BZOJ3159: 决战

方法很简单,树剖,把区间提取出来,打翻转标记,再放回去。本来以为写起来也很简单T_T

注:由于某种原因,我写的是把题目中的r忽略掉的一般情况,否则写起来简单得多。

#include<bits/stdc++.h>
#define N 50005
#define M (s+t>>1)
#define L(t)(t)->c[0]
#define R(t)(t)->c[1]
#define Z(t)(L(t)->s+1)
using namespace std;
typedef long long ll;
struct edge{
	int v;
	edge*s;
}e[N*2];
edge*back=e,*h[N];
void add(int u,int v){
	*back={v,h[u]};
	h[u]=back++;
	*back={u,h[v]};
	h[v]=back++;
}
typedef int arr[N];
arr d,p,num,size,son,top;
void dfs1(int u){
	d[u]=d[p[u]]+1;
	size[u]=1;
	int s=0;
	for(edge*i=h[u];i;i=i->s)
		if(i->v!=p[u]){
			p[i->v]=u;
			dfs1(i->v);
			size[u]+=size[i->v];
			if(s<size[i->v])
				s=size[son[u]=i->v];
		}
}
void dfs2(int u){
	static int tot;
	num[u]=++tot;
	if(size[u]!=1){
		top[son[u]]=top[u];
		dfs2(son[u]);
	}
	for(edge*i=h[u];i;i=i->s)
		if(i->v!=p[u]&&i->v!=son[u])
			dfs2(top[i->v]=i->v);
}
struct node{
	ll a[3];
	int s,u,v;
	bool rev;
	node*c[2];
}mem[N];
node*last=mem,null[1]={0,1<<30},*root;
ll Sum(ll u,ll v){
	return u+v;
}
ll Min(ll u,ll v){
	return u<v?u:v;
}
ll Max(ll u,ll v){
	return u<v?v:u;
}
ll(*f[3])(ll,ll)={
	Sum,Min,Max
};
int devolve(node*t){
	if(t==null)
		return 0;
	if(t->rev){
		L(t)->rev^=1;
		R(t)->rev^=1;
		t->rev=0;
		swap(L(t),R(t));
	}
	if(int u=t->u){
		t->u=0;
		L(t)->u+=u;
		R(t)->u+=u;
		t->a[1]+=u;
		t->a[2]+=u;
		t->v+=u;
		t->a[0]+=u*t->s;
	}
	return Z(t);
}
node*update(node*t){
	devolve(L(t));
	devolve(R(t));
	for(int i=0;i!=3;++i)
		t->a[i]=f[i](f[i](L(t)->a[i],R(t)->a[i]),t->v);
	t->s=R(t)->s+Z(t);
	return t;
}
void link(bool i,node*&t,node*&s){
	node*d=t->c[i];
	t->c[i]=s;
	s=update(t),t=d;
}
node*splay(int v,node*&t=root){
	node*d[]={
		null,null
	};
	while(v!=devolve(t)){
		bool i=v>Z(t);
		v-=i*Z(t);
		if(v!=devolve(t->c[i])&&i==v>Z(t->c[i])){
			v-=i*Z(t->c[i]);
			link(i,t,t->c[i]->c[i^1]);
		}
		link(i,t,d[i]);
	}
	for(int i=0;i!=2;++i){
		node*s=t->c[i^1];
		while(d[i]!=null)
			link(i,d[i],s);
		t->c[i^1]=s;
	}
	return update(t);
}
node*&splay(int s,int t){
	splay(s);
	return L(splay(t-s+2,R(root)));
}
node*build(int s,int t){
	if(s<=t){
		node*i=last++;
		L(i)=build(s,M-1);
		R(i)=build(M+1,t);
		return update(i);
	}
	return null;
}
struct vec{
	int u,v,a;
	vec(){}
	vec(int u,int v):u(u),v(v),a(u){}
	int size(){
		return v-u+1;
	}
}u[N],v[N];
int solve(int i,int j,vec*&u,vec*&v){
	vec**s=&u,**t=&v;
	for(;top[i]!=top[j];i=p[top[i]]){
		if(d[top[i]]<d[top[j]])
			swap(i,j),swap(s,t);
		*(*s)++=vec(num[top[i]],num[i]);
	}
	if(d[i]<d[j])
		swap(i,j),swap(s,t);
	*(*s)++=vec(num[j],num[i]);
	return num[j];
}
void amend(int p,int q,int j){
	vec*s=u,*t=v;
	solve(p,q,s,t);
	while(s--!=u)
		splay(s->u,s->v)->u+=j;
	while(t--!=v)
		splay(t->u,t->v)->u+=j;
}
ll query(int p,int q,int i){
	ll j=i^1?0:1<<30;
	vec*s=u,*t=v;
	solve(p,q,s,t);
	while(s--!=u)
		j=f[i](j,splay(s->u,s->v)->a[i]);
	while(t--!=v)
		j=f[i](j,splay(t->u,t->v)->a[i]);
	return j;
}
void invert(int p,int q){
	node*j=null;
	vec*s=u,*t=v;
	solve(p,q,s,t);
	vec*e=s;
	while(t--!=v)
		*e++=*t;
	for(vec*i=u;i!=e;++i){
		node*&a=splay(i->u,i->v);
		a->rev=i<s;
		L(splay(1,a))=j;
		update(j=a);
		a=null;
		for(vec*z=u;z!=e;++z)
			if(i->a<z->a){
				z->u-=i->size();
				z->v-=i->size();
			}
	}
	j->rev=1;
	for(vec*i=u;i!=e;++i){
		node*&a=splay(i->u,i->u-1);
		a=j;
		j=R(splay(i->size(),a));
		a->rev=i<s;
		R(a)=null;
		update(a);
		for(vec*z=u;z!=e;++z)
			if(i->a<z->a){
				z->u+=i->size();
				z->v+=i->size();
			}
	}
}
int main(){
	int n,m,s,t,u;
	char a[10];
	scanf("%d%d%*d",&n,&m);
	root=build(0,n+1);
	for(int i=1;i!=n;++i){
		scanf("%d%d",&s,&t);
		add(s,t);
	}
	dfs1(1);
	dfs2(top[1]=1);
	while(m--){
		scanf("%s%d%d",a,&s,&t);
		if(a[2]=='v')
			invert(s,t);
		else if(a[2]=='c'){
			scanf("%d",&u);
			amend(s,t,u);
		}else
			printf("%lld\n",query(s,t,a[2]=='j'?2:a[2]!='m'));
	}
}
posted @ 2016-06-29 02:38  f321dd  阅读(236)  评论(0编辑  收藏  举报