Codeforces Round #277 (Div. 2) D. Valid Sets 暴力

D. Valid Sets

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/486/problem/D

Description

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.

Sample Input

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

Sample Output

8

HINT

 

题意

给你一颗树，然后让你找到里面有多少棵子树，满足子树里面最大点减去最小点的差小于等于d

题解：

直接枚举每一个点为根，然后开始dfs，都假设这个根节点是这棵子树的最大值

然后乘法原理搞一搞就好啦

代码

 

#include<iostream>
#include<stdio.h>
#include<vector>
using namespace std;
#define maxn 2005
const int mod = 1e9 + 7;
int a[maxn];
int d,n;
vector<int> G[maxn];
long long dfs(int x,int pre,int k)
{
    long long ans = 1;
    for(int i=0;i<G[x].size();i++)
    {
        int v = G[x][i];
        if(v == pre || (a[k] == a[v]&&v > k))continue;
        if(a[k]>=a[v]&&a[k]-a[v]<=d)
        {
            ans *= ( dfs(v,x,k) + 1 );
            while(ans>=mod)ans%=mod;
        }
    }
    return ans;
}
int main()
{
    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);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    long long ans = 0;
    for(int i=1;i<=n;i++)
    {
        ans += dfs(i,-1,i);
        while(ans>=mod)ans%=mod;
    }
    printf("%lld\n",ans);
}