2017-2018 ACM-ICPC, NEERC, Northern Subregional ContestG - Grand Test
阿新 • • 發佈:2018-11-04
題意:找三條同起點同終點的不相交的路徑
題解:用tarjan的思想,記錄兩個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,那麼說明有兩條返祖邊,那麼就是滿足條件的答案
//#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<bits/stdc++.h> #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<int> #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<ll,ll> #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 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<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;} template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;} inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=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<road.size();i++)printf(" %d",road[i]);puts(""); road.clear(); } void tarjan(int u,int f) { if(ok)return ; dfn[u]=low[u][0]=low[u][1]=++ind; for(int i=0;i<v[u].size();i++) { int x=v[u][i]; if(x==f)continue; if(!dfn[x]) { fa[x]=u; tarjan(x,u); if(low[x][1]<low[u][0]) { low[u][1]=low[u][0]; pos[u][1]=pos[u][0]; low[u][0]=low[x][1]; pos[u][0]=pos[x][1]; } else if(low[x][1]<low[u][1]) { low[u][1]=low[x][1]; pos[u][1]=pos[x][1]; } if(low[x][0]<low[u][0]) { low[u][1]=low[u][0]; pos[u][1]=pos[u][0]; low[u][0]=low[x][0]; pos[u][0]=pos[x][0]; } else if(low[x][0]<low[u][1]) { low[u][1]=low[x][0]; pos[u][1]=pos[x][0]; } } else if(dfn[x]<dfn[u]) { if(dfn[x]<low[u][0]) { low[u][1]=low[u][0]; pos[u][1]=pos[u][0]; low[u][0]=dfn[x]; pos[u][0]=mp(u,x); } else if(dfn[x]<low[u][1]) { low[u][1]=dfn[x]; pos[u][1]=mp(u,x); } } } if(low[u][1]<dfn[u]&&!ok) { ok=1; printf("%d %d\n",u,pos[u][1].se); for(int now=u;now!=pos[u][1].se;now=fa[now])road.pb(now); road.pb(pos[u][1].se); pr(); road.pb(pos[u][1].se); for(int now=pos[u][1].fi;now!=u;now=fa[now])road.pb(now); road.pb(u); reverse(road.begin(),road.end()); pr(); for(int now=pos[u][1].se;now!=pos[u][0].se;now=fa[now])road.pb(now); road.pb(pos[u][0].se); for(int now=pos[u][0].fi;now!=u;now=fa[now])road.pb(now); road.pb(u); reverse(road.begin(),road.end()); pr(); // for(int i=1;i<=8;i++)printf("%d %d\n",i,fa[i]); // printf("%d %d %d %d %d %d %d %d---\n",u,dfn[u],low[u][0],pos[u][0].fi,pos[u][0].se,low[u][1],pos[u][1].fi,pos[u][1].se); return ; } } int main() { freopen("grand.in","r",stdin); freopen("grand.out","w",stdout); int T;scanf("%d",&T); while(T--) { int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) { dfn[i]=low[i][0]=low[i][1]=fa[i]=0; v[i].clear(); } ind=0; for(int i=0,x,y;i<m;i++) { scanf("%d%d",&x,&y); v[x].pb(y),v[y].pb(x); } ok=0; for(int i=1;i<=n;i++) if(!dfn[i]) tarjan(i,-1); if(!ok)puts("-1"); } return 0; } /******************** 1 8 9 1 2 1 3 4 2 5 3 6 4 7 5 8 6 8 7 8 ********************/