P4211 [LNOI2014]LCA

如果出现某些询问一堆东西的和的时候（如LCA的深度和），我们可以考虑不要把这些东西每一个都全部求出来，而应该考虑合并这些询问然后一次性询问他们的和。（例如用树剖转化为区间加和和区间询问）

具体思路见hzw博客

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
inline int r()
{
	int s=0,k=1;char c=getchar();
	while(!isdigit(c))
	{
		if(c=='-')k=-1;
		c=getchar();
	}
	while(isdigit(c))
	{
		s=s*10+c-'0';
		c=getchar();
	}
	return s*k;
}
int deep[1000001],top[1000001],fa[1000001],size[1000001],head[1000001],son[1000001],cnt,w[1000001],tr[1000001],la[1000001],id[1000001],ccnt;
map<pair<int,int>,int>mp;
struct ask
{
	int l,r,z;
}t[1000001];
struct asks
{
	int bh,x;
}f[1000001];
int n,m,times,root;
const int mod=201314;
struct node
{
	int to,next;
}a[1000001];
void add_edge(int from,int to)
{
	a[++cnt].to=to;
	a[cnt].next=head[from];
	head[from]=cnt;
}
void dfs1(int u,int father,int dp)
{
	fa[u]=father;
	deep[u]=dp;
	size[u]=1;
	int maxi=0;
	for(int i=head[u];i;i=a[i].next)
	{
		int v=a[i].to;
		if(v==father)continue;
		dfs1(v,u,dp+1);
		if(size[v]>maxi)
		{
			maxi=size[v];
			son[u]=v;
		}
		size[u]+=size[v];
	}
}
void dfs2(int u,int tp)
{
	id[u]=++times;
	top[u]=tp;
	if(son[u])dfs2(son[u],tp);
	for(int i=head[u];i;i=a[i].next)
	{
		int v=a[i].to;
		if(!top[v])dfs2(v,v);
	}
}
void push_down(int l,int r,int k,int bh)
{
	tr[bh]+=k*(r-l+1);
	tr[bh]%=mod;
	la[bh]+=k;
	la[bh]%=mod;
}
void add(int now_l,int now_r,int l,int r,int k,int bh)
{
	if(l<=now_l&&now_r<=r)
	{
		tr[bh]+=k*(now_r-now_l+1);
		tr[bh]%=mod;
		la[bh]+=k;
		la[bh]%=mod;
		return;
	}
	if(now_l>r||now_r<l)return;
	int mid=(now_l+now_r)/2;
	if(la[bh])
	{
		push_down(now_l,mid,la[bh],bh*2);
		push_down(mid+1,now_r,la[bh],bh*2+1);
		la[bh]=0;
	}
	add(now_l,mid,l,r,k,bh*2);
	add(mid+1,now_r,l,r,k,bh*2+1);
	tr[bh]=tr[bh*2]+tr[bh*2+1];
}
int sum(int now_l,int now_r,int l,int r,int bh)
{
	if(l<=now_l&&now_r<=r)return tr[bh];
	if(now_l>r||now_r<l)return 0;
	int mid=(now_l+now_r)/2;
	if(la[bh])
	{
		push_down(now_l,mid,la[bh],bh*2);
		push_down(mid+1,now_r,la[bh],bh*2+1);
		la[bh]=0;
	}
	return (sum(now_l,mid,l,r,bh*2)+sum(mid+1,now_r,l,r,bh*2+1))%mod;
}
void tr_add(int x,int y,int k)
{
	while(top[x]!=top[y])
	{
		if(deep[top[x]]<deep[top[y]])swap(x,y);
		add(1,n,id[top[x]],id[x],k,1);
		x=fa[top[x]];
	}
	if(deep[x]<deep[y])swap(x,y);
	add(1,n,id[y],id[x],k,1);
}
int tr_sum(int x,int y)
{
	int ans=0;
	while(top[x]!=top[y])
	{
		if(deep[top[x]]<deep[top[y]])swap(x,y);
		ans+=sum(1,n,id[top[x]],id[x],1);
		ans%=mod;
		x=fa[top[x]];
	}
	if(deep[x]<deep[y])swap(x,y);
	ans+=sum(1,n,id[y],id[x],1);
	ans%=mod;
	return ans;
}
bool cmp(asks x,asks y)
{
	return x.bh<y.bh;
}
int main()
{
	n=r();m=r();
	int x,y,z;
	for(int i=2;i<=n;i++)
	{
		x=r()+1;
		add_edge(i,x);
		add_edge(x,i);
	}
	dfs1(1,0,1);
	dfs2(1,1);
	for(int i=1;i<=m;i++)
	{
		t[i].l=r()+1;t[i].r=r()+1;t[i].z=r()+1;
		f[++ccnt].bh=t[i].l-1;
		f[ccnt].x=t[i].z;
		f[++ccnt].bh=t[i].r;
		f[ccnt].x=t[i].z;
	}
	sort(f+1,f+ccnt+1,cmp);
	int lst=1;
	for(int i=1;i<=ccnt;i++)
	{
		int x=f[i].bh,y=f[i].x;
		for(int j=lst;j<=x;j++)
		{
			tr_add(1,j,1);
//			cout<<"add"<<1<<" "<<j<<endl;
		}
		int sm=tr_sum(1,y);
//		cout<<"ask"<<1<<" "<<y<<" "<<sm<<endl;
		mp[make_pair(x,y)]=sm;
		lst=x+1;
	}
	for(int i=1;i<=m;i++)
	{
		int ans=mp[make_pair(t[i].r,t[i].z)]-mp[make_pair(t[i].l-1,t[i].z)];
		cout<<(ans%mod+mod)%mod<<endl;
	}
}

标签：int,void,LNOI2014,bh,1000001,P4211,LCA,include,size
来源： https://www.cnblogs.com/lytql/p/15224563.html