Educational Codeforces Round 42 (Rated for Div. 2)F

http://codeforces.com/contest/962/problem/F

求没有被两个及以上的简单环包含的边

解法：双联通求割顶，在bcc中看这是不是一个简单环，是的话把整个bcc的环加到答案中即可（正确性显然，因为bcc一定是环了，然后如果一个bcc不是简单环，那么所有边一定包含在两个简单环中）

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)

using namespace std;

const double g=10.0,eps=1e-12;
const int N=100000+10,maxn=200000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;

vector<pii>v[N];
int dfn[N],low[N];
int ind,iscut[N],n,m;
vi ans;
struct edge{int u,v,id;};
stack<edge>s;
int bcccnt,bccno[N],ed[N];
vi bcc[N],bb[N];
void tarjan(int u,int f)
{
    dfn[u]=low[u]=++ind;
    int ch=0;
    for(int i=0;i<v[u].size();i++)
    {
        int x=v[u][i].fi;
        if(x==f)continue;
        edge e={u,x,v[u][i].se};
        if(!dfn[x])
        {
            s.push(e);
            ch++;
            tarjan(x,u);
            low[u]=min(low[u],low[x]);
            if(low[x]>=dfn[u])
            {
                iscut[u]=1;
                bcccnt++;
                bcc[bcccnt].clear();
                while(1)
                {
                    edge now=s.top();s.pop();
                    ed[bcccnt]++;bb[bcccnt].pb(now.id);//printf("%d++++%d
",bcccnt,now.id);
                    if(bccno[now.u]!=bcccnt){bcc[bcccnt].pb(now.u);bccno[now.u]=bcccnt;}
                    if(bccno[now.v]!=bcccnt){bcc[bcccnt].pb(now.v);bccno[now.v]=bcccnt;}
                    if(now.u==u&&now.v==x)break;
                }
            }
        }
        else if(dfn[x]<dfn[u])
        {
            s.push(e);
            low[u]=min(low[u],dfn[x]);
        }
    }
    if(f<0&&ch==1)iscut[u]=0;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int a,b;scanf("%d%d",&a,&b);
        v[a].pb(mp(b,i)),v[b].pb(mp(a,i));
    }
    ind=0;
    for(int i=1;i<=n;i++)
        if(!dfn[i])
            tarjan(i,-1);
//    for(int i=1;i<=n;i++)printf("%d ",bccno[i]);puts("");
    memset(dfn,0,sizeof dfn);
    for(int i=1;i<=bcccnt;i++)
    {
        if(ed[i]==bcc[i].size()&&ed[i])
        {
            for(int j=0;j<bb[i].size();j++)ans.pb(bb[i][j]);
        }
    }
    sort(ans.begin(),ans.end());
    printf("%d
",ans.size());
    for(int i=0;i<ans.size();i++)printf("%d ",ans[i]);
    puts("");
    return 0;
}
/***********************
5 6
1 2
1 3
2 3
3 4
4 5
3 5
***********************/