Codeforces Round #277 (Div. 2)D（树形DP计数类）

D. Valid Sets

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ai associated with it.

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

1. S is non-empty.
2. 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.
3. .

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, ..., an(1 ≤ ai ≤ 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.

Sample test(s)

input

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

output

8

input

0 3
1 2 3
1 2
2 3

output

3

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

41

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.

题意：RT

思路：树形DP，dp[u]表示以u为根，且它的所有子树的点v的权值都不超过它，且满足 w[u]-w[v]<=d 的方案数

            对于每个点为根这样算一遍就好了，注意减去算重的情况，即相邻多个点的权值相等，可以用点的代号去重（即规定只能从代号大的往小的DP）

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

typedef long long ll;
const int MAXN = 2005;
const int MOD = (int)1e9+7;
int edge[MAXN<<1];
int next[MAXN<<1];
int head[MAXN];
int esz;
int ww[MAXN];
ll dp[MAXN];
int D;

void addedge(int u,int v)
{
    edge[esz]=v;
    next[esz]=head[u];
    head[u]=esz++;
}

void dfs(int u,int p,int w,int root)
{
    if((w>ww[u]&&w-ww[u]<=D) || (w==ww[u]&&root>=u)){

    dp[u]=1;
    for(int i=head[u];i!=-1;i=next[i]){
        int v=edge[i];
        if(v==p)continue;
        dfs(v,u,w,root);
        dp[u]=dp[u]*((dp[v]+1)%MOD)%MOD;
    }
    }
}

int main()
{
    int n;
    scanf("%d%d",&D,&n);
    memset(head,-1,sizeof(head));
    esz=0;
    for(int i=1;i<=n;i++)
        scanf("%d",&ww[i]);
    for(int i=1;i