2018.8.24(dfs序，线段树，动态树直径的维护)

题目大意：
对于一棵树，每次询问删掉两棵子树的直径

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每次删掉两颗子树相当于在$dfs$序挖掉两段区间再求解
我们在$dfs$序上维护一段区间内点集的直径
每次询问的时候考虑合并
有这么一个结论
合并两个相邻联通块新的直径的两个端点肯定是原来两个联通块的直径的两个直径的端点

#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(int i = j;i <= k;++i)
#define repp(i,j,k) for(int i = j;i >= k;--i)
#define rept(i,x) for(int i = linkk[x];i;i = e[i].n)
#define P pair<int,int>
#define pb push_back
#define pc putchar
#define mp make_pair
#define file(k) memset(k,0,sizeof(k));
#define ll long long
#define ls root * 2
#define rs root * 2 + 1
int n , q;
int linkk[101000] , t , dep[101000];
int ola[401000] , totw , first[101000] , last[101000];
int mn[401000][20];
struct node{
    int n , y;
}e[201000];
struct tree{
    int l,r,len;
}tr[802000];
tree ans;
int read()
{
    int sum = 0;char c = getchar();bool flag = true;
    while(c < '0' || c > '9') {if(c == '-') flag = false;c = getchar();}
    while(c >= '0' && c <= '9') sum = sum * 10 + c - 48 , c = getchar();
    if(!flag) sum = -sum;
    return sum;
} 
void insert(int x,int y)
{
    e[++t].y = y;e[t].n = linkk[x];linkk[x] = t;
    e[++t].y = x;e[t].n = linkk[y];linkk[y] = t;
    return;
}
int Lca(int x,int y)
{
    if(x == y) return x;
    if(x == 0 || y == 0) return 0;
    if(first[x] > first[y]) swap(x,y);
    int k = log2(first[y] - first[x]);
    return dep[ mn[first[x]][k] ] > dep[ mn[first[y] - (1<<k)][k] ] ? mn[first[y] - (1<<k)][k] : mn[first[x]][k];
}
void dfs(int x,int fa)
{
    ola[++totw] = x;first[x] = totw;dep[x] = dep[fa] + 1;
    rept(i,x)
        if(e[i].y != fa)
        {
            dfs(e[i].y,x);
            ola[++totw] = x; 
        }
    last[x] = totw;
}
bool mycmp(tree a,tree b){return a.len > b.len;}
int get(int x,int y){return dep[x] + dep[y] - 2 * dep[Lca(x,y)];}
void update(tree &root,tree a,tree b)
{
    root = a.len > b.len ? a : b;
    tree tmp[4];
    tmp[0].l = a.l;tmp[0].r = b.l;tmp[0].len = get(a.l,b.l);
    tmp[1].l = a.l;tmp[1].r = b.r;tmp[1].len = get(a.l,b.r);
    tmp[2].l = a.r;tmp[2].r = b.l;tmp[2].len = get(a.r,b.l);
    tmp[3].l = a.r;tmp[3].r = b.r;tmp[3].len = get(a.r,b.r);
    sort(tmp,tmp+4,mycmp);
    if(root.len < tmp[0].len) root = tmp[0];
}
void build(int root,int l,int r)
{
    if(l == r)
    {
        tr[root].l = tr[root].r = mn[l][0];
        tr[root].len = 0;
        return;
    }
    int mid = (l + r)>>1;
    build(ls,l,mid);build(rs,mid+1,r);
    update(tr[root],tr[ls],tr[rs]); 
}
void init()
{
    n = read();q = read();
    rep(i,1,n-1)
    {
        int x = read() , y = read();
        insert(x,y);
    }
    dfs(1,0);
    rep(i,1,totw) mn[i][0] = ola[i];
    for(int j = 1;(1<<j) <= totw;++j)
        for(int i = 1;(i + (1<<j)) <= totw;++i)
        {
            mn[i][j] = mn[i][j-1];
            if(dep[ mn[i][j-1] ] > dep[ mn[i+(1<<(j-1))][j-1] ])
                mn[i][j] = mn[i+(1<<(j-1))][j-1];
        }
}
void New(int root,int l,int r,int x,int y)
{
    if(r < x || l > y) return;
    if(x > y) return;
    if(x <= l && r <= y)
    {
        update(ans,ans,tr[root]);
        return;
    }
    int mid = (l + r)>>1;
    New(ls,l,mid,x,y);New(rs,mid+1,r,x,y);
}
void work()
{
    dep[0] = 10000000;
    while(q--)
    {
        int x = read() , y = read();
        ans.l = ans.r = ans.len = 0;
        if(first[x] > first[y]) swap(x,y);
        int a = first[x] , b = last[x] , c = first[y] , d = last[y];
        New(1,1,totw,1,a-1);
        New(1,1,totw,max(b,d)+1,totw);
        if(b < c) New(1,1,totw,b+1,c-1);
        printf("%d\n",ans.len);
    }
}
int main()
{
    freopen("find.in","r",stdin);
    freopen("find.out","w",stdout);
    init();
    build(1,1,totw);
    work();
    return 0;
}