/* *题目大意: *题目为天朝文字,不多说; * *算法思想: *题目很明显要求的是一个LCA问题; *即询问从A到B的需要的步数,即首先从A到达A和B的最近公共祖先需要的步数+1就OK了; * *算法步骤: *由于是有向图,所以开始可以用一个数组ind记录每个顶点的入度; *如果该顶点的入度为0,则可以当做根节点,利用dfs求出树中每个顶点的深度d; *则从u到v的步数ans=d[u]-lca(u,v)+1,当然要考虑几个特殊情况;**/#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<map>#include<algorithm>using namespace std;const int N=400010;int p[N];//并查集的父节点int ind[N];//求顶点的入度,判断根节点int head[N];int qhead[N];//询问bool visit[N];int d[N];struct node{ int to; int w; int next; int lca; int num;};struct query//记录查询{ int u; int v; int lca;} q[N];node edge[N];node qedge[N];//询问边int n,m;int cnt1,cnt2;int cnt;map<string,int> Map;int get_num(string s)//返回每个人对应结点{ if(Map.find(s)==Map.end())//没有搜索到该键值 { Map[s]=++cnt;//对应建图 } // cout<<" Map["<<s<<"]=="<<Map[s]<<endl; return Map[s];}inline void Addedge(int u,int v,int w){ edge[cnt1].w=w; edge[cnt1].to=v; edge[cnt1].next=head[u]; head[u]=cnt1; cnt1++; edge[cnt1].w=w; edge[cnt1].to=u; edge[cnt1].next=head[v]; head[v]=cnt1; cnt1++;}inline void Addqedge(int u,int v,int num){ qedge[cnt2].num=num; qedge[cnt2].to=v; qedge[cnt2].next=qhead[u]; qhead[u]=cnt2; cnt2++; qedge[cnt2].num=num; qedge[cnt2].to=u; qedge[cnt2].next=qhead[v]; qhead[v]=cnt2; cnt2++;}void dfs(int u,int f,int w){ d[u]=w; for(int i=head[u]; i!=-1; i=edge[i].next) { int v=edge[i].to; if(v==f) continue; dfs(v,u,w+edge[i].w); }}int Find(int x){ if(p[x]!=x) p[x]=Find(p[x]); return p[x];}void Tarjan_LCA(int u)//离线LCA算法{ p[u]=u; visit[u]=1; for(int i=head[u]; i!=-1; i=edge[i].next) { if(!visit[edge[i].to]) { Tarjan_LCA(edge[i].to); p[edge[i].to]=u; } } for(int i=qhead[u]; i!=-1; i=qedge[i].next) { if(visit[qedge[i].to]) { qedge[i].lca=Find(qedge[i].to); qedge[i^1].lca=qedge[i].lca; //printf("%d和%d的最近公共祖先为: %d\n",u,qedge[i].to,qedge[i].lca); q[qedge[i].num].lca=qedge[i].lca; } }}void Solve(){ for(int i=0; i<=n; i++) { p[i]=i; } memset(head,-1,sizeof(head)); memset(qhead,-1,sizeof(qhead)); memset(visit,0,sizeof(visit)); memset(ind,0,sizeof(ind)); cnt=cnt1=cnt2=0; int u,v,w; string s1,s2; Map.clear(); for(int i=1; i<n; i++) { cin>>s1>>s2; u=get_num(s1); v=get_num(s2); Addedge(u,v,1); ind[u]++; } for(int i=0; i<m; i++) { cin>>s1>>s2; u=get_num(s1); v=get_num(s2); /*cout<<s1; printf("==%d\n",u); cout<<s2; printf("==%d\n",v);*/ Addqedge(u,v,i); q[i].u=u; q[i].v=v; } int root=0; for (int i=1; i<=n; i++) { if(ind[i]==0) { root=i; } } //printf("root==%d\n",root); dfs(root,-1,0); //for(int i=1; i<=n; i++) //printf("d[%d]==%d\n",i,d[i]); Tarjan_LCA(root);}int main(){ //freopen("C:\\Users\\Administrator\\Desktop\\kd.txt","r",stdin); int tcase; scanf("%d",&tcase); while(tcase--) { scanf("%d%d",&n,&m); Solve(); for (int i=0; i<m; i++) { int ans=0; ans=d[q[i].u]-d[q[i].lca]; if(q[i].lca!=q[i].v) ans++; if(q[i].u==q[i].v) ans=0; printf("%d\n",ans); } } return 0;}
