【NOIP2019模拟赛】test

【NOIP2019模拟赛】test

题目描述

分析

对于操作2-4，显然是树链剖分的裸题，重点是操作1

首先，操作1显然只对操作3产生影响，假设当前根为root，操作3的节点为u

如果(u=root)，显然权值之和为整棵树

如果(u eq root)且(lca(u,root) eq u)那么换根操作不对子树权值和产生影响

如果(u eq root)且(lca(u,root)=u)那么权值和为整棵树减去(u)到(root)的链上最靠近(u)的一点的子树权值和

代码

#include<bits/stdc++.h>
using namespace std;

const int N=1e5+5;
struct Node
{
	int l,r,val;
}tree[4*N];
struct node
{
	int ver,poi;
}edge[2*N];
int n,m,a[N],deep[N],fa[N],siz[N],maxson,son[N],id[N],val[N],cnt,top[N],root,head[N],tot;

void build(int p,int L,int R)
{
	tree[p].l=L; tree[p].r=R;
	if( L==R )
	{
		tree[p].val=val[L];
		return;
	}
	int mid=(L+R)/2;
	build(p*2,L,mid);
	build(p*2+1,mid+1,R);
	tree[p].val=tree[2*p].val+tree[2*p+1].val;
	return;
}

void change(int p,int x,int val)
{
	if( tree[p].l==tree[p].r )
	{
		tree[p].val=val;
		return;
	}
	int mid=(tree[p].l+tree[p].r)/2;
	if( x <= mid ) change(p*2,x,val);
	else change(p*2+1,x,val);
	tree[p].val=tree[2*p].val+tree[2*p+1].val;
	return;
}

int ask(int p,int L,int R)
{
	if( tree[p].l >= L && tree[p].r <= R ) return tree[p].val;
	int mid=(tree[p].l+tree[p].r)/2;
	int tem=0;
	if( L <= mid ) tem+=ask(p*2,L,R);
	if( R > mid ) tem+=ask(p*2+1,L,R);
	return tem; 
}

void add(int x,int y)
{
	edge[++tot].ver=y;
	edge[tot].poi=head[x];
	head[x]=tot;
	return;
}

void Input()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	return;
}

void dfs1(int x,int f,int dep)
{
	deep[x]=dep;
	fa[x]=f;
	siz[x]=1;
	maxson=-1;
	for(int i=head[x];i;i=edge[i].poi)
	{
		int y=edge[i].ver;
		if( y==fa[x] ) continue;
		dfs1(y,x,dep+1);
		siz[x]+=siz[y];
		if( siz[y] > maxson ) son[x]=y,maxson=siz[y];
	}
	return;
}

void dfs2(int x,int topf)
{
	id[x]=++cnt;
	val[cnt]=a[x];
	top[x]=topf;
	if( son[x]==0 ) return;
	dfs2(son[x],topf);
	for(int i=head[x];i;i=edge[i].poi)
	{
		int y=edge[i].ver;
		if( y==son[x] || y==fa[x] ) continue;
		dfs2(y,y);
	}
	return;
}

void Prepare()
{
	dfs1(1,0,1);
	dfs2(1,1);
	build(1,1,n);
	return;
}

void deal1()
{
	scanf("%d",&root);
	return;
}

void deal2()
{
	int x,y;
	scanf("%d%d",&x,&y);
	change(1,id[x],y);
	return;
}

int check(int x,int y)
{
	if( x==y ) return 0;
	if( deep[x] >= deep[y] ) return -1;
	while( deep[x] < deep[y] )
	{
		if( fa[y]==x ) return y;
		y=fa[y];
	}
	return -1;
}

void deal3()
{
	int x,tem=0;
	scanf("%d",&x);
	if( check(x,root)==0 ) tem=ask(1,1,n);
	else
		if( check(x,root)==-1 ) tem=ask(1,id[x],id[x]+siz[x]-1);
		else tem=ask(1,1,n)-ask(1,id[check(x,root)],id[check(x,root)]+siz[check(x,root)]-1);
	printf("%d
",tem);
	return; 
}

int chain(int x,int y)
{
	int tem=0;
	while( top[x]!=top[y] )
	{
		if( deep[top[x]] < deep[top[y]] ) swap(x,y);
		tem+=ask(1,id[top[x]],id[x]);
		x=fa[top[x]];
	}
	if( deep[x] < deep[y] ) swap(x,y);
	tem+=ask(1,id[y],id[x]);
	return tem;
}

void deal4()
{
	int x,y,tem;
	scanf("%d%d",&x,&y);
	tem=chain(x,y);
	printf("%d
",tem);
	return;
}

void work()
{
	for(int i=1;i<=m;i++)
	{
		int que;
		scanf("%d",&que);
		if( que==1 ) deal1();
		else if( que==2 ) deal2();
		else if( que==3 ) deal3();
		else deal4();
	}
	return;
}

int main()
{
	Input();
	Prepare();
	work();
	return 0;
}