SDNU 1492.Problem_A LCA倍增算法

1492.Problem_A

Time Limit: 1000 MS    Memory Limit: 32768 KB

Description

SDNU ACM-ICPC Association is in a mess. If you join it, you should be in the shadow of a person. This person, we call him "dalao". For example, there are two rookies, C and D, rely on A. We can say that A is the closest dalao of rookies C and D.For another example, A have two heelers C and D, D have two heelers E and F.We can know that A is the closest dalao of E and C.

But with the growth of Association, we cannot find the closest dalao between two rookies. So we need your help.

Input

There will be a series of examples.

For each input, the first line contains two integers N. N means the number of people.

Followed by the N-1 lines, each line contains two integers A and B. It means that A is a dalao of B.

Then the next line, contains two integers A and B. It means that we want to know who is the closest dalao between A and B.


Output

For each input, you should output the name of the closest dalao.If you cannot find this dalao, you should output "I am so bad."

Sample Input

7
1 2
1 3
2 5
2 6
6 7
6 8
3 5

Sample Output

1

Hint
Remember that a rookie is also a dalao of itself
For example:
Input
3
1 2
2 3
2 3
Output
2

    选拔赛的一道题目，题意很好理解，简单说就是LCA（最近公共祖先），不过之前没怎么做过这类题目，所以也一直没有整理过自己的模板.....翻到一个dfs+ST的算法就直接敲了，然后....TLE了，才看到后面有更快的算法，不过离比赛结束还剩10分钟已经没时间了，所以就这样放弃了。结果比赛结束下来，敲的后面的倍增算法就A了.....早知道该先敲后面的说，哎呀好气啊。

    下面是AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int MAXN=10010;
const int DEG=20;

struct Edge
{
    int to,next;
}edge[MAXN*2];
int head[MAXN],tot;

void addedge(int u,int v)
{
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void init()
{
    tot=0;
    memset(head,-1,sizeof(head));
}

int fa[MAXN][DEG];
int deg[MAXN];

void BFS(int root)
{
    queue<int>que;
    deg[root]=0;
    fa[root][0]=root;
    que.push(root);
    while(!que.empty())
    {
        int tmp=que.front();
        que.pop();
        for(int i=1;i<DEG;i++)
        {
            fa[tmp][i]=fa[fa[tmp][i-1]][i-1];
        }
        for(int i=head[tmp];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(v==fa[tmp][0])
                continue;
            deg[v]=deg[tmp]+1;
            fa[v][0]=tmp;
            que.push(v);
        }
    }
}

int LCA(int u,int v)
{
    if(deg[u]>deg[v])
        swap(u,v);
    int hu=deg[u],hv=deg[v];
    int tu=u,tv=v;
    for(int det=hv-hu,i=0;det;det>>=1,i++)
    {
        if(det&1)
            tv=fa[tv][i];
    }
    if(tu==tv)
        return tu;
    for(int i=DEG-1;i>=0;i--)
    {
        if(fa[tu][i]==fa[tv][i])
            continue;
        tu=fa[tu][i];
        tv=fa[tv][i];
    }
    return fa[tu][0];
}

bool flag[MAXN];

int main()
{
    int T;
    int n;
    int u,v;
    while(scanf("%d",&n)!=EOF)
    {
        init();
        memset(fa,0,sizeof(fa));
        memset(deg,0,sizeof(deg));
        memset(flag,false,sizeof(flag));
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
            flag[v]=true;
        }
        int root;
        for(int i=1;i<=n;i++)
        {
            if(!flag[i])
            {
                root=i;
                break;
            }
        }
        BFS(root);
        scanf("%d%d",&u,&v);
        if(LCA(u,v)==0)
            cout<<"I am so bad."<<endl;
        else
            cout<<LCA(u,v)<<endl;
    }
    return 0;
}