虚树板子

int n, m, sz[N], dep[N], mi[N];
int L[N], R[N], fa[N], son[N], top[N];
vector<int> g[N], gg[N];
int s[N], cnt, vis[N];

void dfs(int x, int d, int f) {
	L[x]=++*L,sz[x]=1,fa[x]=f,dep[x]=d;
	for (int y:g[x]) if (y!=f) {
		dfs(y,d+1,x),sz[x]+=sz[y];
		if (sz[y]>sz[son[x]]) son[x]=y;
	}
	R[x]=*L;
}
void dfs(int x, int tf) {
	top[x]=tf;
	if (son[x]) dfs(son[x],tf);
	for (int y:g[x]) if (!top[y]) dfs(y,y);
}
int lca(int x, int y) {
	while (top[x]!=top[y]) {
		if (dep[top[x]]<dep[top[y]]) swap(x,y);
		x=fa[top[x]];
	}
	return dep[x]<dep[y]?x:y;
}
bool cmp(int x, int y) {
	return L[x]<L[y];
}
void solve(vector<int> a) {
	sort(a.begin(),a.end(),cmp);
	int sz = a.size();
	REP(i,1,sz-1) a.pb(lca(a[i],a[i-1]));
	sort(a.begin(),a.end(),cmp);
	a.erase(unique(a.begin(),a.end()),a.end());
	s[cnt=1]=a[0],sz=a.size();
	REP(i,1,sz-1) {
		while (cnt>=1) {
			if (L[s[cnt]]<=L[a[i]]&&L[a[i]]<=R[s[cnt]]) {
				gg[s[cnt]].pb(a[i]);
				break;
			}
			--cnt;
		}
		s[++cnt]=a[i];
	}
	DP(s[1]);
	for (int x:a) gg[x].clear();
}

例1 luogu P2495 [SDOI2011]消耗战

大意: 给定有根树树, 边有边权, 根为$1$, $m$个询问, 每次给出$k$个点, 询问使根节点与$k$个点不连通要删除的边权和的最小值.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, m, sz[N], dep[N], mi[N];
int L[N], R[N], fa[N], son[N], top[N];
struct _ {int to,w;};
vector<_> g[N];
vector<int> gg[N];
int s[N], cnt, vis[N];

void dfs(int x, int d, int f, int m) {
    L[x]=++*L,sz[x]=1,fa[x]=f,dep[x]=d,mi[x]=m;
    for (_ e:g[x]) if (e.to!=f) {
        int y=e.to;
        dfs(y,d+1,x,min(m,e.w)),sz[x]+=sz[y];
        if (sz[y]>sz[son[x]]) son[x]=y;
    }
    R[x]=*L;
}
void dfs(int x, int tf) {
    top[x]=tf;
    if (son[x]) dfs(son[x],tf);
    for (_ e:g[x]) if (!top[e.to]) dfs(e.to,e.to);
}
int lca(int x, int y) {
    while (top[x]!=top[y]) {
        if (dep[top[x]]<dep[top[y]]) swap(x,y);
        x=fa[top[x]];
    }
    return dep[x]<dep[y]?x:y;
}
bool cmp(int x, int y) {
    return L[x]<L[y];
}
ll DP(int x) {
    if (vis[x]) return 1e17;
    ll ans = 0;
    for (int y:gg[x]) ans+=min(DP(y),(ll)mi[y]);
    return ans;
}
void solve(vector<int> a) {
    sort(a.begin(),a.end(),cmp);
    int sz = a.size();
    REP(i,1,sz-1) a.pb(lca(a[i],a[i-1]));
    a.pb(1);
    sort(a.begin(),a.end(),cmp);
    a.erase(unique(a.begin(),a.end()),a.end());
    s[cnt=1]=a[0],sz=a.size();
    REP(i,1,sz-1) {
        while (cnt>=1) {
            if (L[s[cnt]]<=L[a[i]]&&L[a[i]]<=R[s[cnt]]) {
                gg[s[cnt]].pb(a[i]);
                break;
            }
            --cnt;
        }
        s[++cnt]=a[i];
    }
    printf("%lld
", DP(1));
    for (int x:a) gg[x].clear();
}
int main() {
    scanf("%d", &n);
    REP(i,1,n-1) {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        g[u].pb({v,w}),g[v].pb({u,w});
    }
    dfs(1,0,0,INF),dfs(1,1);
    scanf("%d", &m);
    REP(i,1,m) {
        int k, t;
        scanf("%d", &k);
        vector<int> v;
        REP(i,1,k) scanf("%d",&t),v.pb(t),vis[t]=1;
        solve(v);
        for (int x:v) vis[x]=0;
    }
}

例2 Kingdom and its Cities CodeForces - 613D 

大意: 给定树, $m$个询问, 每次询问标记$k$个黑点, 求将树划分为$k$个连通块所需要删除的最少白点数, 要求每个连通块内恰好有$1$个黑点.

若两个黑点相邻显然无法划分成功, 否则一定可以. 考虑贪心划分, 若当前点为白点, 并且儿子中有多个未划分的黑点, 显然白点要删除, 若儿子仅有一个黑点可以暂时不删. 若当前点为黑点, 则删除所有未划分的儿子.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, m, vis[N];
int fa[N], dep[N], sz[N], top[N];
int son[N], L[N], R[N];
vector<int> g[N];
int s[N], cnt, ans;

void dfs(int x, int f, int d) {
	sz[x]=1,fa[x]=f,dep[x]=d,L[x]=++*L;
	for (int y:g[x]) if (y!=f) {
		dfs(y,x,d+1),sz[x]+=sz[y];
		if (sz[y]>sz[son[x]]) son[x]=y;
	}
	R[x]=*L;
}
void dfs(int x, int tf) {
	top[x]=tf;
	if (son[x]) dfs(son[x],tf);
	for (int y:g[x]) if (!top[y]) dfs(y,y);
}
int lca(int x, int y) {
	while (top[x]!=top[y]) {
		if (dep[top[x]]<dep[top[y]]) swap(x,y);
		x = fa[top[x]];
	}
	return dep[x]<dep[y]?x:y;
}

int DP(int x) {
	for (int y:g[x]) if (vis[y]&&vis[x]) {
		if (dep[x]+1==dep[y]) return ans=INF;
	}
	int ret = vis[x];
	for (int y:g[x]) { 
		int t = DP(y);
		if (t) {
			if (vis[x]) ++ans;
			else ++ret;
		}
	}
	if (ret>1) ++ans,ret=0;
	return ret;
}

bool cmp(int x,int y) {return L[x]<L[y];}

void solve(vector<int> v) {
	sort(v.begin(),v.end(),cmp);
	int sz=v.size();
	REP(i,1,sz-1) v.pb(lca(v[i],v[i-1]));
	sort(v.begin(),v.end(),cmp);
	v.erase(unique(v.begin(),v.end()),v.end());
	s[cnt=1]=v[0],sz=v.size();
	vis[0] = 1;
	REP(i,1,sz-1) {
		while (cnt>=1) {
			if (L[s[cnt]]<=L[v[i]]&&L[v[i]]<=R[s[cnt]]) {
				g[s[cnt]].pb(v[i]);
				break;
			}
			--cnt;
		}
		s[++cnt] = v[i];
	}
	ans = 0;
	DP(s[1]);
	printf("%d
", ans>=n?-1:ans);
	REP(i,0,sz-1) g[v[i]].clear(),vis[v[i]]=0;
}
int main() {
	scanf("%d", &n);
	REP(i,2,n) {
		int u, v;
		scanf("%d%d", &u, &v);
		g[u].pb(v),g[v].pb(u);
	}
	dfs(1,0,0),dfs(1,1);
	REP(i,1,n) g[i].clear();
	scanf("%d", &m);
	REP(i,1,m) {
		int k, t;
		scanf("%d", &k);
		vector<int> v;
		REP(i,1,k) scanf("%d",&t),v.pb(t),vis[t]=1;
		solve(v);
	}
}