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=900+10;
const int max_log_maxn=10;

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[0][u]=p;
    d[u]=depth;
    int num=G[u].size();
    for (int i=0 ;i<num ;i++)
    {
        int v=G[u][i];
        if (v != p) dfs(v,u,depth+1);
    }
}

void init()
{
    dfs(root,-1,0);
    for (int k=0 ;k+1<max_log_maxn ;k++)
    {
        for (int i=1 ;i<=n ;i++)
        {
            if (father[k][i]<0) father[k+1][i]=-1;
            else father[k+1][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=0 ;k<max_log_maxn ;k++)
    {
        if ((d[v]-d[u])>>k & 1)
            v=father[k][v];
    }
    if (u==v) return u;
    for (int k=max_log_maxn-1 ;k>=0 ;k--)
    {
        if (father[k][u] != father[k][v])
        {
            u=father[k][u];
            v=father[k][v];
        }
    }
    return father[0][u];
}

int main()
{
    while (scanf("%d",&n)!=EOF)
    {
        int u,num,v;
        for (int i=0 ;i<=n ;i++) G[i].clear();
        memset(vis,0,sizeof(vis));
        char str[100];
        memset(str,0,sizeof(str));
        char ch[2],ch2[2],ch3[2];
        for (int i=0 ;i<n ;i++)
        {
            int u=0,num=0;
            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]=1;
            }
        }
        for (int i=1 ;i<=n ;i++) if (!vis[i]) {root=i;break; }
        init();
        memset(vis,0,sizeof(vis));
        scanf("%d",&num);
        memset(str,0,sizeof(str));
        //getchar();
        for (int j=0 ;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=1 ;i<=n ;i++)
            if (vis[i]) printf("%d:%d
",i,vis[i]);
    }
    return 0;
}