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답안 #1023868

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
1023868 2024-07-15T08:13:01 Z caterpillow Burza (COCI16_burza) C++17
0 / 160
 1000 ms 250944 KB 
#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 
    considering the flattened array starting from index i

    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

*/

void sus_assert(bool pred) {
    if (!pred) while (1);
}

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

vt<int> sfx;

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));

    sus_assert(ans <= sfx[i]);

    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));

    sfx.resize(n + 1);
    ROF (i, 0, n) sfx[i] += sfx[i + 1] + val[rpos[i]];
    dbg(sfx);

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

    // sanity check
    int cnt = 0;
    F0R (i, n) if (depth[i] == k) cnt++;
    sus_assert(val[0] == cnt);
    F0R (i, n) if (depth[i] > k) assert(val[i] == 0);

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

Compilation message

burza.cpp:105:1: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
  105 | main() {
      | ^~~~

Subtask #1
0 / 16.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1037 ms 250704 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #2
0 / 16.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1069 ms 250920 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #3
0 / 16.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1054 ms 250760 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #4
0 / 16.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1073 ms 250804 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #5
0 / 16.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1068 ms 250744 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #6
0 / 16.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1072 ms 250708 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #7
0 / 16.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1062 ms 250864 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #8
0 / 16.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1050 ms 250944 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #9
0 / 16.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1061 ms 250708 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #10
0 / 16.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1066 ms 250708 KB Time limit exceeded
2 Halted 0 ms 0 KB -