hdu 5678 ztr loves trees (给一颗有根树，树上的每一个节点有一个权值，每次询问某个子树中所有权值的中位数)

ztr loves trees

Time Limit: 6000/2500 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 296    Accepted Submission(s): 46

Problem Description

Super Deity ztr likes trees from childhood,CCTV-children:"The apple on the apple tree.You and I under the apple tree.Play games in front of the apple tree.So many happiness".

One day,qzh visit ztr to ask some questions.To give a tree with a root,each vertex has a value.Each time query the median of a subtree.

ztr said:this is a water problem,do you do it?But qzh show cannot help but want you who is also a Super Deity to help him.Could you help him?

 

Input

There are T test cases. The first line of input contains an positive integer T indicating the number of test cases.

For each test case:

Each line contains two positive integer n,m.indicating the number of vetrex and the number of query times.
The next line contains n numbers, the ith number indicating the value of vertex i.
The next n-1 lines,each line contains two numbers u and v,indicating there is a edge form u to v.

The next m lines, each line contains a numbers x.indicating query the median of subtree x.

1<=T<=3,1<=n<=105,1<=m<=106,1<=u<=v<=n,1<=val<=109.
The vetrex 1 is the root of the tree.Guarantee input a tree with a root.

 

Output

For each test case:print a line.To avoid huge output,you should hash each answer first,then print it.

The method to hash:a[i] indicates the ith query result, ans=∑a[i]∗10m−imod1,000,000,007  Round to the nearest tenth

 

Sample Input

  
  

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

 

Sample Output

  
  

   
   339.0
  
  

 

给一颗有根树，树上的每一个节点有一个权值，每次询问某个子树中所有权值的中位数

 

思路:

这道题就是在一个子树上询问第k大，于是我们用dfs序将树上询问第k大转化为区间询问第k大，套用主席树即可.

#include<cstdio>
#include<algorithm>
#include<time.h>
#include<iostream>
#include<cstdio>
#include<queue>
#include<string.h>
#include<math.h>
#include<vector>
#include<algorithm>
#include<stdio.h>
#include<map>
using namespace std;
#define maxn 100005
const int MOD=1000000007;
int val[maxn],L[maxn],R[maxn],tot,a[maxn],t[maxn],vis[maxn];
vector<int>G[maxn];
int root[maxn],sz;
double ans[maxn];
struct node{
    int l,r,w;
}T[maxn*30];

void dfs(int u){
    vis[u]=1;
    L[u]=++tot;
    t[tot]=val[u],a[tot]=val[u];
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i];
        if(vis[v]==1)
            continue;
        dfs(v);
    }
    R[u]=tot;
}

void update(int &i,int l,int r,int num){
    T[++sz]=T[i],i=sz;
    T[i].w++;
    if(l==r)
        return ;
    int mid=(l+r)>>1;
    if(num<=mid)
        update(T[i].l,l,mid,num);
    else
        update(T[i].r,mid+1,r,num);
}

int query(int x,int y,int l,int r,int k){
    if(l==r)
        return l;
    int mid=(l+r)>>1;
    int num=T[T[y].l].w-T[T[x].l].w;
    if(num>=k)
        return query(T[x].l,T[y].l,l,mid,k);
    else
        return query(T[x].r,T[y].r,mid+1,r,k-num);
}

int main(){
    int _,n,m,u,v,q;
    scanf("%d",&_);
    while(_--){
        scanf("%d%d",&n,&q);
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++){
            scanf("%d",&val[i]);
            G[i].clear();
        }
        for(int i=1;i<n;i++){
            scanf("%d%d",&u,&v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        tot=0;
        dfs(1);
        sort(t+1,t+n+1);
        int m=unique(t+1,t+n+1)-t-1;
        root[0]=0,sz=0;
        for(int i=1;i<=n;i++){  //有序
            root[i]=root[i-1];
            int num=lower_bound(t+1,t+m+1,a[i])-t;
            update(root[i],1,n,num);
        }
        for(int i=1;i<=n;i++){
            if((R[i]-L[i])%2==1){
                int num1=query(root[L[i]-1],root[R[i]],1,n,(R[i]-L[i]+1)/2);
                int num2=query(root[L[i]-1],root[R[i]],1,n,(R[i]-L[i]+1)/2+1);
                ans[i]=1.0*(t[num1]+t[num2])/2;
            }
            else
                ans[i]=t[query(root[L[i]-1],root[R[i]],1,n,(R[i]-L[i])/2+1)];
        }
        double ANS=0;
        for(int i=1;i<=q;i++){
            scanf("%d",&u);
            ANS=fmod(ans[u]+ANS*10,1.0*MOD);
        }
        printf("%.1f\n",ANS);
    }
    return 0;
}