Codeforces--486D-----Valid Sets思维

最新推荐文章于 2024-07-27 23:44:56 发布

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最新推荐文章于 2024-07-27 23:44:56 发布

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分类专栏： DFS/BFS

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本文介绍了一种算法，用于计算给定树中有多少个子树满足特定条件：子树的最大权值与最小权值之差不大于给定值d。通过DFS深度优先搜索遍历树并计算满足条件的有效子树数量。

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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.

给一棵树，问它有多少子树，子树的最大权值与最小权值之差不大于d，而且非空

依次选取每个节点作为根节点，默认根节点为权值最大（每个子树都可看作权值最大点为根节点），由题意知，父节点与子节点只有两种关系

连接或不连接，若不连接，则该子节点之后的节点都无法连接（题目规则2），两节点相等的情况可能重复考虑，规定只能从小标号到达标号或

者相反即可避免重复，暴力搜索dfs

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<map>
#include<queue>
#include<algorithm>
#include<cmath>
#include<vector>
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CL(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long LL;
const int maxn = 1e4+10;
const int MOD = 1e9+7;
int val[maxn];
int d;
vector <int > G[maxn];
LL dfs(int i, int re, int t){
    LL cnt = 1;
    for(int j = 0; j < G[i].size(); j++){
        int c = G[i][j];
        if(c == re || val[t] < val[c] || (val[t] == val[c] && c < t) || val[t] - val[c] > d) continue;
        cnt = cnt * (dfs(c, i, t)+1) % MOD;
    }
    return cnt;
}
int main(){
    int a, b, n, kcase = 1;
    scanf("%d%d", &d, &n);
    for(int i = 1; i <= n; i++) scanf("%d", &val[i]);
    for(int i = 1; i < n; i++){
        scanf("%d%d", &a, &b);
        G[a].push_back(b); G[b].push_back(a);
    }
    LL ans = 0;
    for(int i = 1; i <= n; i++){
        ans = (ans + dfs(i, 0, i)) % MOD;
    }
    printf("%lld\n", ans);
    return 0;
}