Codeforces 486D Valid Sets(暴力)

题目链接：Codeforces 486D Valid Sets

题目大意：给定一棵树，每个节点有个权值，现在要选定一些节点，要求非空，并且maxVal-minVal不大于d。问说有多

少种选择方法。

解题思路：枚举每个节点作为根节点，默认根节点为权值最大的节点，然后各个孩子节点用乘法原理即可。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

typedef long long ll;
const ll mod = 1e9 + 7;
const int maxn = 2005;

int N, D, W[maxn];
vector<int> G[maxn];

ll dfs(int u, int f, int rt) {
    int n = G[u].size();
    ll ret = 1;

    for (int i = 0; i < n; i++) {
        int v = G[u][i];

        if (v == f || W[rt] < W[v] || (W[rt] == W[v] && v > rt) || W[rt] - W[v] > D)
            continue;
        ret = ret * (dfs(v, u, rt) + 1) % mod;
    }
    return ret;
}

int main () {
    scanf("%d%d", &D, &N);
    for (int i = 1; i <= N; i++)
        scanf("%d", &W[i]);
    int u, v;
    for (int i = 1; i < N; i++) {
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }

    ll ans = 0;
    for (int i = 1; i <= N; i++)
        ans = (ans + dfs(i, -1, i)) % mod;
    printf("%lld\n", ans);
    return 0;
}