wikioi1036 商务旅行 挺水的lca

链接：http://wikioi.com/problem/1036/

题意不写了。

思路：很明显找到lca然后用两个点的深度相加-lca的深度就是这一步的最近步数。

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <stdlib.h>
 #include <vector>
 #include <queue>
 #define loop(s,i,n) for(i = s;i < n;i++)
 #define cl(a,b) memset(a,b,sizeof(a))
 using namespace std;
 const int maxn = ;
 int low[maxn],dfn[maxn],set[maxn],father[maxn],dfsclock,cut;
 int depth[maxn];

 vector<int >g[maxn];
 int find(int x)
 {
     if(set[x] != x)
     set[x] = find(set[x]);

     return set[x];
 }
 int merge(int x,int y)
 {
     x = find(x);
     y = find(y);
     if(y != x)
     {
         set[y] = x;
         return ;
     }
     return ;
 }
 void tarjan(int u,int pre,int deep)
 {
     int v,i,j;
     dfn[u] = low[u] = ++dfsclock;
     depth[u] = deep;
     loop(,i,g[u].size())
     {
         v = g[u][i];
         if(!dfn[v])
         {
             tarjan(v,u,deep+);
             father[v] = u;
             low[u] = min(low[v],low[u]);
             if(low[v] > dfn[u])
             cut++;
             else
             merge(u,v);
         }
         else if(v != pre)
         low[u] = min(low[u],dfn[v]);
     }
 }
 int lca(int u,int v)
 {

     while(u != v)
     {
         while(dfn[u] >= dfn[v] && u != v)
         {
             if(merge(u,father[u]))
             cut--;
             u = father[u];
         }
         while(dfn[v] >= dfn[u] && u != v)
         {
             if(merge(v,father[v]))
             cut--;
             v =  father[v];
         }
     }
     return u;
 }
 int main()
 {
     int n,m;
     int i,x,y;
     scanf("%d",&n);
     {
         int u,v;
         loop(,i,n+)
         {
             g[i].clear();
             set[i] = i;
             father[i] = ;
         }
         loop(,i,n)
         {
             scanf("%d %d",&u,&v);
             g[u].push_back(v);
             g[v].push_back(u);

         }
         cl(dfn,);
         cl(low,);
         cut = dfsclock = ;

         int k;
         scanf("%d",&k);
         tarjan(,-,);
         u = ;
         int ans;
         ans = ;
         while(k--)
         {
             scanf("%d",&v);
             int f;
             f = lca(u,v);
             ans += depth[u]+depth[v]-*depth[f];
             u = v;
         }
         cout<<ans<<endl;
     }
     return ;
 }