poj 1470 Closest Common Ancestors LCA

题目链接：http://poj.org/problem?id=1470

Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the closest common ancestor of u and v in the tree. The closest common ancestor of two nodes u and v is the node w that is an ancestor of both u and v and has the greatest depth in the tree. A node can be its own ancestor (for example in Figure 1 the ancestors of node 2 are 2 and 5)

题目描述：给出一棵树的节点之间的信息，然后给出一些询问，每次询问是求出两个节点的LCA。统计每个节点被作为询问中LCA的次数并输出。

算法分析：LCA离线算法，从算法思想和思维上并不难，可以说是一道模板题，但这道题的输入有点麻烦，处理一下输入就可以了。

另外ZOJ1141也是这道题，我在ZOJ上面AC了

但是在POJ上WA了，最后处理输入之后AC了。

看看程序想了想，应该POJ上面形如5:(5)这样的输入中有可能不是冒号和括号，而是其他的符号吧。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #include<vector>
 #define inf 0x7fffffff
 using namespace std;
 const int maxn=+;
 const int max_log_maxn=;

 int n,root;
 vector<int> G[maxn];
 int father[max_log_maxn][maxn],d[maxn];
 int vis[maxn];

 void dfs(int u,int p,int depth)
 {
     father[][u]=p;
     d[u]=depth;
     int num=G[u].size();
     for (int i= ;i<num ;i++)
     {
         int v=G[u][i];
         if (v != p) dfs(v,u,depth+);
     }
 }

 void init()
 {
     dfs(root,-,);
     for (int k= ;k+<max_log_maxn ;k++)
     {
         for (int i= ;i<=n ;i++)
         {
             if (father[k][i]<) father[k+][i]=-;
             else father[k+][i]=father[k][father[k][i] ];
         }
         father[k][root]=root;
     }
 }

 int LCA(int u,int v)
 {
     if (d[u]>d[v]) swap(u,v);
     for (int k= ;k<max_log_maxn ;k++)
     {
         if ((d[v]-d[u])>>k & )
             v=father[k][v];
     }
     if (u==v) return u;
     for (int k=max_log_maxn- ;k>= ;k--)
     {
         if (father[k][u] != father[k][v])
         {
             u=father[k][u];
             v=father[k][v];
         }
     }
     return father[][u];
 }

 int main()
 {
     while (scanf("%d",&n)!=EOF)
     {
         int u,num,v;
         for (int i= ;i<=n ;i++) G[i].clear();
         memset(vis,,sizeof(vis));
         char str[];
         memset(str,,sizeof(str));
         char ch[],ch2[],ch3[];
         for (int i= ;i<n ;i++)
         {
             int u=,num=;
             scanf("%d%1s%1s%d%1s",&u,ch,ch2,&num,ch3);
 //            scanf("%s",str);
 //            int u=0,num=0;
 //            int len=strlen(str);
 //            int j=0;
 //            for (j=0 ;j<len && str[j]!=':';j++) u=u*10+str[j]-'0';
 //            for (j=j+2 ;j<len && str[j]!=')';j++) num=num*10+str[j]-'0';
             while (num--)
             {
                 scanf("%d",&v);
                 G[u].push_back(v);
                 vis[v]=;
             }
         }
         for (int i= ;i<=n ;i++) if (!vis[i]) {root=i;break; }
         init();
         memset(vis,,sizeof(vis));
         scanf("%d",&num);
         memset(str,,sizeof(str));
         //getchar();
         for (int j= ;j<num ;j++)
         {
             scanf("%1s%d%d%1s",ch,&u,&v,ch2);
 //            gets(str);
 //            int u=0,v=0;
 //            int len=strlen(str);
 //            int i=0;
 //            for (i=1 ;i<len && str[i]>='0' && str[i]<='9' ;i++)
 //                u=u*10+str[i]-'0';
 //            while (i<len)
 //            {
 //                if (str[i]>='0' && str[i]<='9') break;
 //                i ++ ;
 //            }
 //            while (i<len)
 //            {
 //                if (str[i]==')') break;
 //                v=v*10+str[i]-'0';
 //                i ++ ;
 //            }
             int ans=LCA(u,v);
             vis[ans]++;
         }
         for (int i= ;i<=n ;i++)
             if (vis[i]) printf("%d:%d\n",i,vis[i]);
     }
     return ;
 }