HDU4547 CD操作 LCA

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4547

　　去年区预赛网络赛的时候写过一到LCA的题目<HDU4409>，本来想用那份代码修改下来交这道题的，不过貌似难得修改，就重新写了ST+DFS的代码当模板用。跑了2078ms，应该是map的效率低了。。。

//STATUS:C++_AC_2078MS_26116KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N= 200010;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

struct Edge{
    int u,v;
}e[N];
int first[N],next[N],f[N][20],d[N],E[N],R[N],vis[N];
int T,n,m,mt,all,dis,root;
map<string,int> q;

void adde(int a,int b)
{
    e[mt].u=a;e[mt].v=b;
    next[mt]=first[a],first[a]=mt++;
}

int Minele(int i,int j)
{
    return d[i]<d[j]?i:j;
}

void rmq_init(int n)
{
    int i,j;
    for(i=1;i<=n;i++)f[i][0]=i;
    for(j=1;(1<<j)<=n;j++){
        for(i=1;i+(1<<j)-1<=n;i++){
            f[i][j]=Minele(f[i][j-1],f[i+(1<<(j-1))][j-1]);
        }
    }
}

int rmq(int l,int r)
{
    int k=0;
    while((1<<(k+1))<=r-l+1)k++;
    return Minele(f[l][k],f[r-(1<<k)+1][k]);
}

void dfs(int u,int deep)
{
    d[dis]=deep;
    E[dis]=u;
    R[u]=dis++;
    int i;
    for(i=first[u];i!=-1;i=next[i]){
        dfs(e[i].v,deep+1);
        d[dis]=deep;
        E[dis++]=u;
    }
}

void LCA_Init()
{
    dis=1;
    dfs(root,0);
    rmq_init(dis-1);
}

int LCA(int a,int b)
{
    int left=R[a],right=R[b];
    if(left>right)swap(left,right);
    return E[rmq(left,right)];
}

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,fa,a,b;
    char s1[42],s2[42];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        q.clear();all=1;
        mem(first,-1);mt=0;
        mem(vis,0);
        while(--n){
            scanf("%s%s",s1,s2);
            if(!q[s1])q[s1]=all++;
            if(!q[s2])q[s2]=all++;
            vis[q[s1]]=1;
            adde(q[s2],q[s1]);
        }
        for(i=1;i<all;i++)if(!vis[i]){root=i;break;}

        LCA_Init();
        while(m--){
            scanf("%s%s",s1,s2);
            a=q[s1];b=q[s2];
            fa=LCA(a,b);
            printf("%d\n",d[R[a]]-d[R[fa]]+(d[R[b]]>d[R[fa]]));
        }
    }
    return 0;
}