题意：每次询问给个点集，问图里有几个点删除后点集中有点对不能互相到达。
对原图建广义圆方树，答案即这些点在树上两两间链的并的圆点个数。建虚树统计即可。
说起来很容易，写起来也很容易... 超休闲的
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#include <cmath> #include <queue> #include <cstdio> #include <vector> #include <iomanip> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define N 200005 #define ll long long using namespace std; char xB[1<<15],*xS=xB,*xT=xB; #define getchar() (xS==xT&&(xT=(xS=xB)+fread(xB,1,1<<15,stdin),xS==xT)?0:*xS++) inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } #define clr(x) memset(x,0,sizeof(x)) int T; int n,m,q; struct zgz { int next,to; }edge[N<<1],Edge[N<<2]; int head[N],cnt=1; int Head[N<<1],Cnt=1; void add(int from,int to) { edge[cnt].to=to; edge[cnt].next=head[from]; head[from]=cnt++; } void Add(int from,int to) { Edge[Cnt].to=to; Edge[Cnt].next=Head[from]; Head[from]=Cnt++; } inline int Mn(int a,int b) {return a<b?a:b;} int dfn[N],low[N],tot; int stk[N],top,scc; void tarjan(int x,int f) { dfn[x]=low[x]=++tot; stk[++top]=x; int ch=0; for(int i=head[x];i;i=edge[i].next) { int to=edge[i].to; if(!dfn[to]) { ch++; tarjan(to,x); low[x]=Mn(low[x],low[to]); if((low[to]>=dfn[x]&&f)||(!f&&ch>1)) { int y; Add(++scc,x),Add(x,scc); do{ y=stk[top--]; Add(scc,y),Add(y,scc); }while(y!=to); } } else low[x]=Mn(low[x],dfn[to]); } } #define M N<<1 int sum[M],deep[M]; int sz[M],fa[M],son[M]; int st[M],ed[M],tim; void dfs1(int x,int f) { st[x]=++tim; fa[x]=f,sz[x]=1; sum[x]=sum[f]+(x<=n); for(int i=Head[x];i;i=Edge[i].next) { int to=Edge[i].to; if(to==f)continue ; deep[to]=deep[x]+1; dfs1(to,x); sz[x]+=sz[to]; if(sz[to]>sz[son[x]])son[x]=to; } ed[x]=tim; } int Top[M]; void dfs2(int x,int tp) { Top[x]=tp; if(son[x])dfs2(son[x],tp); for(int i=Head[x];i;i=Edge[i].next) { int to=Edge[i].to; if(to!=fa[x]&&to!=son[x])dfs2(to,to); } } int lca(int a,int b) { while(Top[a]!=Top[b]) { if(deep[Top[a]]<deep[Top[b]])swap(a,b); a=fa[Top[a]]; } return deep[a]<deep[b]?a:b; } void init() { clr(head),clr(dfn),clr(low),clr(Head),clr(stk); clr(sum),clr(deep),clr(sz),clr(fa),clr(son),clr(Top); tot=top=tim=0; cnt=Cnt=1; scc=n; } int s[N]; bool cmp(const int &a,const int &b) {return st[a]<st[b];} bool in_subtree(int a,int b)// a in b's subtree {return st[b]>=st[a]&&ed[b]<=ed[a];} int main() { T=read(); while(T--) { n=read(),m=read(); init(); for(int i=1;i<=m;i++) { int x=read(),y=read(); add(x,y),add(y,x); } tarjan(1,0); if(top) { scc++; while(top) { int y=stk[top--]; Add(scc,y),Add(y,scc); } } dfs1(1,0); dfs2(1,1); q=read(); while(q--) { int k=read(); if(k==2)//sub 2 { int a=read(),b=read(); int c=lca(a,b); printf("%d\n",sum[a]+sum[b]-sum[c]-sum[fa[c]]-2); } else { for(int i=1;i<=k;i++)s[i]=read(); int ans=-k; sort(s+1,s+k+1,cmp); int now=k; for(int i=2;i<=k;i++)s[++now]=lca(s[i],s[i-1]); k=now; sort(s+1,s+k+1); k=unique(s+1,s+k+1)-(s+1); sort(s+1,s+k+1,cmp); vector<int> v; v.clear(),v.push_back(s[1]); for(int i=2;i<=k;i++) { while(!v.empty()&&!in_subtree(v.back(),s[i]))v.pop_back(); ans+=sum[fa[s[i]]]-sum[v.back()]; v.push_back(s[i]); } for(int i=1;i<=k;i++)ans+=(s[i]<=n); printf("%d\n",ans); } } } } |
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