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 '\n'
#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\n", 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 '\n'
#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\n", 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);
}
}
