - 2, fst[i] = -1;
for(RG int i = 1, u, v; i <= m; ++i) u = Read(), v = Read(), Add(u, v), Add(v, u);
Dfs(1, 0);
for(RG int i = 1; i <= n; ++i) printf("%lld\n", ans[i]);
return 0;
}