#58. 【WC2013】糖果公园

 

给出一棵n个节点的树，每个节点都有一种颜色c[i]。

q次询问[l,r]路径上的点的权值和vjwi。每一个节点的权值就是当前节点的颜色值v[c[i]]和从l到该节点经过该颜色的次数num[c[i]]对应的w值w[num[c[i]]之间的乘积（很绕，看题目描述吧，懒得描述了)

 

这是树上莫队。分块是对子树进行。如果一个某一个子树的size满足块大小，则分成一块。

void dfs1(int t, int f){
	sz[t]=1;
	fa[t] = f;
	dp[t] = dp[f]+1;
	dfn[t]=++clk;
	st.push(t);
	int cur_p = st.size();
	for(int v : g[t]){
		if(v==f)continue;
		dfs1(v,t);
		if(sz[v] > sz[hs[t]]) hs[t] = v;
		sz[t]+=sz[v];
		if(st.size() - cur_p >= block_size){
			//get one block
			while(st.size() > cur_p){
				block_num[st.top()] = cur_block;
				st.pop();
			}
			cur_block++;
		}
	}
	if(t==1){
		while(!st.empty()){
			block_num[st.top()] = cur_block;
			st.pop();
		}
		cur_block++;
	}
}

最后就是正常的莫队操作了。

 

树上莫队的难点在于[l,r]的边界移动。

假设从[l,r]移动到[l',r'], 将l移动到l'或者r移动到r'时，对树上的标记取反（异或）来判断是否加入或者删除。

但是这里会有重复操作，见https://www.luogu.com.cn/blog/dedicatus545/solution-p4074

因此需要当l走到l'的时候，lca(l,l')是不操作的。r到r'同理。

void move(int old_x, int new_x){
	if(dp[old_x] < dp[new_x]) swap(old_x,new_x);
	while(dp[old_x]!=dp[new_x]){
		change(old_x);
		old_x = fa[old_x];
	}
	while(old_x != new_x){
		change(old_x);
		change(new_x);
		old_x = fa[old_x];
		new_x = fa[new_x];
	}
}

最后将(l,r)的lca和[l',r']的lca取反保证他们只操作一次。

for(int i = 0; i < q.size();++i){
	    if(dfn[q[i].l]>dfn[q[i].r]) swap(q[i].l,q[i].r);
		int f = lca(l,r);
		change(f);
		while(cur<q[i].t){
			cur++;
			int pos = modi[cur].pos;
			int old_c = modi[cur].old_c;
			int new_c = modi[cur].c;
			change(pos, old_c, new_c);
		}
		while(cur>q[i].t){
			int pos = modi[cur].pos;
			int old_c = modi[cur].old_c;
			int new_c = modi[cur].c;
			change(pos, new_c,old_c);
			cur--;
		}

		if(l!=q[i].l) {
			move(l,q[i].l);
			l = q[i].l;
		}
		if(r!=q[i].r){
			move(r,q[i].r);
			r = q[i].r;
		}

		f = lca(l,r);
		change(f);
		ans[q[i].id] = sum;
	}

另外(l,r)需要保证他们的dfs序。

完整代码

int block_size;

struct query{
	int l,r,id;
	int block;
	int r_block;
	int t;
	bool operator<(const query &q) const{
		if(block != q.block) return block<q.block;
		if(r_block == q.r_block) return id < q.id;
		return r_block<q.r_block;
	}
}q[N];

struct Revise{ 
	int pos, c, old_c; 
};

int v[N],w[N];
int c[N], num[N],old[N];
ll ans[N];
ll sum;
stack<int> st;
int cur_block,clk;
int hs[N],sz[N], block_num[N];
int fa[N],top[N],dp[N],vis[N],dfn[N];
vector<int> g[N];

void dfs1(int t, int f){
	sz[t]=1;
	fa[t] = f;
	dp[t] = dp[f]+1;
	dfn[t]=++clk;
	st.push(t);
	int cur_p = st.size();
	for(int v : g[t]){
		if(v==f)continue;
		dfs1(v,t);
		if(sz[v] > sz[hs[t]]) hs[t] = v;
		sz[t]+=sz[v];
		if(st.size() - cur_p >= block_size){
			//get one block
			while(st.size() > cur_p){
				block_num[st.top()] = cur_block;
				st.pop();
			}
			cur_block++;
		}
	}
	if(t==1){
		while(!st.empty()){
			block_num[st.top()] = cur_block;
			st.pop();
		}
		cur_block++;
	}
}

void dfs2(int t, int f){
	top[t] = f;
	if(!hs[t]) return;
	dfs2(hs[t],f);
	for(int v : g[t]){
		if(v==fa[t] || v == hs[t])continue;
		dfs2(v,v);
	}
}

int lca(int x, int y){
	if(dp[x] < dp[y]) swap(x,y);
	while(top[x] != top[y]){
		if(dp[top[x]] >dp[top[y]]) x = fa[top[x]];
		else y = fa[top[y]];
	}
	return dp[x] < dp[y]? x:y;
}

void del(int idx){
	sum -= 1ll*w[num[idx]] * v[idx];
	num[idx]--;
}

void add(int idx){
	num[idx]++;
	sum += 1ll*w[num[idx]] * v[idx];
}

void change(int pos, int old_c, int new_c){
	if(vis[pos]){
		del(old_c);
		add(new_c);
	}
	c[pos] = new_c;
}

void change(int idx){
	if(vis[idx]){
		del(c[idx]);
	}
	else {
		add(c[idx]);
	}
	vis[idx]^=1;
}

void move(int old_x, int new_x){
	if(dp[old_x] < dp[new_x]) swap(old_x,new_x);
	while(dp[old_x]!=dp[new_x]){
		change(old_x);
		old_x = fa[old_x];
	}
	while(old_x != new_x){
		change(old_x);
		change(new_x);
		old_x = fa[old_x];
		new_x = fa[new_x];
	}
}

int main(){
	int n,m,Q;
	scanf("%d%d%d",&n,&m,&Q);
	block_size = pow(n,2.0/3);
	clk = 0;
	fr(i,1,m+1) {
		sf("%d",&v[i]);
	}
	fr(i,1,n+1) {
		sf("%d",&w[i]);
	}
	fr(i,0,n-1){
		int u,v;
		sf("%d%d",&u,&v);
		g[u].pb(v);
		g[v].pb(u);
	}
	fr(i,1,n+1){
		sf("%d",&c[i]);
		old[i] = c[i];
	}
	dfs1(1,1);
	dfs2(1,1);
	vector<query> q;
	vector<Revise> modi;
	fr(i,0,Q){
		int t,x,y;
		sf("%d%d%d",&t,&x,&y);
		if(t==1){
			query qy;
			qy.l = x;qy.r = y;
			qy.id = q.size();
			qy.block = block_num[qy.l];
			qy.r_block = block_num[qy.r];
			qy.t = modi.size()-1;
			q.pb(qy);
		}
		else {
			Revise r;
			r.pos = x;r.c = y;
			r.old_c = old[r.pos];
			modi.pb(r);
			old[r.pos] = r.c;
		}
	}

    sort(q.begin(),q.end());
	sum = 0;
	int l = 1, r = 1;
	change(l);
	int cur = -1;
	for(int i = 0; i < q.size();++i){
	    if(dfn[q[i].l]>dfn[q[i].r]) swap(q[i].l,q[i].r);
		int f = lca(l,r);
		change(f);
		while(cur<q[i].t){
			cur++;
			int pos = modi[cur].pos;
			int old_c = modi[cur].old_c;
			int new_c = modi[cur].c;
			change(pos, old_c, new_c);
		}
		while(cur>q[i].t){
			int pos = modi[cur].pos;
			int old_c = modi[cur].old_c;
			int new_c = modi[cur].c;
			change(pos, new_c,old_c);
			cur--;
		}

		if(l!=q[i].l) {
			move(l,q[i].l);
			l = q[i].l;
		}
		if(r!=q[i].r){
			move(r,q[i].r);
			r = q[i].r;
		}

		f = lca(l,r);
		change(f);
		ans[q[i].id] = sum;
	}
	
	fr(i,0,q.size()){
		printf("%lld\n",ans[i]);
	}
}