[BZOJ1787 Meet]

[关键字]：LCA

[题目大意]：求出一个集合点使树中三个点到这个点距离总和最小

//=============================================================

[分析]：无论怎么建树，这个集合点一定在这三个点两两的LCA之间，所以就转化成求LCA的问题，但这道题用离线的tarjan算法十分麻烦，所以就用的LCA转RMQ的在线算法。自我感觉代码写的相当给力：非递归+STL，可惜MLE……

[代码]：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
#include<cmath>
using namespace std;

const int MAXN=500005;
const int INF=0x7fffffff;

int n,m,ans,pos;
int E[MAXN*2],R[MAXN],D[MAXN];
int f[MAXN*2][20],d[MAXN*2][20];
stack<int> s;
vector<int> e[MAXN];
vector<int>::iterator first[MAXN];

void Init()
{
     int i,x,y;
     scanf("%d%d",&n,&m);
     for (i=1;i<n;i++)
     {
         scanf("%d%d",&x,&y);
         e[x].push_back(y);
         e[y].push_back(x);
     }
     for (i=1;i<=n;i++) first[i]=e[i].begin();
}

void DFS()
{
     s.push(1);
     E[++E[0]]=1;
     R[1]=1,D[1]=0;
     while (!s.empty())
     {
           int u=s.top();
           vector<int>::iterator i;
           for (i=first[u];i!=e[u].end();++i)
               if (!R[*i])
               {
                          s.push(*i);
                          D[*i]=D[u]+1;
                          E[++E[0]]=*i;
                          R[*i]=E[0];
                          first[u]=i+1;
                          break;
               }
           if (i==e[u].end()) 
           {
                              s.pop();
                              if (!s.empty()) E[++E[0]]=s.top();
           }
     }
/*     printf("%d\n",E[0]);
     for (int i=1;i<=E[0];++i) printf("%d ",E[i]);
     printf("\n");
     for (int i=1;i<=n;++i)
         printf("%d\n",R[i]);*/
}

void RMQ()
{
     int len=(int)floor(log((double)E[0])/log(2.0));
     for (int i=1;i<=E[0];++i)
         f[i][0]=D[E[i]],d[i][0]=E[i];
     for (int j=1;j<=len;++j)
         for (int i=1;i<=E[0]-(1<<j)+1;++i)
             if (f[i][j-1]<f[i+(1<<(j-1))][j-1]) 
                f[i][j]=f[i][j-1],d[i][j]=d[i][j-1];
             else
                 f[i][j]=f[i+(1<<(j-1))][j-1],d[i][j]=d[i+(1<<(j-1))][j-1];
}

int Find(int x,int y)
{
    if (x>y) swap(x,y);
    int len=(int)floor(log((double)y-x+1)/log(2.0));
    if (f[x][len]<f[y-(1<<len)+1][len])
       return d[x][len];
    else
        return d[y-(1<<len)+1][len];
}

void Work(int x,int y,int z)
{
     int LCA=Find(R[x],R[y]);
     int temp=D[x]+D[y]-2*D[LCA];
     int lca=Find(R[LCA],R[z]);
     temp+=D[LCA]+D[z]-2*D[lca];
     if (ans>temp) ans=temp,pos=LCA;
}

void Solve()
{
     int i,x,y,z;
     for (i=1;i<=m;i++)
     {
         scanf("%d%d%d",&x,&y,&z);
         ans=INF;
         Work(x,y,z);
         Work(x,z,y);
         Work(y,z,x);
         printf("%d %d\n",pos,ans);
     }
}

int main()
{
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    Init();
    DFS();
    RMQ();
    Solve();
    return 0;
}