【BZOJ1803】Spoj1487 Query on a tree III 主席树+DFS序

最新推荐文章于 2020-09-11 21:07:44 发布

转载最新推荐文章于 2020-09-11 21:07:44 发布 · 52 阅读

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原文链接：http://www.cnblogs.com/CQzhangyu/p/6952518.html

本文介绍了一种解决BZOJ1803树上查询问题的方法，利用主席树结合DFS序实现对树中节点标签的第k大查询。通过具体的代码示例展示了如何构建树结构，进行DFS遍历以及实现主席树的相关操作。

【BZOJ1803】Spoj1487 Query on a tree III

Description

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.

Input

The first line contains one integer n (1 <= n <= 10^5). The next line contains n integers li (0 <= li <= 109) 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) 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)

Output

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.

Sample Input

5
1 3 5 2 7
1 2
2 3
1 4
3 5
4
2
3
4 1
3 2
3 2

Sample Output

5 4 5 5

题解：这样例你告诉我是求子树第k大？（the k_th largest）分明是第k小啊！

直接主席树+DFS序搞定。

#include <cstdio> 
#include <cstring> 
#include <iostream> 
#include <algorithm> 
using namespace std; 
const int maxn=100010; 
int n,m,cnt,tot; 
int to[maxn<<1],next[maxn<<1],head[maxn],v[maxn],p[maxn],q[maxn],rt[maxn]; 
int ls[maxn*30],rs[maxn*30],siz[maxn*30],s[maxn*30]; 
struct number 
{ 
    int val,org; 
}num[maxn]; 
bool cmp(number a,number b) 
{ 
    return a.val<b.val; 
} 
void insert(int x,int &y,int l,int r,int a,int b) 
{ 
    if(l>r)  return ; 
    y=++tot,siz[y]=siz[x]+1; 
    if(l==r) 
    { 
        s[y]=b; 
        return ; 
    } 
    int mid=l+r>>1; 
    if(a<=mid)   rs[y]=rs[x],insert(ls[x],ls[y],l,mid,a,b); 
    else    ls[y]=ls[x],insert(rs[x],rs[y],mid+1,r,a,b); 
} 
int query(int a,int b,int l,int r,int c) 
{ 
    if(l==r)    return s[a]; 
    int mid=l+r>>1,sm=siz[ls[a]]-siz[ls[b]]; 
    if(c<=sm)    return query(ls[a],ls[b],l,mid,c); 
    else    return query(rs[a],rs[b],mid+1,r,c-sm); 
} 
void dfs(int x,int fa) 
{ 
    p[x]=++p[0]; 
    insert(rt[p[0]-1],rt[p[0]],1,n,v[x],x); 
    for(int i=head[x];i!=-1;i=next[i])  if(to[i]!=fa)   dfs(to[i],x); 
    q[x]=p[0]; 
} 
void add(int a,int b) 
{ 
    to[cnt]=b,next[cnt]=head[a],head[a]=cnt++; 
} 
int main() 
{ 
    scanf("%d",&n); 
    int i,a,b; 
    for(i=1;i<=n;i++)    scanf("%d",&num[i].val),num[i].org=i; 
    sort(num+1,num+n+1,cmp); 
    for(i=1;i<=n;i++)    v[num[i].org]=i; 
    memset(head,-1,sizeof(head)); 
    for(i=1;i<n;i++) 
    { 
        scanf("%d%d",&a,&b); 
        add(a,b),add(b,a); 
    } 
    dfs(1,0); 
    scanf("%d",&m); 
    for(i=1;i<=m;i++) 
    { 
        scanf("%d%d",&a,&b); 
        printf("%d\n",query(rt[q[a]],rt[p[a]-1],1,n,b)); 
    } 
    return 0; 
}

转载于:https://www.cnblogs.com/CQzhangyu/p/6952518.html