给你一棵nn个结点的树。

qq次询问,每次询问给出kk和一个大小为kk的点集SS,求最少删掉多少个不在SS中的点可以使得SS中的点两两不联通。

无解输出1-1

n,q,(k)105n, q, (\sum k)\le 10^5

CF613D

Solution

又是一道虚树的板子题

首先考虑单次查询怎么做,显然可以直接贪心

f[x]f[x]表示xx的子树内要满足条件至少删掉几个点,g[x]g[x]表示xx的子树内还剩下多少个SS中的点与xx联通(为被阻断)

分类讨论一下,然后用虚树优化即可

这道题甚至比消耗战还要简单

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
T sum = 0, fl = 1; char ch = getchar();
for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
return sum * fl;
}

inline void proc_status ()
{
ifstream t ("/proc/self/status");
cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 1e5 + 100;

int N, Q;
int e, Begin[Maxn], To[Maxn << 1], Next[Maxn << 1];
int fa[Maxn], dep[Maxn], dfn[Maxn], dfs_clock, size[Maxn], son[Maxn], top[Maxn];
int K, A[Maxn << 1];

inline void add_edge (int x, int y) { To[++e] = y; Next[e] = Begin[x]; Begin[x] = e; }

inline void dfs0 (int x)
{
dep[x] = dep[fa[x]] + 1, size[x] = 1;

for (int i = Begin[x]; i; i = Next[i])
{
int y = To[i];
if (y == fa[x]) continue;
fa[y] = x;

dfs0(y);

if (size[y] > size[son[x]]) son[x] = y;
size[x] += size[y];
}
}

inline void dfs1 (int x, int now)
{
dfn[x] = ++dfs_clock, top[x] = now;

if (son[x]) dfs1 (son[x], now);

for (int i = Begin[x]; i; i = Next[i])
{
int y = To[i];
if (y == fa[x] || y == son[x]) continue;

dfs1(y, y);
}
}

inline int get_lca (int x, int y)
{
while (top[x] != top[y])
{
if (dep[top[x]] < dep[top[y]]) swap(x, y);
x = fa[top[x]];
}
if (dep[x] < dep[y]) swap(x, y);
return y;
}

int f[Maxn], g[Maxn], Vis[Maxn];

namespace VTREE
{
int Begin[Maxn], To[Maxn << 2], Next[Maxn << 2];
int top, Stack[Maxn << 1];

inline void add_edge (int x, int y) { To[++e] = y; Next[e] = Begin[x]; Begin[x] = e; }

inline void get_dp (int x)
{
f[x] = g[x] = 0;
int cnt = 0, sum = 0;

for (int i = Begin[x]; i; i = Next[i])
{
int y = To[i];
if (y == fa[x]) continue;

get_dp (y);

f[x] += f[y];
if (g[y]) ++cnt, sum = g[y];
}

if (Vis[x])
{
f[x] += cnt;
g[x] = 1;
}
else
{
if (cnt <= 1) g[x] = sum;
else ++f[x], g[x] = 0;
}
}

inline int cmp_by_dfn (int x, int y) { return dfn[x] < dfn[y]; }

inline void build ()
{
sort(A + 1, A + K + 1, cmp_by_dfn);
/**/ for (int i = K; i > 1; --i) A[++K] = get_lca (A[i], A[i - 1]); A[++K] = 1;
sort(A + 1, A + K + 1, cmp_by_dfn);
K = unique (A + 1, A + K + 1) - A - 1;

for (int i = 1; i <= K; ++i)
{
while (top && dfn[Stack[top]] + size[Stack[top]] - 1< dfn[A[i]]) --top;
if (top) add_edge (Stack[top], A[i]);
Stack[++top] = A[i];
}

}

inline int work ()
{
get_dp (1);
return f[1];
}

}

inline void Init () { for (int i = 1; i <= K; ++i) Vis[A[i]] = VTREE :: Begin[A[i]] = 0; e = VTREE :: top = 0; }

inline void Solve ()
{
dfs0(1), dfs1(1, 1);

Q = read<int>();
while (Q--)
{
K = read<int>(); for (int i = 1; i <= K; ++i) A[i] = read<int>(), Vis[A[i]] = 1;

int fl = 0;
for (int i = 1; i <= K; ++i) if (Vis[fa[A[i]]]) {fl = 1; break; }

if (fl) puts("-1");
else
{
VTREE :: build ();
printf("%d\n", VTREE :: work());
}

Init();
}
}

inline void Input ()
{
N = read<int>();
for (int i = 1; i < N; ++i)
{
int x = read<int>(), y = read<int>();
add_edge (x, y);
add_edge (y, x);
}
}

int main()
{

#ifndef ONLINE_JUDGE
freopen("D.in", "r", stdin);
freopen("D.out", "w", stdout);
#endif

Input ();
Solve ();

return 0;
}