Tarjan

顯然不是割點的點答案都是$(n-1)*2$，我們只需要算一算割點的答案就好了。而割點的額外答案為DFS樹上各個子樹之間兩兩配對的貢獻加上子樹內和子樹外的貢獻。這些在Tarjan求割點的同時都可以順帶出來。

#include<cctype>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 100005
#define M 500005
#define F inline
using namespace std;
typedef long long
 
 LL;
struct edge{ int nxt,to; }ed[M<<2];
int n,m,k,ti,h[N],dfn[N],low[N],sz[N]; LL ans[N];
F char readc(){
	static char buf[100000],*l=buf,*r=buf;
	if (l==r) r=(l=buf)+fread(buf,1,100000,stdin);
	return l==r?EOF:*l++;
}
F int _read(){
	int x=0; char ch=readc();
	while (!isdigit(ch)) ch=readc();
	while (isdigit
 
(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=readc();
	return x;
}
void Tarjan(int x){
	dfn[x]=low[x]=++ti,sz[x]=1; LL t=0;
	for (int i=h[x],v;i;i=ed[i].nxt)
		if (!dfn[v=ed[i].to]){
			Tarjan(v),sz[x]+=sz[v],low[x]=min(low[x],low[v]);
			if (low[v]>=dfn[x]) ans[x]+=t*sz[v],t+=sz[v];
		}
		else low[
 
x]=min(low[x],dfn[v]);
	(ans[x]+=t*(n-t-1)+n-1)<<=1;
}
#define add(x,y) ed[++k]=(edge){h[x],y},h[x]=k
int main(){
	n=_read(),m=_read(),k=1;
	for (int i=1,x,y;i<=m;i++)
		x=_read(),y=_read(),add(x,y),add(y,x);
	Tarjan(1);
	for (int i=1;i<=n;i++) printf("%lld\n",ans[i]);
	return 0;
}