BZOJ1095 [ZJOI2007]Hide 捉迷藏 【动态点分治 + 堆】

题目链接

BZOJ1095

题解

传说中的动态点分治,一直不敢碰
今日一会,感觉其实并不艰涩难懂

考虑没有修改,如果不用树形dp的话,就得点分治
对于每个重心,我们会考虑其分治的子树内所有点到它的距离,然后取所有不同子树中最大的两个相加来更新答案

如果带修改怎么办呢?
考虑一个点\(u\)被修改了,会对哪些点产生影响。显然是包含它的分治子树之中

点分树##

我们在点分治过程中,将每个重心与上一层分治重心相连,形成一棵树结构,叫做点分树
一个点被修改,仅影响其在点分树中所有祖先的答案

那么这样我们就可以维护了
对于每个点\(u\),我们开两个堆分别储存一下信息:
\(C\):以\(C\)为根的分治子树中所有点到点分树中\(u\)的父亲的距离
\(B\)\(u\)在点分树中所有儿子\(C\)堆堆顶

对于全局开一个堆\(A\),储存所有堆\(B\)的最大值和次大值之和

那么堆\(A\)堆顶就是当前的答案

对于堆来说,插入是很方便的,但是删除就没那么简单,我们可以对每个堆再开一个堆,储存所有被删除的值,每次访问堆顶时,如果删除堆堆顶与当前堆顶相同,那么都\(pop\)掉,直到不相同位置

还要注意的就是堆间操作的顺序
我们要操作\(C\),就得先将父亲\(B\)堆中该\(C\)堆堆顶\(pop\)掉,操作完再加回去
同样,我们要操作\(B\),也得先从\(A\)中将对应贡献删除

感觉这题难点并不在动态点分治。。。而是在堆的维护吧。。。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 100005,maxm = 200005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
struct heap{
	priority_queue<int> a,b;
	void ins(int x){if (x >= 0) a.push(x);}
	void del(int x){if (x >= 0) b.push(x);}
	int size(){return a.size() - b.size();}
	int top(){
		if (!size()) return -1;
		while (!b.empty() && a.top() == b.top()) a.pop(),b.pop();
		return a.top();
	}
	int sum(){
		if (size() < 2) return -1;
		int x = top(); a.pop();
		int y = top(); ins(x);
		return x + y;
	}
}C[maxn],B[maxn],A;
int bin[50],Log[maxm],mn[maxm][19];
int n,fa[maxn],dep[maxn],dfn[maxn],cnt;
int h[maxn],ne = 1;
struct EDGE{int to,nxt;}ed[maxn << 1];
void build(int u,int v){
	ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;
	ed[++ne] = (EDGE){u,h[v]}; h[v] = ne;
}
void dfs(int u){
	mn[++cnt][0] = dep[u]; dfn[u] = cnt;
	Redge(u) if ((to = ed[k].to) != fa[u]){
		fa[to] = u; dep[to] = dep[u] + 1;
		dfs(to);
		mn[++cnt][0] = dep[u];
	}
}
int dis(int u,int v){
	int l = dfn[u],r = dfn[v];
	if (l > r) swap(l,r);
	int t = Log[r - l + 1];
	int d = min(mn[l][t],mn[r - bin[t] + 1][t]);
	return dep[u] + dep[v] - (d << 1);
}
int F[maxn],Siz[maxn],vis[maxn],Fa[maxn],sum,rt;
int pre[maxn];
void getrt(int u){
	F[u] = 0; Siz[u] = 1;
	Redge(u) if (!vis[to = ed[k].to] && to != Fa[u]){
		Fa[to] = u; getrt(to);
		Siz[u] += Siz[to];
		F[u] = max(F[u],Siz[to]);
	}
	F[u] = max(F[u],sum - Siz[u]);
	if (F[u] < F[rt]) rt = u;
}
int c[maxn],ci;
void dfs1(int u){
	Siz[u] = 1; c[++ci] = u;
	Redge(u) if (!vis[to = ed[k].to] && to != Fa[u]){
		Fa[to] = u; dfs1(to);
		Siz[u] += Siz[to];
	}
}
void solve(int u){
	vis[u] = true; Siz[u] = 1; ci = 0;
	Redge(u) if (!vis[to = ed[k].to]){
		Fa[to] = u; dfs1(to);
		Siz[u] += Siz[to];
	}
	if (pre[u]){
		C[u].ins(dis(u,pre[u]));
		REP(i,ci) C[u].ins(dis(c[i],pre[u]));
	}
	Redge(u) if (!vis[to = ed[k].to]){
		F[rt = 0] = INF; sum = Siz[to];
		getrt(to);
		pre[rt] = u; to = rt;
		solve(rt);
		B[u].ins(C[to].top());
	}
	B[u].ins(0);
	A.ins(B[u].sum());
}
int light[maxn];
void Insert(int x){
	int v;
	A.del(B[x].sum());
	B[x].ins(0);
	A.ins(B[x].sum());
	for (int u = x; pre[u]; u = pre[u]){
		v = pre[u];
		A.del(B[v].sum());
		B[v].del(C[u].top());
		C[u].ins(dis(x,v));
		B[v].ins(C[u].top());
		A.ins(B[v].sum());
	}
}
void Delete(int x){
	int v;
	A.del(B[x].sum());
	B[x].del(0);
	A.ins(B[x].sum());
	for (int u = x; pre[u]; u = pre[u]){
		v = pre[u];
		A.del(B[v].sum());
		B[v].del(C[u].top());
		C[u].del(dis(x,v));
		B[v].ins(C[u].top());
		A.ins(B[v].sum());
	}
}
int main(){
	bin[0] = 1; for (int i = 1; i <= 25; i++) bin[i] = bin[i - 1] << 1;
	Log[0] = -1; for (int i = 1; i < maxm; i++) Log[i] = Log[i >> 1] + 1;
	n = read();
	for (int i = 1; i < n; i++)
		build(read(),read());
	dfs(1);
	for (int j = 1; j <= 18; j++)
		for (int i = 1; i <= cnt; i++){
			if (i + bin[j] - 1 > cnt) break;
			mn[i][j] = min(mn[i][j - 1],mn[i + bin[j - 1]][j - 1]);
		}
	F[rt = 0] = INF; sum = n;
	getrt(1);
	solve(rt);
	int m = read(),v; char opt;
	while (m--){
		opt = getchar();
		while (opt != 'G' && opt != 'C') opt = getchar();
		if (opt == 'G'){
			v = A.top();
			printf("%d\n",v >= 0 ? v : -1);
		}
		else {
			v = read();
			light[v] ^= 1;
			if (!light[v]) Insert(v);
			else Delete(v);
		}
	}
	return 0;
}

posted @ 2018-05-18 16:13  Mychael  阅读(78)  评论(0编辑  收藏