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

链接：

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

题意：

给你一棵树，从中选一棵子树，满足

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.

.

题解：

从每个节点dfs，但是要避免重复，代码中的continue就是避免重复

代码：

int d, n;
int a[MAXN];
VI G[MAXN];
int dp[MAXN];

void dfs(int u, int fa, int rt) {
    dp[u] = 1;
    rep(i, 0, G[u].size()) {
        int v = G[u][i];
        if (v == fa) continue;
        if (a[v] > a[rt] + d) continue;
        if (a[v] < a[rt]) continue;
        if (a[v]==a[rt] && v < rt) continue;
        dfs(v, u, rt);
        dp[u] = 1LL * dp[u] * (dp[v] + 1) % MOD;
    }
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    cin >> d >> n;
    rep(i, 1, n + 1) cin >> a[i];
    rep(i, 1, n) {
        int u, v;
        cin >> u >> v;
        G[u].pb(v);
        G[v].pb(u);
    }
    int ans = 0;
    rep(i, 1, n + 1) {
        memset(dp, 0, sizeof(dp));
        dfs(i, 0, i);
        ans = (ans + dp[i]) % MOD;
    }
    cout << ans << endl;
    return 0;
}