CF - 629(div3) -- E - Tree Queries【LCA + 思维】

题意

给出一棵树，问是否存在1到某个结点的一条路径，使得k个结点到这条路径的距离<=1.

思路

在k个结点中，找深度最大的这个结点mx，然后与每个结点求ui = LCA(mx, ki)，判断dis(ui, ki) <= 1即可。

时间复杂度m * k * logn

AC代码

#include
#include
#include
#include
#include
using namespace std;
#define IOS ios::sync_with_stdio(false)
const int maxn = 4e5 + 100;
struct Edge{
	int to, w, next;
}edge[maxn];
int dis[maxn], fa[21][maxn], head[maxn], path[maxn], node[maxn], deep[maxn], cnt;
bool vis[maxn];
void init(){
	memset(head, -1, sizeof(head));
	memset(fa, 0, sizeof(fa));
	memset(deep, 0, sizeof(deep));
	memset(vis, false, sizeof(vis));
	memset(dis, 0, sizeof(dis));
	cnt = 0;
}
void add(int u, int v, int w){
	edge[cnt].to = v;
	edge[cnt].w = w;
	edge[cnt].next = head[u];
	head[u] = cnt++;
}
void dfs(int u, int fno){
	for (int i = 1; i <= 20; ++i){
		fa[i][u] = fa[i - 1][fa[i - 1][u]];
	}
	for (int i = head[u]; ~i; i = edge[i].next){
		int v = edge[i].to;
		if (v == fno)	continue;
		dis[v] = dis[u] + edge[i].w;
		fa[0][v] = u;
		deep[v] = deep[u] + 1;
		dfs(v, u);
	}
}
int LCA(int x, int y){
	if (deep[x] < deep[y]){
		swap(x, y);
	}
	for (int i = 20; i >= 0; --i){
		if (deep[x] -deep[y] >> i){
			x = fa[i][x];
		}
	}
	if (x == y)		return x;
	for (int i = 20; i >= 0; --i){
		if (fa[i][x] != fa[i][y]){
			x = fa[i][x];
			y = fa[i][y];
		}
	}
	return fa[0][x];
}
void solve(){
	int n, m, u, v;
	scanf("%d%d", &n, &m);
	init();
	for (int i = 1; i < n; ++i){
		scanf("%d%d", &u, &v);
		add(u, v, 1);
		add(v, u, 1);
	}
	dfs(1, -1);
	int k;
	while (m--){
		scanf("%d", &k);
		int mx = 0, base = 0;
		for (int i = 0; i < k; ++i){
			scanf("%d", &node[i]);
			if (mx < deep[node[i]]){
				mx = deep[node[i]];
				base = node[i];
			}
		}
		bool flag = true;
		for (int i = 0; i < k; ++i){
			int lca = LCA(base, node[i]);
			if (dis[lca] + dis[node[i]] - dis[lca] * 2 > 1){
				flag = false;
				break;
			}
		}
		if (flag)	puts("YES");
		else puts("NO");
	}
}
int main(){
	solve();
	return 0;
}