[LNOI2014]LCA(树剖+线段树)

(\%\%\% Fading) 此题是他第一道黑题(我的第一道黑题是蒲公英)

一直不敢开,后来发现是差分一下,将询问离线,树剖+线段树维护即可

(Code Below:)

#include <bits/stdc++.h>
#define pii pair<int,int>
#define mp make_pair
#define F first
#define S second
#define lson (rt<<1)
#define rson (rt<<1|1)
using namespace std;
const int maxn=50000+10;
const int p=201314;
int n,m,ans[maxn],sum[maxn<<2],lazy[maxn<<2];
int head[maxn],to[maxn<<1],nxt[maxn<<1],tot;
int top[maxn],dep[maxn],siz[maxn],son[maxn],fa[maxn],id[maxn],tim;
vector<pii> in[maxn],out[maxn];

inline int read(){
	register int x=0,f=1;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return (f==1)?x:-x;
}

inline void pushup(int rt){
	sum[rt]=sum[lson]+sum[rson];
}

inline void pushdown(int rt,int len){
	if(lazy[rt]){
		sum[lson]=(sum[lson]+lazy[rt]*(len-(len>>1)))%p;
		sum[rson]=(sum[rson]+lazy[rt]*(len>>1))%p;
		lazy[lson]=(lazy[lson]+lazy[rt])%p;
		lazy[rson]=(lazy[rson]+lazy[rt])%p;
		lazy[rt]=0;
	}
}

void update(int L,int R,int C,int l,int r,int rt){
	if(L <= l && r <= R){
		sum[rt]=(sum[rt]+(r-l+1)*C)%p;
		lazy[rt]=(lazy[rt]+C)%p;
		return ;
	}
	pushdown(rt,r-l+1);
	int mid=(l+r)>>1;
	if(L <= mid) update(L,R,C,l,mid,lson);
	if(R > mid) update(L,R,C,mid+1,r,rson);
	pushup(rt);
}

int query(int L,int R,int l,int r,int rt){
	if(L <= l && r <= R){
		return sum[rt];
	}
	pushdown(rt,r-l+1);
	int mid=(l+r)>>1,ans=0;
	if(L <= mid) ans=(ans+query(L,R,l,mid,lson))%p;
	if(R > mid) ans=(ans+query(L,R,mid+1,r,rson))%p;
	return ans;
}

inline void addedge(int x,int y){
	to[++tot]=y;
	nxt[tot]=head[x];
	head[x]=tot;
}

void dfs1(int x,int f){
	siz[x]=1;fa[x]=f;
	dep[x]=dep[f]+1;
	int maxson=-1;
	for(int i=head[x],y;i;i=nxt[i]){
		y=to[i];
		if(y==f) continue;
		dfs1(y,x);
		siz[x]+=siz[y];
		if(siz[y]>maxson){
			maxson=siz[y];
			son[x]=y;
		}
	}
}

void dfs2(int x,int topf){
	id[x]=++tim;
	top[x]=topf;
	if(son[x]) dfs2(son[x],topf);
	for(int i=head[x],y;i;i=nxt[i]){
		y=to[i];
		if(y==fa[x]||y==son[x]) continue;
		dfs2(y,y);
	}
}

void modify(int x){
	while(top[x]!=1){
		update(id[top[x]],id[x],1,1,n,1);
		x=fa[top[x]];
	}
	update(1,id[x],1,1,n,1);
}

int ask(int x){
	int ans=0;
	while(top[x]!=1){
		ans=(ans+query(id[top[x]],id[x],1,n,1))%p;
		x=fa[top[x]];	
	}
	ans=(ans+query(1,id[x],1,n,1))%p;
	return ans;
}

int main()
{
	n=read(),m=read();
	int l,r,x;
	for(int i=2;i<=n;i++){
		x=read()+1;addedge(x,i);
	}
	dfs1(1,0);dfs2(1,1);
	for(int i=1;i<=m;i++){
		l=read()+1,r=read()+1,x=read()+1;
		if(l>1) in[l-1].push_back(mp(x,i));
		out[r].push_back(mp(x,i));
	}
	for(int i=1;i<=n;i++){
		modify(i);
		for(int j=0;j<in[i].size();j++) ans[in[i][j].S]=(p-ask(in[i][j].F))%p;
		for(int j=0;j<out[i].size();j++) ans[out[i][j].S]=(ans[out[i][j].S]+ask(out[i][j].F))%p;
	}
	for(int i=1;i<=m;i++) printf("%d
",ans[i]);
	return 0;
}
原文地址:https://www.cnblogs.com/owencodeisking/p/10225960.html