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[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>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;
}