求出支配树输出到father的和即可
支配树见：https://blog.csdn.net/a710128/article/details/49913553

#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;
}