pog loves szh III

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 470    Accepted Submission(s): 97

Problem Description

Pog and Szh are playing games. Firstly Pog draw a tree on the paper. Here we define 1 as the root of the tree.Then Szh choose some nodes from the tree. He wants Pog helps to find the least common ancestor (LCA) of these node.The question is too difficult for Pog.So he decided to simplify the problems.The nodes picked are consecutive numbers from li to ri ([li,ri]).

Hint : You should be careful about stack overflow !

Input

Several groups of data (no more than 3 groups,n≥10000 or Q≥10000).

The following line contains ans integers,n(2≤n≤300000).

AT The following n−1 line, two integers are bi and ci at every line, it shows an edge connecting bi and ci.

The following line contains ans integers,Q(Q≤300000).

AT The following Q line contains two integers li and ri(1≤li≤ri≤n).

Output

For each case,output Q integers means the LCA of [li,ri].

Sample Input

5

1 2

1 3

3 4

4 5

5

1 2

2 3

3 4

3 5

1 5

Sample Output

1

1

3

3

1

Hint

Be careful about stack overflow.

这个题有点裸，以为LCA是满足结合律的，所以直接拿个线段树来搞就可以了~

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cstdio>
using namespace std ;
typedef long long LL ;
const int N = 400030 ;
const int DEG = 30 ;

int n , lca[N<<2] ;

struct Edge
{
    int to,next;
}edge[N*2];

int head[N],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[N][DEG] , deg[N];

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];
}

#define root 1,n,1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define lr rt<<1
#define rr rt<<1|1

void Up( int rt ) {
    lca[rt] = LCA( lca[lr] , lca[rr] ) ;
}

void build( int l , int r , int rt ) {
    if( l == r ) {
        lca[rt] = l ;
        return ;
    }
    int mid = (l+r) >> 1 ;
    build(lson),build(rson);
    Up(rt);
}

int query( int l , int r , int rt , int L , int R ) {
    if( L == l && r == R ) return lca[rt] ;
    int mid = (l+r) >> 1 ;
    if( R <= mid ) return query( lson , L , R ) ;
    else if( L > mid ) return query( rson , L , R ) ;
    else return LCA( query( lson , L , mid ) , query( rson , mid + 1 , R ) ) ;
}

int main() {
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    while( ~scanf("%d",&n) ) {
        init();
        for( int i = 1 ; i < n ; ++i ) {
            int u , v ; scanf("%d%d",&u,&v);
            addedge( u , v ) ;
            addedge( v , u ) ;
        }
        BFS(1);
        build(root) ;
        int q ; scanf("%d",&q);
        while( q-- ) {
            int x , y ; scanf("%d%d",&x,&y);
            printf("%d\n",query(root,x,y));
        }
    }
    return 0 ;
}