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：
31 int d, n; 32 int a[MAXN]; 33 VI G[MAXN]; 34 int dp[MAXN]; 35 36 void dfs(int u, int fa, int rt) { 37 dp[u] = 1; 38 rep(i, 0, G[u].size()) { 39 int v = G[u][i]; 40 if (v == fa) continue; 41 if (a[v] > a[rt] + d) continue; 42 if (a[v] < a[rt]) continue; 43 if (a[v]==a[rt] && v < rt) continue; 44 dfs(v, u, rt); 45 dp[u] = 1LL * dp[u] * (dp[v] + 1) % MOD; 46 } 47 } 48 49 int main() { 50 ios::sync_with_stdio(false), cin.tie(0); 51 cin >> d >> n; 52 rep(i, 1, n + 1) cin >> a[i]; 53 rep(i, 1, n) { 54 int u, v; 55 cin >> u >> v; 56 G[u].pb(v); 57 G[v].pb(u); 58 } 59 int ans = 0; 60 rep(i, 1, n + 1) { 61 memset(dp, 0, sizeof(dp)); 62 dfs(i, 0, i); 63 ans = (ans + dp[i]) % MOD; 64 } 65 cout << ans << endl; 66 return 0; 67 }