BZOJ 4448 主席树+树链剖分(在线)

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为什么题解都是离线的…… (抄都没法抄
搞一棵主席树
1 操作 新树上的当前节点设成1
2 操作 查max(i-xx-1,0)那棵树上这条路径上有多少个点是1
让你找经过了多少个点 查的时候用deep 搞出来就好了

//By SiriusRen#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define N 444444#define mem(x,y) memset(x,y,sizeof(x))int n,q,op,xx,yy,zz,first[N],v[N],next[N],tot,root,rt[N],ans;int size[N],fa[N],son[N],deep[N],top[N],change[N],back[N],cnt;struct Tree{int l,r,num;}tree[N*30];void Add(int x,int y){v[tot]=y,next[tot]=first[x],first[x]=tot++;}void add(int x,int y){Add(x,y),Add(y,x);}void dfs(int x){    size[x]=1;    for(int i=first[x];~i;i=next[i])if(v[i]!=fa[x]){        deep[v[i]]=deep[x]+1,fa[v[i]]=x,dfs(v[i]);        size[x]+=size[v[i]];        if(size[son[x]]<size[v[i]])son[x]=v[i];    }}void dfs2(int x,int tp){    change[x]=++cnt,back[cnt]=x;top[x]=tp;    if(son[x])dfs2(son[x],tp);    for(int i=first[x];~i;i=next[i])        if(v[i]!=fa[x]&&v[i]!=son[x])dfs2(v[i],v[i]);}void push_up(int pos){tree[pos].num=tree[tree[pos].l].num+tree[tree[pos].r].num;}int build(int l,int r){    int pos=++cnt;    if(l==r){return pos;}    int mid=(l+r)>>1;    tree[pos].l=build(l,mid),tree[pos].r=build(mid+1,r);    return pos;}int insert(int o,int l,int r,int num){    int pos=++cnt;tree[pos]=tree[o];    if(l==r){tree[pos].num=1;return pos;}    int mid=(l+r)>>1;    if(mid>=num)tree[pos].l=insert(tree[o].l,l,mid,num);    else tree[pos].r=insert(tree[o].r,mid+1,r,num);    push_up(pos);return pos;}int query(int o,int l,int r,int L,int R){    if(l>=L&&r<=R)return tree[o].num;    int mid=(l+r)>>1;     if(mid<L)return query(tree[o].r,mid+1,r,L,R);    else if(mid>=R)return query(tree[o].l,l,mid,L,R);    else return query(tree[o].l,l,mid,L,R)+query(tree[o].r,mid+1,r,L,R);}int find(int x,int y,int wei){    int fx=top[x],fy=top[y],temp=0;    while(fx!=fy){        if(deep[fx]<deep[fy])swap(x,y),swap(fx,fy);        temp+=query(rt[wei],1,n,change[fx],change[x]),ans+=deep[x]-deep[fx]+1;        x=fa[fx],fx=top[x];    }    if(deep[x]<deep[y])swap(x,y);    ans+=deep[x]-deep[y]+1;    return temp+query(rt[wei],1,n,change[y],change[x]);}int main(){    mem(first,-1);scanf("%d",&n);    for(int i=1;i<=n;i++){        scanf("%d",&fa[i]);        if(!fa[i])root=i;else add(i,fa[i]);    }    dfs(root),dfs2(root,root),cnt=0,rt[0]=build(1,n);    scanf("%d",&q);    for(int i=1;i<=q;i++){        scanf("%d",&op);        if(op==1){            rt[i]=rt[i-1],ans=0;            scanf("%d%d%d",&xx,&yy,&zz);            zz=max(i-zz-1,0);            printf("%d %d\n",ans,find(xx,yy,zz));        }        else scanf("%d",&xx),rt[i]=insert(rt[i-1],1,n,change[xx]);    }}

这里写图片描述

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