Count on the path

Count on the path

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Problem Description

bobo has a tree, whose vertices are conveniently labeled by 1,2,…,n.

Let f(a,b) be the minimum of vertices not on the path between vertices a and b.

There are q queries (u_i,v_i) for the value of f(u_i,v_i). Help bobo answer them.

 

Input

The input consists of several tests. For each tests:

The first line contains 2 integers n,q (4≤n≤10⁶,1≤q≤10⁶). Each of the following (n - 1) lines contain 2 integers a_i,b_i denoting an edge between vertices a_i and b_i (1≤a_i,b_i≤n). Each of the following q lines contains 2 integer u′_i,v′_i (1≤u_i,v_i≤n).

The queries are encrypted in the following manner.

u₁=u′₁,v₁=v′₁.
For i≥2, u_i=u′_i⊕f(u_{i - 1},v_{i - 1}),v_i=v′_i⊕f(u_i-1,v_i-1).

Note ⊕ denotes bitwise exclusive-or.

It is guaranteed that f(a,b) is defined for all a,b.

The task contains huge inputs. `scanf` in g++ is considered too slow to get accepted. You may (1) submit the solution in c++; or (2) use hand-written input utilities.

 

Output

For each tests:

For each queries, a single number denotes the value.

 

Sample Input

4 1 1 2 1 3 1 4 2 3 5 2 1 2 1 3 2 4 2 5 1 2 7 6

 

Sample Output

4 3 1

分析：参考http://www.lai18.com/content/8107004.html；

　　　学到的东西很多，比如求子树的第二小，在当前根下却不在其中一个儿子的子树的最小值；

　　　bel[i]表示i在1的哪个儿子下，fa[i]表示i的父亲，son[i]表示i的所有子树里面最小的节点，sec_son[i]对所有i的儿子j的son[j]排序后取第二小的；

　　　dp[i]表示不是i，却是i的父亲的儿子j下的最小son[j]，dp1[i]表示从1走到i所有分支子树下的最小节点；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
const int maxn=1e6+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,q,tot,h[maxn],bel[maxn],fa[maxn],son[maxn],sec_son[maxn],dp[maxn],dp1[maxn],ans;
struct node
{
    int to,nxt;
}e[maxn<<1];
void add(int x,int y)
{
    tot++;
    e[tot].to=y;
    e[tot].nxt=h[x];
    h[x]=tot;
}
void dfs(int now,int pre)
{
    son[now]=sec_son[now]=inf;
    for(int i=h[now];i;i=e[i].nxt)
    {
        int to=e[i].to;
        if(to!=pre)
        {
            fa[to]=now;
            if(now!=1)bel[to]=bel[now];
            dfs(to,now);
            if(son[now]>min(son[to],to))
            {
                sec_son[now]=son[now];
                son[now]=min(son[to],to);
            }
        }
    }
}
void dfs1(int now,int pre)
{
    if(now!=1)dp1[now]=min(dp1[fa[now]],dp[now]);
    else dp1[now]=inf;
    for(int i=h[now];i;i=e[i].nxt)
    {
        int to=e[i].to;
        if(to!=pre)
        {
            dfs1(to,now);
        }
    }
}
void solve(int x,int y)
{
    int now_ans=inf;
    for(int i=h[1];i;i=e[i].nxt)
    {
        int to=e[i].to;
        if(to!=bel[x]&&to!=bel[y])now_ans=min(min(now_ans,son[to]),to);
    }
    if(x!=1)now_ans=min(now_ans,son[x]);
    if(y!=1)now_ans=min(now_ans,son[y]);
    now_ans=min(now_ans,dp1[x]);
    now_ans=min(now_ans,dp1[y]);
    ans=now_ans;
}
int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&q))
    {
        ans=0;
        tot=0;
        memset(h,0,sizeof(h));
        rep(i,1,n)bel[i]=i;
        rep(i,1,n-1)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            add(a,b);
            add(b,a);
        }
        dfs(1,0);
        rep(i,2,n)
        {
            if(fa[i]==1)
            {
                dp[i]=inf;
                continue;
            }
            if(min(son[i],i)!=son[fa[i]])dp[i]=son[fa[i]];
            else dp[i]=sec_son[fa[i]];
        }
        dfs1(1,0);
        while(q--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            a^=ans,b^=ans;
            if(bel[a]==bel[b])ans=1;
            else solve(a,b);
            printf("%d
",ans);
        }
    }
    //system("Pause");
    return 0;
}