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 vshould 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 modulo1000000007 (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 42 1 3 21 21 33 4
output
8
input
0 31 2 31 22 3
output
3
input
4 87 8 7 5 4 6 4 101 61 25 81 33 56 73 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：
给定一棵树，树有点权，现在有树中有多少个有效的集合
有效的集合：
1.集合非空
2.集合是连通的，也就是说集合组成的还是一棵树
3.集合中，最大点权-最下点权<=d
这道题暑假的时候有想过，没有想出来
今天一想，其实就是一道简单的计数问题
由于n很小，O(n^2)是可以的
要max-min<=d
也就是要max<=min+d
dp[i]:i在集合里面，并且集合的最小点权就是i的点权的有效集合的个数
则：ans=sigma(dp[i])
对于一个节点root，我们考虑这个点的点权是他所在的有效集合中的最小点权,并且以root为根开始进行树形DP
如果节点i的点权>=a[root]&& 点权<=a[root]+d
我们就认为root可以扩展到i，不断扩展
并且有dp[u]=dp[u]*(1LL+dp[v])%mod
这样dfs一遍就可以在O(n)算出dp[root]了
以每一个点作为root 来dfs一遍，累加就可以得到ans了
注意一个问题：
有可能a[u]==a[v]
我们以root=u时扩展到v，并且加入了v，算了一遍
然后以root=v时扩展到u，这个时候我们如果把u加入，就会重复计算了
那么在有多个点的点权相等时，我们怎么避免重复计算，只算一次呢？
其实只要我们设一个数组vis[i][j],算第一次的时候我们把数组标记为true，后面就不再加入了
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define LL long long
using namespace std;
const int maxn=2010;
const int mod=1e9+7;
LL dp[maxn];
int a[maxn];
bool vis[maxn][maxn];
int sum;
int root;
struct Edge
{
int to,next;
};
Edge edge[maxn<<1];
int head[maxn];
int tot;
void init()
{
memset(head,-1,sizeof head);
tot=0;
}
void addedge(int u,int v)
{
edge[tot].to=v;
edge[tot].next=head[u];
head[u]=tot++;
}
void solve(int ,int d);
int main()
{
init();
int n,d;
scanf("%d %d",&d,&n);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
}
for(int i=1;i<n;i++){
int u,v;
scanf("%d %d",&u,&v);
addedge(u,v);
addedge(v,u);
}
solve(n,d);
return 0;
}
void dfs(int u,int pre)
{
dp[u]=1;
for(int i=head[u];~i;i=edge[i].next){
int v=edge[i].to;
if(v==pre || a[v]<a[root] || a[v]>sum)
continue;
if(a[v]==a[root]){
if(!vis[v][root]){
vis[v][root]=true;
vis[root][v]=true;
dfs(v,u);
}
else
continue;
}
else{
dfs(v,u);
}
dp[u]=dp[u]*(1LL+dp[v])%mod;
}
}
void solve(int n,int d)
{
memset(vis,false,sizeof vis);
LL ans=0;
for(int i=1;i<=n;i++){
sum=a[i]+d;
root=i;
dfs(root,root);
ans=(ans+dp[root])%mod;
ans=(ans+mod)%mod;
//cout<<dp[root]<<endl;
}
printf("%I64d\n",ans);
return ;
}