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+5,DEG=18;
int t,n,m,a,b,ed,g
,nxt
,v
,w
,fa
[DEG],dep
,idx;
bool fg
;
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>0?a:-a;}
void dfs(int u,int pre){
dep[u]=dep[pre]+1,fa[u][0]=pre;
F(i,1,DEG-1)fa[u][i]=fa[fa[u][i-1]][i-1];
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,0,DEG-1)if((dep[b]-dep[a])&(1<<i))b=fa[b][i];
if(a!=b)for(int i=DEG-1;i<0?a=fa[a][0]:0,i>=0;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,1,n)g[i]=-1,fg[i]=0;ed=0,idx=1;
F(i,1,n-1){
cin>>s1>>s2;
a=cnt[s1],cnt[s1]=(a==0)?idx++:a,a=(a==0?idx-1:a);
b=cnt[s2],cnt[s2]=(b==0)?idx++:b,b=(b==0?idx-1:b);
adg(b,a),fg[a]=1;
}
int root;
F(i,1,n)if(!fg[i]){root=i;break;}
dep[root]=0;
dfs(root,root);
F(i,1,m){
cin>>s1>>s2;
a=cnt[s1],b=cnt[s2];
int lca=LCA(a,b),ans=0;
if(lca==a&&lca!=b)ans=1;
else if(lca==b&&lca!=a)ans=Abs(dep[a]-dep[b]);
else if(lca!=a&&lca!=b)ans=Abs(dep[a]-dep[lca])+1;
printf("%d\n",ans);
}
}
return 0;
}