CF613D Kingdom and its Cities

发表于 | 阅读数

给定一棵树，每次询问$k_i$个点，询问最少需要截断多少个点，可以使得这些点两两联通。

$\sum k_i \leq 10^5$

显然是虚树，然后考虑如何$dp$。

首先特胖一下是否有相邻的两个点。

$f[x]$为以$x$为根的子树的答案.

$g[x]$为此时有没有关键点直连到 $x$ 的父亲，即把$x$去掉。

  • 如果$x$不是关键点，并且旗下关键点$>1$,那么显然可以$f[x]=\sum dp[y]+1$，把自己去了并且$g[x]=0$
  • 如果是关键点,子树里有顶部存在关键点去掉中间的点,$dp[x]=\sum (g[x] +dp[y])$,
  • 如果$x$不是关键点，并且旗下关键点$g[x]=1$,无所谓，那么显然可以$f[x]=\sum dp[y]$，$g[x]=(g[x]=1)$
代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 2e5 + 10;
const int mod = 1e9 + 7;
const int INF = 1e9;
int read()
{
    int x = 0, F = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            F = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * F;
}

namespace Tree
{

    int st[N][22], dfn[N], dep[N], tot, lg2[N];
    vector<int> G[N];
    void dfs(int x, int fa)
    {
        st[++tot][0] = x;
        dep[x] = dep[fa] + 1;
        dfn[x] = tot;
        for (int to : G[x])
        {
            if (to == fa)
                continue;
            dfs(to, x);
            st[++tot][0] = x;
        }
    }
    int lower(int u, int v)
    {
        return (dep[u] < dep[v]) ? u : v;
    }

    void Lca_init()
    {
        dfs(1, 0);
        for (int i = 2; i <= tot; i++)
            lg2[i] = lg2[i >> 1] + 1;
        for (int l = 1; (1 << l) <= tot; l++)
        {
            for (int i = 1; i + (1 << l) - 1 <= tot; i++)
                st[i][l] = lower(st[i][l - 1], st[i + (1 << (l - 1))][l - 1]);
        }
    }
    int lca(int u, int v)
    {
        u = dfn[u];
        v = dfn[v];
        if (u > v)
            swap(u, v);
        int i = lg2[v - u + 1], w = (1 << i);
        return lower(st[u][i], st[v - w + 1][i]);
    }
    int dis(int u, int v)
    {
        int lc = lca(u, v);
        return dep[u] + dep[v] - 2 * dep[lc];
    }
}; // namespace Tree
using namespace Tree;

int vis[N];
int b[N];
namespace XTree
{
    stack<int> s;
    int F[N], G[N];
    int flag;

    vector<int> g[N];

    bool cmp(int x, int y)
    {
        return Tree::dfn[x] < Tree::dfn[y];
    }

    void addEdge(int u, int v)
    {

        int d = Tree::dis(u, v);

        if (vis[u] && vis[v] && d == 1)
            flag = 0;
        g[u].push_back(v);
        g[v].push_back(u);
    }

    void build(int k)
    {
        flag = 1;

        while (!s.empty())

            s.pop();
        sort(b + 1, b + 1 + k, cmp);

        s.push(1);
        for (int i = 1; i <= k; ++i)
        {
            int x = b[i];
            int lca = Tree::lca(x, s.top());
            while (s.top() != lca)
            {
                int tmp = s.top();
                s.pop();
                if (dfn[s.top()] < dfn[lca])
                    s.push(lca);
                addEdge(s.top(), tmp);
            }
            s.push(x);
        }
        while (s.top() != 1)
        {
            int tmp = s.top();
            s.pop();
            addEdge(s.top(), tmp);
        }
    }
    void dfs(int x, int fa)
    {
        F[x] = 0, G[x] = 0;
        for (int to : g[x])
        {
            if (to == fa)
                continue;
            dfs(to, x);
            F[x] += F[to];
            G[x] += G[to];
        }
        if (!vis[x])
            F[x] += (G[x] > 1), G[x] = (G[x] == 1);
        else
            F[x] += G[x], G[x] = 1;
        g[x].clear();
    }
    void solve()
    {
        dfs(1, 0);
        if (!flag)
            puts("-1");
        else
        {
            cout << F[1] << endl;
        }
    }
} // namespace XTree

int main()
{
    int n = read();
    for (int i = 1; i < n; i++)
    {
        int u = read(), v = read();
        G[u].push_back(v);
        G[v].push_back(u);
    }
    Lca_init();
    int q = read();

    for (int i = 1; i <= q; i++)
    {
        int k = read();

        for (int j = 1; j <= k; j++)
        {
            int x = read();
            b[j] = x, vis[x] = 1;
        }

        XTree::build(k);
        XTree::solve();

        for (int j = 1; j <= k; j++)
            vis[b[j]] = 0;
    }
}