hdu_4547_CD操作(在线LCA)

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=4547

题意：中文，不解释

题解：很裸的LCA，注意父目录打开子目录一次就够了，这里我才用倍增在线LCA+map过

 #include<cstdio>
 #include<iostream>
 #include<map>
 #include<string>
 #define F(i,a,b) for(int i=a;i<=b;i++)
 #pragma comment(linker, "/STACK:1024000000,1024000000")
 using namespace std;
 map<string,int>cnt;
 const int N=1e5+,DEG=;
 int t,n,m,a,b,ed,g[N],nxt[N],v[N],w[N],fa[N][DEG],dep[N],idx;
 bool fg[N];
 string s1,s2;
 inline void adg(int x,int y){v[++ed]=y,nxt[ed]=g[x],g[x]=ed;}

 int Abs(int a){return a>?a:-a;}
 void dfs(int u,int pre){
     dep[u]=dep[pre]+,fa[u][]=pre;
     F(i,,DEG-)fa[u][i]=fa[fa[u][i-]][i-];
     for(int i=g[u];~i;i=nxt[i])if(v[i]!=pre)dfs(v[i],u);
 }

 int LCA(int a,int b){
     if(dep[a]>dep[b])a^=b,b^=a,a^=b;
     if(dep[a]<dep[b])F(i,,DEG-)if((dep[b]-dep[a])&(<<i))b=fa[b][i];
     if(a!=b)for(int i=DEG-;i<?a=fa[a][]:,i>=;i--)
     if(fa[a][i]!=fa[b][i])a=fa[a][i],b=fa[b][i];
     return a;
 }

 int main(){
     scanf("%d",&t);
     while(t--){
         scanf("%d%d",&n,&m);
         cnt.clear();
         F(i,,n)g[i]=-,fg[i]=;ed=,idx=;
         F(i,,n-){
             cin>>s1>>s2;
             a=cnt[s1],cnt[s1]=(a==)?idx++:a,a=(a==?idx-:a);
             b=cnt[s2],cnt[s2]=(b==)?idx++:b,b=(b==?idx-:b);
             adg(b,a),fg[a]=;
         }
         int root;
         F(i,,n)if(!fg[i]){root=i;break;}
         dep[root]=;
         dfs(root,root);
         F(i,,m){
             cin>>s1>>s2;
             a=cnt[s1],b=cnt[s2];
             int lca=LCA(a,b),ans=;
             if(lca==a&&lca!=b)ans=;
             else if(lca==b&&lca!=a)ans=Abs(dep[a]-dep[b]);
             else if(lca!=a&&lca!=b)ans=Abs(dep[a]-dep[lca])+;
             printf("%d\n",ans);
         }
     }
     return ;
 }