匹配
哈希能A
水到爆炸
回家
事实上我做过一个原题,甚至比这个回家难的多,而且那个题多组询问必经点
然后我做一组询问就打炸了
大约就是删了很多东西,然后自己想的太简单了
直接统计了割点,懒得打lca和树上差分,懒得打dfs,偷懒让我付出很大代价
最后只有10,
打代码一定不能偷懒,一定不能偷懒
#include<bits/stdc++.h> using namespace std; #define ll long long #define A 810000 ll dfn[A],low[A],ver[A],nxt[A],head[A],s[A],fa[A],belong[A],kx[A]; ll Head[A],Nxt[A],Ver[A],To[A],vst[A]; ll tot=0,tot2=0,num=0,n,m,t,root; vector<ll> dcc[A],ans; bool cut[A]; void add(ll x,ll y){nxt[++tot]=head[x],head[x]=tot,ver[tot]=y; } void Add(ll x,ll y){Nxt[++tot2]=Head[x],Head[x]=tot2,Ver[tot2]=y; } inline ll read(){ll f=1,x=0;char c=getchar();while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}while(isdigit(c)){x=(x<<1)+(x<<3)+(c-'0');c=getchar();}return f*x; } void tarjan(ll x) {ll flag=0;dfn[x]=low[x]=++tot;s[++s[0]]=x;for(ll i=head[x];i;i=nxt[i]){ll y=ver[i];if(!dfn[y]){tarjan(y);low[x]=min(low[x],low[y]);if(low[y]>=dfn[x]){flag++;num++;if(flag>1||x!=root)cut[x]=1;while(s[0]){ll p=s[s[0]--];dcc[num].push_back(p);if(p==y)break;}dcc[num].push_back(x);}}else low[x]=min(low[x],dfn[y]);} } void dfs(ll x) {if(x==belong[n])return ;vst[x]=1;for(ll i=Head[x];i;i=Nxt[i]){ll y=Ver[i];if(vst[y])continue;fa[y]=x;dfs(y);}return ; } void re(){num=0,tot=0;ans.clear();tot2=0;for(ll i=0;i<=800000;i++) dcc[i].clear();memset(nxt,0,sizeof(nxt));memset(Head,0,sizeof(Head));memset(Nxt,0,sizeof(Nxt));memset(fa,0,sizeof(fa));memset(vst,0,sizeof(vst));memset(head,0,sizeof(head));memset(ver,0,sizeof(ver));memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));memset(cut,0,sizeof(cut));memset(kx,0,sizeof(kx)); } int main() {t=read();while(t--){re();n=read(),m=read();for(ll i=1;i<=m;i++){ll x=read(),y=read();add(x,y);add(y,x);}for(ll i=1;i<=n;i++)if(!dfn[i]) root=i,tarjan(i);ll nu=num;for(ll i=1;i<=n;i++)if(cut[i])belong[i]=++nu,kx[nu]=i;for(ll i=1;i<=num;i++)for(ll j=0;j<dcc[i].size();j++){ll x=dcc[i][j];if(cut[x]) Add(belong[x],i),Add(i,belong[x]);else belong[x]=i;} /* for(ll i=1;i<=n;i++){cout<<"belong="<<belong[i]<<endl;} */ dfs(belong[1]);ll x=fa[belong[n]];while(x!=belong[1]){if(x==0) break;if(x>num) ans.push_back(kx[x]);x=fa[x];}printf("%lld\n",1ll*ans.size());sort(ans.begin(),ans.end());for(ll i=0;i<ans.size();i++)printf("%lld ",ans[i]);cout<<endl;} }