#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e4 + 7;

struct Edge{
	int to;
	int next;
}edge[MAXN * 4];
int head[MAXN],cnt;
int node[MAXN],fa[MAXN],son[MAXN];

void init(){
	for(int i = 1;i < MAXN;i++){
		head[i] = -1;
		edge[i].next = -1;
		node[i] = 1;
		son[i] = 0;
	}
	cnt = 0;
}

void addedge(int u,int v){
	edge[cnt].to = v;
	edge[cnt].next = head[u];
	head[u] = cnt++;
}

int dfs(int x,int father){
	fa[x] = father;
	for(int i = head[x];~i;i = edge[i].next){
		if(edge[i].to != father){
			son[x]++;
			node[x] += dfs(edge[i].to,x);
		}
	}
	return node[x];
}

int n,m,p;
int main(){
	cin >> n >> m >> p;
	init();
	for(int i = 1;i < n;i++){
		int x,y;
		cin >> x >> y;
		addedge(x,y);
		addedge(y,x);
	}
	dfs(m,0);
	/*
	cout << endl;
	for(int i = 1;i <= n;i++){
		cout << i << " " << fa[i] << endl;
	}
	cout << endl;
	*/
	for(int i = 1;i <= p;i++){
		int x;
		cin >> x;
		int ans;
		if(son[x] <= 1){
			ans = 0;
		}else{
			ans = 2;
		}
		for(int j = head[x];~j;j = edge[j].next){
			if(edge[j].to != fa[x]){
				ans *= node[edge[j].to];
				//cout << edge[j].to << " " << node[edge[j].to] << endl;
			}
		}
		ans = ans + 2 * node[x] - 1;
		cout << ans << endl;
	}
	return 0;
} 

代码如上

思路是当 询问的节点的子节点数 小于等于 $1$ 时，$ans$ $=$ $($以该节点为根的子树大小 $-$ $1$ $)$ $*$ $2$ $+$ 该节点本身

否则 $ans$ $=$ 以该节点的不同儿子为根的子树中的节点 的排列数 $+$ $($以该节点为根的子树大小 $-$ $1$ $)$ $*$ $2$ $+$ 该节点本身