树链剖分——复习计划

树链剖分

应用

一.

• 题目链接： https://codeforces.com/problemset/problem/1403/B
• 解题思路： 该题需要考虑每条边给答案带来的贡献，通过简单的画图和思考我们可以发现，当一个节点下有偶数个叶子节点时，该节点到父亲节点的边的贡献为2，若为奇数则为1，那么最后的答案为$N-2+sigma(du[i]mod2==0)$，利用树链剖分维护即可

int n, q, rt, sum;
const int N = 2e5+5;
struct node {
    int v0, v1, l, r, tag;
}tr[N<<2];
vector <int> edge[N];
int deg[N];
int value[N]; //节点值
int fa[N] , deep[N] , son[N] , siz[N];
int dfn[N] ,top[N], id[N], tot = 0;
void dfs1 (int u)
{
    siz[u] = 1;
    for (auto v : edge[u])
    {
        if (v != fa[u])
        {
            fa[v] = u;
            deep[v] = deep[u] + 1;
            dfs1 (v);
            value[u] += value[v];
            siz[u] += siz[v];
            if (siz[v] > siz[son[u]])
                son[u] = v;
        }
    }
}
void dfs2 (int u, int t)
{
    top[u] = t;
    dfn[u] = ++ tot;
    id[tot] = u;
    if (son[u]) dfs2 (son[u], t);
    for (auto v : edge[u])
        if (v != son[u] && v != fa[u])
            dfs2 (v, v);
}
void push_up (int i)
{
    tr[i].v0 = tr[i<<1].v0 + tr[i<<1|1].v0;
    tr[i].v1 = tr[i<<1].v1 + tr[i<<1|1].v1;
}
void push_down (int i)
{
    if (tr[i].tag)
    {
        tr[i<<1].tag ^= tr[i].tag;
        tr[i<<1|1].tag ^= tr[i].tag;
        swap (tr[i<<1].v0, tr[i<<1].v1);
        swap (tr[i<<1|1].v0, tr[i<<1|1].v1);
        tr[i].tag = 0;
    }
}
void build (int i , int l , int r)
{
    tr[i].l = l, tr[i].r = r, tr[i].tag = 0;
    if (l == r)
    {
        if (value[id[l]] % 2) tr[i].v1 ++;
        else tr[i].v0 ++;
        return ;
    }
    int mid = (l + r) >> 1;
    build (i << 1, l , mid);
    build (i << 1 | 1 , mid + 1 ,r);
    push_up (i);
}
void update (int i , int l ,int r, int L , int R)
{
    if (l >= L && r <= R)
    {
        swap (tr[i].v0, tr[i].v1);
        tr[i].tag ^= 1;
        return ;
    }
    push_down(i);
    int mid = (l + r) >> 1;
    if (L <= mid) update (i<<1,l,mid,L,R);
    if (R >  mid) update (i<<1|1,mid+1,r,L,R);
    push_up(i);
}
void update2 (int x, int y)
{
    while(top[x] != top[y])
    {
        if(deep[top[x]] < deep[top[y]])
            swap(x,y);
        update(1, 1, n, dfn[top[x]], dfn[x]);
        x = fa[top[x]];
    }
    if(deep[x] > deep[y]) swap(x,y);
    update(1, 1, n, dfn[x], dfn[y]);
}
int main(void)
{
    CLOSE;
    cin >> n >> q;
    for (int i = 1 ; i < n ; i ++)
    {
        int u, v;
        cin >> u >> v;
        edge[u].pb (v);
        edge[v].pb (u);
        deg[u] ++ , deg[v] ++;
    }
    for (int i = 1 ; i <= n ; i ++)
        if (deg[i] == 1)
            value[i] = 1, sum ++;
    for (int i = 1 ; i <= n ; i ++)
        if (sz(edge[i]))
        {
            rt = i;
            break ;
        }
    dfs1 (rt), dfs2 (rt, rt), build (1,1,n);
    vector <int> ve;
    while (q --)
    {
        int now = sum;
        ve.clear();
        int x, v;
        cin >> x;
        for (int i = 1 ; i <= x ; i ++)
        {
            cin >> v;
            ve.pb(v);
            if (deg[v] != 1) update2 (rt, v), now ++;
            deg[v] ++;
        }
        if (now % 2) cout << -1 << endl;
        else cout << n + x - 2 + tr[1].v0 << endl;
        for (auto it : ve)
        {
            deg[it] --;
            if (deg[it] != 1) update2 (rt, it);
        }
    }
}