BZOJ4196 NOI2015 软件包管理器 树链剖分

题意：给定一颗树，维护：1、给定一个节点，求该节点到根的路径上总点数-点权和，并将路径上的所有点的权值赋为1  2、给定一个节点，求该节点子树的点权和，并将子树中所有点的权值赋为0

题解：链剖裸题

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <iostream>
#include <algorithm>
using namespace std;

const int MAXN=100000+2;
struct HASH{
    int u;
    HASH *next;
    HASH(){}
    HASH(int _u,HASH *_next):u(_u),next(_next){}
}mem[MAXN*2];
struct NODE{
    int f,b,c,s,m,l,r;
    HASH *child;
}node[MAXN];
typedef struct TREE{
    int l,r,s,a;
    TREE *lchild,*rchild;
    TREE(){}
    TREE(int _l,int _r):l(_l),r(_r),s(0),a(-1),lchild(0),rchild(0){}
} *ROOT;
ROOT root;
int N,Q,cnt;
char S[20+2];

void Insert(int u,int v){ node[u].child=&(mem[cnt++]=HASH(v,node[u].child));}

void DFS1(int f,int x){
    node[x].c=1,node[x].s=-1;
    for(HASH *p=node[x].child;p;p=p->next)
        if(p->u!=f){
            DFS1(x,p->u);
            node[x].c+=node[p->u].c;
            if(node[x].s==-1 || node[p->u].c>node[node[x].s].c) node[x].s=p->u;
        }
}

void DFS2(int f,int x,int b){
    node[x].f=f,node[x].b=b,node[x].m=node[x].l=node[x].r=++cnt;
    if(node[x].s!=-1) DFS2(x,node[x].s,b);
    for(HASH *p=node[x].child;p;p=p->next){
        if(p->u!=f && p->u!=node[x].s) DFS2(x,p->u,p->u);
        node[x].r=max(node[x].r,node[p->u].r);
    }
}

void Build(ROOT &x,int l,int r){
    x=new TREE(l,r);
    if(l==r) return;

    int m=(l+r)>>1;
    Build(x->lchild,l,m),Build(x->rchild,m+1,r);
}

void Pushup(ROOT &x){ x->s=x->lchild->s+x->rchild->s;}

void Pushdown(ROOT &x,int m){
    if(x->a==1){
        x->lchild->a=x->rchild->a=1;
        x->lchild->s=m-(m>>1);
        x->rchild->s=m>>1;
        x->a=-1;
    }
    else if(!x->a){
        x->lchild->a=x->rchild->a=0;
        x->lchild->s=x->rchild->s=0;
        x->a=-1;
    }
}

int Summation(ROOT &x,int l,int r,bool a){
    if(l<=x->l && x->r<=r){
        if(a) return x->r-x->l+1-x->s;
        else return x->s;
    }

    Pushdown(x,x->r-x->l+1);

    int m=(x->l+x->r)>>1,ret=0;
    if(l<=m) ret+=Summation(x->lchild,l,r,a);
    if(r>m) ret+=Summation(x->rchild,l,r,a);

    return ret;
}

void Update(ROOT &x,int l,int r,bool a){
    if(l<=x->l && x->r<=r){
        if(a) x->a=1,x->s=x->r-x->l+1;
        else x->a=0,x->s=0;
        return;
    }

    Pushdown(x,x->r-x->l+1);

    int m=(x->l+x->r)>>1;
    if(l<=m) Update(x->lchild,l,r,a);
    if(m<r) Update(x->rchild,l,r,a);

    Pushup(x);
}

int Query1(int x){
    int ret=0;
    for(int i=x;i!=-1;i=node[node[i].b].f)
        ret+=Summation(root,node[node[i].b].m,node[i].m,1);
    for(int i=x;i!=-1;i=node[node[i].b].f)
        Update(root,node[node[i].b].m,node[i].m,1);
    return ret;
}

int Query2(int x){
    int ret=Summation(root,node[x].l,node[x].r,0);
    Update(root,node[x].l,node[x].r,0);
    return ret;
}

int main(){
    scanf("%d",&N);
    for(int i=1,u;i<N;i++){
        scanf("%d",&u);
        Insert(i,u),Insert(u,i);
    }

    DFS1(-1,0),cnt=0,DFS2(-1,0,0);
    Build(root,1,N);

    scanf("%d",&Q);
    for(int i=1,x;i<=Q;i++){
        scanf("%s %d",S,&x);
        if(S[0]=='i') printf("%d\n",Query1(x));
        else printf("%d\n",Query2(x));
    }

    return 0;
}