Codeforces 486D Valid Sets【树型Dp】

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本文介绍了一种在给定约束条件下寻找有效子树的方法。主要讨论了如何通过枚举节点并遍历树来解决子树计数问题，同时提供了一个AC代码示例。

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As you know, an undirected connected graph with n nodes and n - 1 edges is called a tree. You are given an integer d and a tree consisting of n nodes. Each node i has a value a_i associated with it.

We call a set S of tree nodes valid if following conditions are satisfied:

S is non-empty.
S is connected. In other words, if nodes u and v are in S, then all nodes lying on the simple path between u and v should also be presented in S.
$\text{[math]}$ .

Your task is to count the number of valid sets. Since the result can be very large, you must print its remainder modulo 1000000007 (10⁹ + 7).

Input

The first line contains two space-separated integers d (0 ≤ d ≤ 2000) and n (1 ≤ n ≤ 2000).

The second line contains n space-separated positive integers a₁, a₂, ..., a_n(1 ≤ a_i ≤ 2000).

Then the next n - 1 line each contain pair of integers u and v (1 ≤ u, v ≤ n) denoting that there is an edge between u and v. It is guaranteed that these edges form a tree.

Output

Print the number of valid sets modulo 1000000007.

Examples

Input

Output

Input

Output

Input

4 8
7 8 7 5 4 6 4 10
1 6
1 2
5 8
1 3
3 5
6 7
3 4

Output

Note

In the first sample, there are exactly 8 valid sets: {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {3, 4} and {1, 3, 4}. Set {1, 2, 3, 4} is not valid, because the third condition isn't satisfied. Set {1, 4} satisfies the third condition, but conflicts with the second condition.

题目大意：

给你N个点，一个限制D.

现在让你在找到一共有多少颗子树，使得其子树中最大权值点-最小权值点<=D.

思路：

1、如果我们直接考虑设定dp状态进行计数稍微有些困难，假设设定dp【i】【j】表示以i为根的子树中，最大权值点的值为j的方案数的话，假设在维护一个最小，ans并不能累加dp*dp2.所以我们这里直接设定各种维度都稍有计数难度或者是时间复杂度难度。

2、所以我们不妨枚举出来一个点作为最大权值点，那么其能够走到的点权值一定小于等于这个点，并且每个点和这个枚举出来的点的差的绝对值小于等于D.

那么枚举加遍历树的时间复杂度是O（n^2）；时间上是允许的。

那么有：dp【u】*=（dp【v】+1）；

然而这里要有一个去重问题，就是如果a【i】==a【root】.我们不能重复去计算，所以这里确保相等权值的点有一个编号大小限制即可。

Ac代码：

#include<stdio.h>
#include<iostream>
#include<vector>
#include<string.h>
using namespace std;
#define mod 1000000007
#define ll __int64
vector<ll >mp[2005];
ll a[2500];
ll dp[2500];
ll d,n;
void Dfs(ll u,ll from,ll val,ll num)
{
    dp[u]=1;
    for(ll i=0;i<mp[u].size();i++)
    {
        ll v=mp[u][i];
        if(v==from)continue;
        if(a[v]<=val&&val-a[v]<=d)
        {
            if(a[v]==val&&v>num)continue;
            Dfs(v,u,val,num);
            dp[u]*=(dp[v]+1);
            dp[u]%=mod;
        }
    }
}
int main()
{
    while(~scanf("%I64d%I64d",&d,&n))
    {
        for(ll i=1;i<=n;i++)mp[i].clear();
        for(ll i=1;i<=n;i++)scanf("%I64d",&a[i]);
        for(ll i=0;i<n-1;i++)
        {
            ll x,y;
            scanf("%I64d%I64d",&x,&y);
            mp[x].push_back(y);
            mp[y].push_back(x);
        }
        ll output=0;
        for(ll i=1;i<=n;i++)
        {
            Dfs(i,-1,a[i],i);
            output+=dp[i];
            output%=mod;
        }
        printf("%I64d\n",output);
    }
}