8VC Venture Cup 2016 - Final Round D. Preorder Test 二分 树形dp

Preorder Test

题目连接：

http://www.codeforces.com/contest/627/problem/D

Description

For his computer science class, Jacob builds a model tree with sticks and balls containing n nodes in the shape of a tree. Jacob has spent ai minutes building the i-th ball in the tree.

Jacob's teacher will evaluate his model and grade Jacob based on the effort he has put in. However, she does not have enough time to search his whole tree to determine this; Jacob knows that she will examine the first k nodes in a DFS-order traversal of the tree. She will then assign Jacob a grade equal to the minimum ai she finds among those k nodes.

Though Jacob does not have enough time to rebuild his model, he can choose the root node that his teacher starts from. Furthermore, he can rearrange the list of neighbors of each node in any order he likes. Help Jacob find the best grade he can get on this assignment.

A DFS-order traversal is an ordering of the nodes of a rooted tree, built by a recursive DFS-procedure initially called on the root of the tree. When called on a given node v, the procedure does the following:

Print v.

Traverse the list of neighbors of the node v in order and iteratively call DFS-procedure on each one. Do not call DFS-procedure on node u if you came to node v directly from u.

Input

The first line of the input contains two positive integers, n and k (2 ≤ n ≤ 200 000, 1 ≤ k ≤ n) — the number of balls in Jacob's tree and the number of balls the teacher will inspect.

The second line contains n integers, ai (1 ≤ ai ≤ 1 000 000), the time Jacob used to build the i-th ball.

Each of the next n - 1 lines contains two integers ui, vi (1 ≤ ui, vi ≤ n, ui ≠ vi) representing a connection in Jacob's tree between balls ui and vi.

Output

Print a single integer — the maximum grade Jacob can get by picking the right root of the tree and rearranging the list of neighbors.

Sample Input

5 3

3 6 1 4 2

1 2

2 4

2 5

1 3

Sample Output

3

Hint

题意

给你一棵树，带点权

让你找到一个dfs搜索的顺序中，至少大于k个点，且这k个点的最小值最大

题解：

二分答案，然后我们进行check

我们把大于mid的点标为1

然后我们就可以开始树dp了

显然对于某个点来说，除了他儿子那棵子树的所有点都是满足条件的，否则他最多选择两个儿子的不完整子树。

然后我们通过这个进行dp就好了，记录最大值和次大值。

对了，还得check一下他的父亲，看看这个点是否能够往上延展。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+7;
int n,k;
int a[maxn],w[maxn],sz[maxn],up[maxn],dp[maxn];
vector<int> E[maxn];
int flag = 1;
void dfs(int x,int fa,int m)
{
    sz[x]=1;
    for(int i=0;i<E[x].size();i++)
    {
        int v = E[x][i];
        if(v==fa)continue;
        dfs(v,x,m);
        w[x]+=w[v];
        sz[x]+=sz[v];
    }
}
void solve(int x,int fa,int m)
{
    int Max1=0,Max2=0,tot=0;
    for(int i=0;i<E[x].size();i++)
    {
        int v=E[x][i];
        if(v==fa)continue;
        if(w[x]-w[v]==sz[x]-sz[v]&&up[x])up[v]=1;
        solve(v,x,m);
        if(a[v]<m)continue;
        if(sz[v]==w[v])tot+=sz[v];
        else
        {
            if(dp[v]>Max1)Max2=Max1,Max1=dp[v];
            else if(dp[v]>Max2)Max2=dp[v];
        }
    }
    if(a[x]<m)return;
    if(tot+Max1+Max2+up[x]*(n-sz[x])+1>=k)flag=1;
    dp[x]=tot+Max1+1;
}
int check(int x)
{
    for(int i=1;i<=n;i++)
        dp[i]=0,up[i]=0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]>=x)
            w[i]=1;
        else
            w[i]=0;
    }
    flag = 0;
    dfs(1,-1,x);
    up[1]=w[1];
    solve(1,-1,x);
    return flag;
}
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<n;i++)
    {
        int x,y;scanf("%d%d",&x,&y);
        E[x].push_back(y);
        E[y].push_back(x);
    }
    int l = 0,r = 1e6+5,ans = 1e6+5;
    while(l<=r)
    {
        int mid = (l+r)/2;
        if(check(mid))l=mid+1,ans=mid;
        else r=mid-1;
    }
    cout<<ans<<endl;
}
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
int n,m,t,i,x[11][11];
int mo=10007;
void multiply(int a[11][11],int b[11][11])
{
    int i,j,k,c[11][11];
    for (i=0;i<=t;i++)
        for (j=0;j<=t;j++)
        {
            c[i][j]=0;
            for (k=0;k<=t;k++) c[i][j]=(c[i][j]+a[i][k]*b[k][j])%mo;
        }

    for (i=0;i<=t;i++)
        for (j=0;j<=t;j++)
            a[i][j]=c[i][j];
}
void binary(int x[11][11],int a)
{
    int i,j,y[11][11];
    if (a==1) return;
    for (i=0;i<=t;i++)
        for (j=0;j<=t;j++)
            y[i][j]=x[i][j];
    multiply(x,x);
    binary(x,a/2);
    if (a%2==1) multiply(x,y);
}
int main()
{
    scanf("%d%d%d",&m,&n,&t);
    for (i=0;i<=t;i++)
	{
        if (i!=0) x[i-1][i]=t-i+1;
        x[i][i]=m-t;
        if (i!=t) x[i+1][i]=i+1;
	}
	binary(x,n);
	printf("%d\n",x[0][0]);
}