​​http://www.elijahqi.win/archives/1287​​​
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:
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.
.
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.
Examples
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.

题目要求我们求一个连续的区间 满足这个区间内最大值和最小值小于等于给定的d 然后去求 这样的区间的数量 考虑一种树形dp的方法 设定dp[x] 为认为我当前x号的值是最大的然后去求一下可以到哪里 我们去dp[y] y这个点可以选 那么就递归dfs去搜索y这个点的子树求出结果 如果不选y这个点 那么y的子树也一定不会被选 所以直接略过就好 但是我们可能有重复的 怎么办 规定一下方向保证只算一次 然后统计答案自己画画图就好

#include<cstdio>
#define N 2200
#define mod 1000000007
#define ll long long
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0;char ch=gc();
    while (ch<'0'||ch>'9') ch=gc();
    while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc();}
    return x;
}
int h[N],w[N],fa[N],n,d,num,dp[N];
struct node{
    int y,next;
}data[N<<1];
void dfs(int x,int top){
    dp[x]=1;
    for (int i=h[x];i;i=data[i].next){
        int y=data[i].y;if (fa[x]==y) continue;int tmp=w[top]-d;if (tmp<0) tmp=0; 
        if (w[y]<tmp) continue;
        if (w[y]==w[top]&&y<top) continue;
        if (w[y]>w[top]) continue;fa[y]=x;
        dfs(y,top);dp[x]=(ll)dp[x]*(dp[y]+1)%mod;
        fa[y]=0;
    }
}
int main(){
    freopen("cf.in","r",stdin);
    d=read();n=read();
    for (int i=1;i<=n;++i) w[i]=read();
    for (int i=1;i<n;++i){
        int x=read(),y=read();
        data[++num].y=y;data[num].next=h[x];h[x]=num;
        data[++num].y=x;data[num].next=h[y];h[y]=num;
    }int ans=0;
    for (int i=1;i<=n;++i){
        dfs(i,i);
        ans=(ll)(ans+dp[i])%mod;
    }
    printf("%d",ans);
    return 0;
}