题解 P4074 【[WC2013]糖果公园】

upd:2020.10.16 修复了一些错别字

题目链接

Solution [WC2013]糖果公园

题目大意:多次询问树上路径权值(计算方法题目已给出),带修改可离线

分析:这题可谓集各种莫队算法于一身(反正我会的莫队全在里面了)

首先,观察计算路径权值

(sum_cval[c] sum_{i =1}^{cnt[c]}w[c])

(val)表示美味指数,(cnt)表示出现次数,(w)表示新鲜指数

显然无论增加还是减少点,对答案的影响都是可以(O(1))求的

然后树上莫队就好了

既然是莫队,本质上还是一个暴力,朴素莫队对序列分块,这里对树进行分块

[SCOI2005]王室联邦这题就是给树分块的板子题

然后套用莫队思想,只不过和序列上的莫队有点区别

朴素莫队可以直接在序列上跳(l),(r)指针,这里跳的方式有点不同

设当前的记录(u,v)这条路径上的答案,我们要求(tu,tv)路径上的答案,假如直接消除(u,v)再加入(tu,tv)就和暴力没有任何区别

我们可以翻转(u,tu)路径和(v,tv)路径,也就是把出现过的删掉,没出现过的加进去((LCA)这玩意儿比较烦,单独处理),具体证明可以看这位巨佬的(blog)

带修莫队记个时间戳就好,块大小(n^{frac{2}{3}})

#include <cstdio>
#include <cctype>
#include <algorithm>
#include <vector>
#include <stack>
#include <cmath>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 100,maxdep = 25;
inline int read(){
	int x = 0;char c = getchar();
	while(!isdigit(c))c = getchar();
	while(isdigit(c))x = x * 10 + c - '0',c = getchar();
	return x;
}
vector<int> G[maxn];
stack<int> stk;
inline void addedge(int from,int to){G[from].push_back(to);};
int dep[maxn],faz[maxn][maxdep + 1],col[maxn],v[maxn],w[maxn],vis[maxn],cnt[maxn],val[maxn],n,m,q,siz,col_tot,ques_tot,nowu,nowv,nowtim;
ll ans[maxn],tmpans;
inline void dfs(int u){
	int tmp = stk.size();
	dep[1] = 1;
	for(int i = 1;i <= maxdep;i++)
		faz[u][i] = faz[faz[u][i - 1]][i - 1];
	for(int v : G[u]){
		if(v == faz[u][0])continue;
		dep[v] = dep[u] + 1;
		faz[v][0] = u;
		dfs(v);
		if(stk.size() - tmp >= siz){
			col_tot++;
			while(stk.size() != tmp)col[stk.top()] = col_tot,stk.pop();
		}
	}
	stk.push(u);
}
inline int lca(int x,int y){
	if(dep[x] < dep[y])swap(x,y);
	for(int i = maxdep;i >= 0;i--)
		if(dep[faz[x][i]] >= dep[y])x = faz[x][i];
	if(x == y)return x;
	for(int i = maxdep;i >= 0;i--)
		if(faz[x][i] != faz[y][i])
			x = faz[x][i],y = faz[y][i];
	return faz[x][0];
}
inline void rev(int u){//翻转点
	if(vis[u]) tmpans -= (ll)v[val[u]] * w[cnt[val[u]]--];
	else tmpans += (ll)v[val[u]] * w[++cnt[val[u]]];
	vis[u] ^= 1;
}
inline void modify(int x,int y){//翻转路径
	if(dep[x] < dep[y])swap(x,y);
	while(dep[faz[x][0]] >= dep[y])
		rev(x),x = faz[x][0];
	while(x != y)
		rev(x),rev(y),x = faz[x][0],y = faz[y][0];
}
struct Modi{
	int pos,from,to;
}modi[maxn];
inline void add_modi(int tim){
	Modi tmp = modi[tim];bool flag = false;
	if(vis[tmp.pos])flag = true,rev(tmp.pos);
	val[tmp.pos] = tmp.to;
	if(flag)rev(tmp.pos);
}
inline void del_modi(int tim){
	Modi tmp = modi[tim];bool flag = false;
	if(vis[tmp.pos])flag = true,rev(tmp.pos);
	val[tmp.pos] = tmp.from;
	if(flag)rev(tmp.pos);
}
struct Ques{
	int u,v,tim,id;
	bool operator < (const Ques &rhs)const{
		if(col[u] != col[rhs.u])return col[u] < col[rhs.u];
		if(col[v] != col[rhs.v])return col[v] < col[rhs.v];
		return tim < rhs.tim;
	}
}ques[maxn];
int main(){
	n = read(),m = read(),q = read();
	siz = pow(n,0.66);
	for(int i = 1;i <= m;i++)v[i] = read();
	for(int i = 1;i <= n;i++)w[i] = read();
	for(int u,v,i = 1;i < n;i++)
		u = read(),v = read(),addedge(u,v),addedge(v,u);
	dfs(1);
	for(int i = 1;i <= n;i++)
		val[i] = read();
	for(int opt,x,y,i = 1;i <= q;i++){
		opt = read(),x = read(),y = read();
		if(!opt)modi[++nowtim] = Modi{x,val[x],y},val[x] = y;
		else ques_tot++,ques[ques_tot] = Ques{x,y,nowtim,ques_tot};
	}
	sort(ques + 1,ques + 1 + ques_tot);
	nowu = nowv = 1;
	for(int i = 1;i <= ques_tot;i++){
		while(nowtim < ques[i].tim)add_modi(++nowtim);
		while(nowtim > ques[i].tim)del_modi(nowtim--);
		modify(nowu,ques[i].u);
		modify(nowv,ques[i].v);
		nowu = ques[i].u,nowv = ques[i].v;
		int lc = lca(ques[i].u,ques[i].v);//LCA单独处理
		rev(lc);
		ans[ques[i].id] = tmpans;
		rev(lc);
	}
	for(int i = 1;i <= ques_tot;i++)printf("%lld
",ans[i]);
	return 0;
}