求出支配树输出到father的和即可

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#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
const int N=300005;
int n,m,h[N],cnt,se[N],id[N],rl[N],fa[N],dfn[N],tot,f[N],v[N];
long long ans[N];
vector<int>p[N],d[N];
struct qwe
{
int ne,to;
}e[N];
int read()
{
int r=0,f=1;
char p=getchar();
while(p>'9'||p<'0')
  {
if(p=='-')
f=-1;
p=getchar();
  }
while(p>='0'&&p<='9')
  {
r=r*10+p-48;
p=getchar();
  }
return r*f;
}
void add(int u,int v)
{
cnt++;
e[cnt].ne=h[u];
e[cnt].to=v;
h[u]=cnt;
}
void dfs(int u)
{
dfn[u]=++tot;
rl[tot]=u;
for(int i=h[u];i;i=e[i].ne)
  {
p[e[i].to].push_back(u);
if(!dfn[e[i].to])
    {
fa[e[i].to]=u;
dfs(e[i].to);
    }
  }
}
int zhao(int x)
{
if(f[x]==x)
return x;
int fa=zhao(f[x]);
if(dfn[se[v[f[x]]]]<dfn[se[v[x]]])
v[x]=v[f[x]];
return f[x]=fa;
}
int main()
{
while(~scanf("%d%d",&n,&m))
  {
memset(h,0,sizeof(h));
memset(id,0,sizeof(id));
memset(dfn,0,sizeof(dfn));
memset(ans,0,sizeof(ans));
cnt=0;tot=0;
for(int i=1;i<=n;i++)
p[i].clear(),d[i].clear(),f[i]=v[i]=se[i]=i;
for(int i=1;i<=m;i++)
    {
int x=read(),y=read();
add(x,y);
    }
dfs(n);
for(int i=tot;i>=2;i--)
    {
for(int j=0;j<p[rl[i]].size();j++)
      {
if(dfn[p[rl[i]][j]]<dfn[rl[i]])
se[rl[i]]=(dfn[se[rl[i]]]<dfn[p[rl[i]][j]])?se[rl[i]]:p[rl[i]][j];
else
        {
zhao(p[rl[i]][j]);
se[rl[i]]=(dfn[se[rl[i]]]<dfn[se[v[p[rl[i]][j]]]])?se[rl[i]]:se[v[p[rl[i]][j]]];
        }
      }
f[rl[i]]=fa[rl[i]];
d[se[rl[i]]].push_back(rl[i]);
for(int j=0;j<d[fa[rl[i]]].size();j++)
      {
zhao(d[fa[rl[i]]][j]);
id[d[fa[rl[i]]][j]]=(dfn[se[v[d[fa[rl[i]]][j]]]]<dfn[se[d[fa[rl[i]]][j]]])?v[d[fa[rl[i]]][j]]:fa[rl[i]];
      }
    }
for(int i=2;i<=tot;i++)
if(id[rl[i]]!=se[rl[i]])
id[rl[i]]=id[id[rl[i]]];
for(int i=1;i<=n;i++)
ans[i]=0;
for(int i=1;i<=tot;i++)
    {
ans[rl[i]]+=rl[i];
if(id[rl[i]])
ans[rl[i]]+=ans[id[rl[i]]];
    }
for(int i=1;i<n;i++)
printf("%lld ",ans[i]);
printf("%lld\n",ans[n]);
  }
return 0;
}