SP1487 PT07J - Query on a tree III

题意翻译

你被给定一棵带点权的n个点的有根数，点从1到n编号。

定义查询 query(x,k): 寻找以x为根的k大点的编号（从小到大排序第k个点）

假设没有两个相同的点权。

输入格式：第一行为整数n，第二行为点权，接下来n-1行为树边，接下来一行为整数m，下面m行为两个整数x,k，代表query(x,k)

输出格式： m行，输出每次查询的结果。

Translated by @vegacx

题目描述

You are given a node-labeled rooted tree with n nodes.

Define the query (x, k): Find the node whose label is k-th largest in the subtree of the node x. Assume no two nodes have the same labels.

输入输出格式

输入格式：

The first line contains one integer n (1 <= n <= 10 ^{5}5 ). The next line contains n integers _l {i}i (0 <= _l {i}i <= 10 ^{9}9 ) which denotes the label of the i-th node.

Each line of the following n - 1 lines contains two integers u, v. They denote there is an edge between node _u_and node v. Node 1 is the root of the tree.

The next line contains one integer m (1 <= m <= 10 ^{4}4 ) which denotes the number of the queries. Each line of the next m contains two integers x, k. (k <= the total node number in the subtree of x)

输出格式：

For each query (x, k), output the index of the node whose label is the k-th largest in the subtree of the node x.

输入输出样例

输入样例#1：

5

1 3 5 2 7

1 2

2 3

1 4

3 5

4

2 3

4 1

3 2

3 2

输出样例#1：

5

4

5

5

Solution

传送门

树上静态主席树...

显然我们要根据dfs序来建立主席树,也正因如此,所以我们在dfs的过程中要重新赋一下值,给当前dfs序为i的节点另开两个存值的数组,一个是原数组,一个离散数组,这样值才能与主席树对上号



void dfs(int u,int fa) {

    id[u]=++cnt, a[cnt]=b[cnt]=w[u], inv[cnt]=u, size[u]=1;//inv[]存的是通过dfs序找到节点

    for (int i=head[u];i;i=nex[i]) {

        if (to[i]==fa) continue;

        dfs(to[i],u); size[u]+=size[to[i]];

    }

}

在加入节点的过程中,因为题目问的是节点编号而不是值,我们还要把当前值赋为节点编号

至于查询,就是以dfs序和子树大小来查,因为是连续的...

其实如果会静态主席树,这道题稍微难一点的主要还是建树,会建树就会查询了

Code

#include<bits/stdc++.h>

#define Min(a,b) ((a)<(b)?(a):(b))

#define Max(a,b) ((a)>(b)?(a):(b))

#define rg register

#define mid ((l+r)>>1)

#define in(i) (i=read())

using namespace std;

const int N=1e5+10;

int read() {

    int ans=0,f=1; char i=getchar();

    while(i<'0' || i>'9') {if(i=='-') f=-1; i=getchar();}

    while(i>='0' && i<='9') ans=(ans<<1)+(ans<<3)+i-'0',i=getchar();

    return ans*=f;

}

int T,n,m,tot,cur,cnt;

int w[N],a[N],b[N],id[N],inv[N],size[N],ans[N];

int to[N<<1],nex[N<<1],head[N];

int rt[N<<5],lc[N<<5],rc[N<<5],sum[N<<5];

void add(int a,int b) {

    to[++cur]=b,nex[cur]=head[a],head[a]=cur;

    to[++cur]=a,nex[cur]=head[b],head[b]=cur;

}

void build(int &u,int l,int r) {

    u=++tot,sum[u]=0;

    if (l==r) return;

    build(lc[u],l,mid);

    build(rc[u],mid+1,r);

}

void update(int &u,int l,int r,int pre,int x) {

    u=++tot;

    sum[u]=sum[pre]+1,lc[u]=lc[pre],rc[u]=rc[pre];

    if (l==r) return;

    if(x<=mid) update(lc[u],l,mid,lc[pre],x);

    else update(rc[u],mid+1,r,rc[pre],x);

}

int query(int u,int v,int l,int r,int k) {

    if(l==r) return ans[l];

    int rest=sum[lc[v]]-sum[lc[u]];

    if(k<=rest) return query(lc[u],lc[v],l,mid,k);

    else return query(rc[u],rc[v],mid+1,r,k-rest);

}

void dfs(int u,int fa) {

    id[u]=++cnt, a[cnt]=b[cnt]=w[u], inv[cnt]=u, size[u]=1;

    for (int i=head[u];i;i=nex[i]) {

        if (to[i]==fa) continue;

        dfs(to[i],u); size[u]+=size[to[i]];

    }

}

int main()

{

    in(n);

    for (int i=1;i<=n;i++) in(w[i]);

    for (int i=1,x,y;i<n;i++) in(x), in(y), add(x,y);

    dfs(1,0);

    sort(b+1,b+1+n); build(rt[0],1,n);

    for (int i=1;i<=n;i++) {

        a[i]=lower_bound(b+1,b+1+n,a[i])-b;

        ans[a[i]]=inv[i];

        update(rt[i],1,n,rt[i-1],a[i]);

    }

    in(m);

    for (int i=1,x,k;i<=m;i++) {

        in(x), in(k);

        printf("%d\n",query(rt[id[x]-1],rt[id[x]+size[x]-1],1,n,k));

    }

}