Tarjan
显然不是割点的点答案都是 ( n − 1 ) ∗ 2 (n-1)*2 (n−1)∗2,我们只需要算一算割点的答案就好了。而割点的额外答案为DFS树上各个子树之间两两配对的贡献加上子树内和子树外的贡献。这些在Tarjan求割点的同时都可以顺带出来。
#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;
}