CF893F Subtree Minimum Query

输入输出格式

输入格式：

The first line contains two integers $n$ and $r$ ( $1<=r<=n<=100000$ ) — the number of vertices in the tree and the index of the root, respectively.

The second line contains n integers $a_{1},a_{2},...,a_{n}$ ( $1<=a_{i}<=10^{9}$ ) — the numbers written on the vertices.

Then $n-1$ lines follow, each containing two integers $x$ and $y$ ( $1<=x,y<=n$ ) and representing an edge between vertices $x$ and $y$ . It is guaranteed that these edges form a tree.

Next line contains one integer $m$ ( $1<=m<=10^{6}$ ) — the number of queries to process.

Then m lines follow, $i$ -th line containing two numbers $p_{i}$ and $q_{i}$ , which can be used to restore $i$ -th query ( $1<=p_{i},q_{i}<=n$ ).

$i$ -th query can be restored as follows:

Let last last be the answer for previous query (or $0$ if $i=1$ ). Then $x_{i}=((p_{i}+last) \bmod n)+1$, and $k_{i}=(q_{i}+last) \bmod n$ .

输出格式：

Print $m$ integers. $i$ -th of them has to be equal to the answer to $i$ -th query.

题意大概就是给你一个有跟有点权的树，边权均为$1$，每次询问一个点子树中距离Ta不超过$k$距离的点的最小点权。

发现$dfs$序限定子树是一个区间，可以放在线段树上，然后深度确定另一个区间，套一颗平衡树，就可以了。

事实上这道题还有一个高妙的做法，我并不会。

说不定以后会看一看呐

Code:

#include <cstdio>

#include <cstdlib>

#include <cstring>

#define ls ch[now][0]

#define rs ch[now][1]

const int N=1e5+10;

int ch[N*30][2],dep[N*30],dat[N*30],mx[N*30],val[N*30],root[N<<2],tot;

int min(int x,int y){return x<y?x:y;}

void updata(int now){mx[now]=min(dat[now],min(mx[ls],mx[rs]));}

void split(int now,int k,int &x,int &y)

{

    if(!now){x=y=0;return;}

    if(dep[now]<=k)

        x=now,split(rs,k,rs,y);

    else

        y=now,split(ls,k,x,ls);

    updata(now);

}

int Merge(int x,int y)

{

    if(!x||!y) return x+y;

    if(val[x]<val[y])

    {

        ch[x][1]=Merge(ch[x][1],y);

        updata(x);

        return x;

    }

    else

    {

        ch[y][0]=Merge(x,ch[y][0]);

        updata(y);

        return y;

    }

}

int New(int d,int de)

{

    val[++tot]=rand(),dat[tot]=mx[tot]=d,dep[tot]=de;

    return tot;

}

void Insert(int id,int d,int de)

{

    int x,y;

    split(root[id],de,x,y);

    root[id]=Merge(x,Merge(New(d,de),y));

}

int ask(int id,int de)

{

    int x,y,z;

    split(root[id],de,x,y);

    z=mx[x];

    root[id]=Merge(x,y);

    return z;

}

int query(int id,int L,int R,int l,int r,int de)

{

    if(l==L&&r==R)

        return ask(id,de);

    int Mid=L+R>>1;

    if(r<=Mid) return query(id<<1,L,Mid,l,r,de);

    else if(l>Mid) return query(id<<1|1,Mid+1,R,l,r,de);

    else return min(query(id<<1,L,Mid,l,Mid,de),query(id<<1|1,Mid+1,R,Mid+1,r,de));

}

int Next[N<<1],to[N<<1],head[N],cnt;

void add(int u,int v)

{

    to[++cnt]=v,Next[cnt]=head[u],head[u]=cnt;

}

int dfn[N],low[N],Dep[N],ha[N],dfs_clock,n,m,rt,a[N];

void dfs(int now,int fa)

{

    dfn[now]=++dfs_clock;

    ha[dfs_clock]=now;

    for(int i=head[now];i;i=Next[i])

    {

        int v=to[i];

        if(v!=fa)

            Dep[v]=Dep[now]+1,dfs(v,now);

    }

    low[now]=dfs_clock;

}

void build(int id,int l,int r)

{

    for(int i=l;i<=r;i++)

        Insert(id,a[ha[i]],Dep[ha[i]]);

    if(l==r) return;

    int mid=l+r>>1;

    build(id<<1,l,mid),build(id<<1|1,mid+1,r);

}

int main()

{

    memset(dat,0x3f,sizeof(dat));

    memset(mx,0x3f,sizeof(mx));

    scanf("%d%d",&n,&rt);

    for(int i=1;i<=n;i++) scanf("%d",a+i);

    for(int u,v,i=1;i<n;i++)

    {

        scanf("%d%d",&u,&v);

        add(u,v),add(v,u);

    }

    dfs(rt,0);

    build(1,1,n);

    scanf("%d",&m);

    int las=0;

    for(int p,q,i=1;i<=m;i++)

    {

        scanf("%d%d",&p,&q);

        p=(p+las)%n+1,q=(q+las)%n;

        printf("%d\n",las=query(1,1,n,dfn[p],low[p],Dep[p]+q));

    }

    return 0;

}

2018.10.13