CodeForces - 1328 E. Tree Queries
原题地址:
http://codeforces.com/contest/1328/problem/E
基本题意:
给你一棵根为1的有根树,然后每次查询,给出k个点,如果这k个点与从根连下来的某一条链的距离不超过1就输出YES否则输出NO。
基本思路:
我们观察可以想到,如果要满足 k个点与从根连下来的某一条链的距离不超过1 这个条件,那么这k个顶点中的任意一个顶点 i 与这k个顶点里最深的顶点的 lca 一定是这条链上离这第 i 个顶点最近的那个点,所以只要这个距离不超过1,即存在这样一条链,否则即不存在。
具体就是只要这k个顶点中的每一个顶点 a[i] 都满足 dist( lca ( MaxDepthNode, a[i] ) , a[i] ) <= 1
则说明这样一条链存在。
参考代码:
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false)
#define ll long long
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define INF 0x3f3f3f3f
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
const int maxn = 2e5 + 10;
int n,m;
struct edge {
int next, v;
}edges[maxn*2];
int cnt;
int head[maxn];
void init() {
memset(head, -1, sizeof(head));
cnt = 0;
}
void add_edge(int u,int v) {
edges[cnt].next = head[u];
edges[cnt].v = v;
head[u] = cnt++;
}
int dep[maxn];
int f[maxn][21];
void dfs(int u,int fa) {
dep[u] = dep[fa] + 1;
for (int i = 0; i <= 19; i++)
f[u][i + 1] = f[f[u][i]][i];
for (int i = head[u]; i != -1; i = edges[i].next) {
int v = edges[i].v;
if (v == fa)
continue;
f[v][0] = u;
dfs(v, u);
}
}
int lca(int x,int y) {
if (dep[x] < dep[y])
swap(x, y);
for (int i = 20; i >= 0; i--) {
if (dep[f[x][i]] >= dep[y])
x = f[x][i];
if (x == y)
return x;
}
for (int i = 20; i >= 0; i--) {
if (f[x][i] != f[y][i]) {
x = f[x][i];
y = f[y][i];
}
}
return f[x][0];
}
int dist(int a,int b){
return dep[a] + dep[b] - 2 * dep[lca(a, b)];
}
int k,a[maxn];
signed main() {
// IO;
n = read(), m = read();
init();
rep(i, 1, n - 1) {
int u, v;
u = read(), v = read();
add_edge(u, v);
add_edge(v, u);
}
dfs(1, 0);
rep(cas, 1, m) {
k = read();
rep(i, 1, k) a[i] = read();
pii mx;
mx.first = -1;
rep(i, 1, k) {
if (dep[a[i]] > mx.first) {
mx.first = dep[a[i]];
mx.second = a[i];
}
}
bool v = true;
rep(i, 1, k) {
if (dist(lca(mx.second, a[i]), a[i]) > 1) {
v = false;
break;
}
}
if (v) cout << "YES" << endl;
else cout << "NO" << endl;
}
return 0;
}