牛客 4C Alliances (dfs序)

大意: 给定树, 有$k$个帮派, 第$i$个帮派所占据点为$c_i$, 以及$c_i$两两相连路径上的所有点. 一个点可能被多个帮派占领. $q$个询问, 第$i$个询问给定$t_i$个帮派, 给定点$u$, 求将$t_i$个帮派合并后, 点$u$到帮派的最近距离.

先求出帮派合并后的$lca$, 若$u$不在$lca$所在子树内, 那么最短距离就是$dis(u,lca)$, 否则在帮派占据的点中, 找到$dfs$序与$u$最接近的点$x$, 显然$lca(u,x)$属于帮派, 所以最短距离就为$dis(u,lca(u,x))$.

#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, P2 = 998244353, 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

const int N = 1e6+10;
int n, k, q, qry[N], Top[N];
vector<int> g[N], f[N];
int no[N], son[N], L[N], R[N];
int sz[N], top[N], dep[N], fa[N];

void dfs(int x, int f, int d) {
	no[L[x]=++*L]=x,fa[x]=f,dep[x]=d,sz[x]=1;
	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 dis(int x, int y) {
	return dep[x]+dep[y]-2*dep[lca(x,y)];
}
int main() {
	scanf("%d", &n);
	REP(i,2,n) {
		int u=rd(),v=rd();
		g[u].pb(v),g[v].pb(u);
	}
	dfs(1,0,0),dfs(1,1);
	scanf("%d", &k);
	REP(i,1,k) {
		int c=rd();
		REP(j,1,c) {
			int t=rd();
			f[i].pb(L[t]);
			if (j==1) Top[i]=t;
			else Top[i]=lca(Top[i],t);
		}
		sort(f[i].begin(),f[i].end());
	}
	scanf("%d", &q);
	while (q--) {
		int u=rd(),sz=rd(),Lca=0,ans=1e9;
		REP(i,1,sz) qry[i]=rd(),Lca=i==1?Top[qry[i]]:lca(Lca,Top[qry[i]]);
		if (L[Lca]<=L[u]&&L[u]<=R[Lca]) {
			REP(i,1,sz) {
				auto t = lower_bound(f[qry[i]].begin(),f[qry[i]].end(),L[u]);
				if (t!=f[qry[i]].end()) ans = min(ans, dis(u,lca(u,no[*t])));
				if (t!=f[qry[i]].begin()) ans = min(ans, dis(u,lca(u,no[*--t])));
			}
		}
		ans = min(ans,dis(u,Lca));
		printf("%d\n",ans);
	}
}