Submission #1023865

#	Time	Username	Problem	Language	Result	Execution time	Memory
1023865		caterpillow	Burza (COCI16_burza)	C++17	0 / 160	342 ms	508536 KiB

burza

#include <bits/stdc++.h>
#include <random>

using namespace std;

using ll = long long;
using pl = pair<ll, ll>;
using pi = pair<int, int>;
#define vt vector
#define f first
#define s second
#define pb push_back
#define all(x) x.begin(), x.end() 
#define size(x) ((int) (x).size())
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define F0R(i, b) FOR (i, 0, b)
#define endl '\n'
const ll INF = 1e18;
const int inf = 1e9;

template<template<typename> class Container, typename T>
ostream& operator<<(ostream& os, Container<T> o) {
    os << "{"; 
    int g = size(o); 
    for (auto i : o) os << i << ((--g) == 0 ? "" : ", "); 
    os << "}";
    return os;
}

void _print() {
    cerr << "\n";
}

template<typename T, typename ...V>
void _print(T t, V... v) {
    cerr << t; if (sizeof...(v)) cerr << ", "; _print(v...);
}

#ifdef LOCAL
#define dbg(x...) cerr << #x << " = "; _print(x);
#else
#define dbg(x...)
#define cerr if (0) std::cerr
#endif

/*

you want to stop the coin from taking a path of length k
you only block off nodes at the same depth as the turn number
    ie you block off one of the immediate subtrees
hence, you block off one node at depths 1, 2, 3, ... k

take dfs order

let dp[i][j][k] be the max # of depth k nodes u can block off 
    in i's subtree using depths only in the range [j, k]

    if you take the current node (and its allowed) you split the range and take the sum of the dps in its next sibling (or next node, if non exists)
    otherwise, you go into the next node

*/

int n, k, t;
vt<vt<int>> adj;
vt<int> depth, pos, jump, val, rpos;

void dfs(int u, int p = -1) {
    pos[u] = t++;
    if (depth[u] == k) val[u] = 1;
    for (int v : adj[u]) {
        if (v == p) continue;
        depth[v] = depth[u] + 1;
        dfs(v, u);
        val[u] += val[v];
    }
    jump[u] = t;
}

int memo[400][400][400];

int dp(int i, int l, int r) {
    if (i == n) return 0;
    if (l > r) return 0;
    if (memo[i][l][r] != -1) return memo[i][l][r];
    int u = rpos[i]; // get the actual node index at this position

    int ans = 0;
    if (l <= depth[i] && depth[i] <= r) { // allowed to take yourself
        ans = val[u] + dp(jump[u], l, depth[u] - 1) + dp(jump[u], depth[u] + 1, r);
    }
    ans = max(ans, dp(i + 1, l, r));

    return memo[i][l][r] = ans;
}

main() {
    cin.tie(0)->sync_with_stdio(0);
    
    cin >> n >> k;
    adj.resize(n);
    depth = pos = jump = val = rpos = vt<int>(n, 0);
    F0R (i, n - 1) {
        int u, v; cin >> u >> v;
        u--, v--;
        adj[u].pb(v);
        adj[v].pb(u);
    }
    dfs(0);
    F0R (i, n) rpos[pos[i]] = i;
    memset(memo, -1, sizeof(memo));

    dbg(depth);
    dbg(pos);
    dbg(jump);
    dbg(val);
    dbg(rpos);

    // sanity check
    int cnt = 0;
    F0R (i, n) if (depth[i] == k) cnt++;
    if (cnt != val[0]) while (1);
    

    int ans = dp(0, 1, k);
    assert(ans <= val[0]);
    if (ans == val[0]) cout << "DA\n";
    else cout << "NE\n";
}

Compilation message (stderr)

burza.cpp:97:1: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
   97 | main() {
      | ^~~~

• Subtask 1: 0 / 16
• Subtask 2: 0 / 16
• Subtask 3: 0 / 16
• Subtask 4: 0 / 16
• Subtask 5: 0 / 16
• Subtask 6: 0 / 16
• Subtask 7: 0 / 16
• Subtask 8: 0 / 16
• Subtask 9: 0 / 16
• Subtask 10: 0 / 16