题意:找三条同起点同终点的不相交的路径

题解:用tarjan的思想,记录两个low表示最小和次小的dfs序,以及最小和次小的位置,如果次小的dfs序比dfn小,那么说明有两条返祖边,那么就是满足条件的答案

//#pragma GCC optimize(2)//#pragma GCC optimize(3)//#pragma GCC optimize(4)//#pragma GCC optimize("unroll-loops")//#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")#include
    
     #define fi first#define se second#define db double#define mp make_pair#define pb push_back#define pi acos(-1.0)#define ll long long#define vi vector
     
      #define mod 1000000007#define ld long double#define C 0.5772156649#define ls l,m,rt<<1#define rs m+1,r,rt<<1|1#define pll pair
      
       #define pil pair
       
        #define pli pair
        
         #define pii pair
         
          //#define cd complex
          
           #define ull unsigned long long//#define base 1000000000000000000#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))#define fin freopen("a.txt","r",stdin)#define fout freopen("a.txt","w",stdout)#define fio ios::sync_with_stdio(false);cin.tie(0)inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}template
           
            inline T const& MAX(T const &a,T const &b){return a>b?a:b;}template
            
             inline T const& MIN(T const &a,T const &b){return a
             
              >=1;}return ans;}inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}using namespace std;const double eps=1e-8;const ll INF=0x3f3f3f3f3f3f3f3f;const int N=100000+10,maxn=50000+10,inf=0x3f3f3f3f;vi v[N],road;int dfn[N],low[N][2],fa[N];pii pos[N][2];int ind;bool ok;void pr(){ printf("%d",road.size()); for(int i=0;i