제출 #1261786

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1261786		euclid	Burza (COCI16_burza)	C++20	160 / 160	74 ms	844 KiB

burza

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
#define pb push_back
#define fi first
#define se second
#define vi vector<int>
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vl vector<ll>

int n, k, cnt = 0;
const int MN = 403, MS = (1<<20)+3, INF = 1e9 + 7;
vi G[MN], deps[23];
pii range[MN]; //range of leaves that node i can cover
int dep[MN];
int dp[MS]; //dp[state]: picking a node from each depth in state,
//the maximum leaf i s.t. leaves 1-i are all covered

void dfs1(int node, int fa) {
//    cout << node << '\n';
    range[node].fi = INF;
    dep[node] = dep[fa]+1;
    deps[dep[node]].pb(node);
    if(dep[node] == k) {
        range[node] = {++cnt, cnt};
        return;
    }
    for(int child : G[node]) {
        if(child == fa) continue;
        dfs1(child, node);
        range[node].fi = min(range[node].fi, range[child].fi);
        range[node].se = max(range[node].se, range[child].se);
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> n >> k;
    for(int i = 1; i < n; i++) {
        int u, v; cin >> u >> v;
        G[u].pb(v); G[v].pb(u);
    }
    if(k > 20) {
        cout << "DA" << '\n';
        return 0;
    }
    dep[0] = -1;
    dfs1(1, 0);
//    cout << range[1].fi << ' ' << range[1].se << '\n';
    for(int s = 1; s < (1<<k); s++) {
        for(int j = 0; j < k; j++) {
            int d = j+1;
            for(int node : deps[d]) {
                if((s&(1<<j))==0) continue;
                int ss = s^(1<<j);
                dp[s] = max(dp[s], dp[ss]);
//                if(s == 3 && j == 1) cout << node << ' ' << range[node].fi << ' ' << range[node].se << ' ' << dp[ss] << '\n';
                if(range[node].fi-1 <= dp[ss]) dp[s] = max(dp[s], range[node].se);
            }
        }
    }
//    cout << dp[3] << '\n';
    cout << (dp[(1<<k)-1]==cnt ? "DA":"NE") << '\n';

//    6 2
//    1 2
//    2 3
//    3 4
//    1 5
//    5 6

}

• Subtask 1: 16 / 16
• Subtask 2: 16 / 16
• Subtask 3: 16 / 16
• Subtask 4: 16 / 16
• Subtask 5: 16 / 16
• Subtask 6: 16 / 16
• Subtask 7: 16 / 16
• Subtask 8: 16 / 16
• Subtask 9: 16 / 16
• Subtask 10: 16 / 16