割點

非割點且度數=m-n+2的就是答案。

割點竟然沒一次寫對，智商已降低。

設根為1。對於非1節點，一定有父親的邊，則它是割點的充要條件是存在一個兒子的low>=自己的dfn。對於1節點，它是割點的充要條件是有至少兩個兒子。

MD，對於非1節點忘記判斷 x != 1了

#include<cstdio>
#define R register 
#define N 100005
#define cmin(u,v) ((u)>(v)?(u)=(v):0)
using namespace std;
namespace runzhe2000
{
    int last[N], ecnt, n, m
 
, deg[N], cut[N], low[N], dfn[N], timer, out[N], cnt;
    struct edge{int next, to;}e[N<<1];
    inline void addedge(R int a, R int b)
    {
        e[++ecnt] = (edge){last[a], b};
        last[a] = ecnt;
    }
    void tarjan(R int x)
    {
        low[x] = dfn[x] = ++timer;
        R int siz = 0
 
;
        for(R int i = last[x]; i; i = e[i].next)
        {
            R int y = e[i].to;
            if(!dfn[y])
            {
                tarjan(y); siz++;
                cmin(low[x], low[y]);
                if(x != 1 && low[y] >= dfn[x]) cut[x] = 1;
            }
            else
 
 cmin(low[x], dfn[y]);
        }
        if(x == 1 && siz > 1) cut[1] = 1; 
    }
    void main()
    {
        scanf("%d%d",&n,&m);
        for(R int i = 1, x , y; i <= m; i++)
        {
            scanf("%d%d",&x,&y);
            deg[x]++; deg[y]++;
            addedge(x, y);
            addedge(y, x);
        }
        tarjan(1);
        for(R int i = 1; i <= n; i++)
            if(!cut[i] && deg[i] == m-n+2)
                out[++cnt] = i;
        printf("%d\n",cnt);
        for(R int i = 1; i <= cnt; i++) printf("%d ",out[i]);
    }
}
int main()
{
    runzhe2000::main();
}