A - Horizon

原创

whiteawll 2022-08-16 14:49:27 博主文章分类：练习题目 ©著作权

文章标签 深度优先图论算法 #define #include 文章分类 运维

©著作权归作者所有：来自51CTO博客作者whiteawll的原创作品，请联系作者获取转载授权，否则将追究法律责任

// Problem: A - Horizon
// Contest: AtCoder - Denso Create Programming Contest 2022(AtCoder Beginner Contest 239)
// URL: https://atcoder.jp/contests/abc239/tasks/abc239_a
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 2022-02-19 20:00:54
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cmath>
#include<vector>
#includ<algorithm>
using namespace std;

#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 ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fir first
#define pb push_back
#define sec second
#define sortall(x) sort((x).begin(),(x).end())
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}

bool check (int tar) {
    if (tar < 2) return 0;
    for (int i = 2; i <= tar / i; i ++)
        if (tar % i ==0) return 0;
    return 1;
}
const int N = 1e5 + 10;

int n, q;
int a[N];
vector<int> edge[N];
vector<int> num[N];

void dfs (int u, int fa) {
    num[u].pb(a[u]);
    for (auto ver : edge[u]) {
        if (ver != fa) {
            dfs(ver, u);
            for (auto x : num[ver]) {
                num[u].pb(x);
            }
        }
    }
    sort(num[u].begin(), num[u].end());
    reverse(num[u].begin(), num[u].end());
    if (num[u].size() > 20) {
        num[u].resize(20);
    }
}
void solve() {
    cin >> n>> q;
    for (int i = 1; i <= n; i ++)
        cin >> a[i];
    for (int i = 1; i <= n - 1; i ++) {
        int a, b;
        cin >> a >> b;
        edge[a].pb(b);
        edge[b].pb(a);
    }
    dfs(1, 0);
    while (q --) {
        int v, k;
        cin >> v>> k;
        k --;
        cout << num[v][k] << endl;
    }
}   
int main () {
    int t;
    t =1;
    while (t --) solve();
    return 0;
}