Submission #1153098

#	Time	Username	Problem	Language	Result	Execution time	Memory
1153098		tfgs	Fireworks (APIO16_fireworks)	C++17	19 / 100	49 ms	1096 KiB

fireworks

#ifndef LOCAL
#include <bits/stdc++.h>
#endif
using namespace std;
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector")
#pragma GCC target("avx2")

#define int int64_t
const int inf=1e18;

#ifdef LOCAL
#include "algo/debug.h"
#else
template <typename... Args>
void dummy(Args&&... args){}
#define ps dummy
#endif

#define f first
#define s second
template<class T> using V = vector<T>; 
using vi = V<int>;
using vb = V<bool>;
using vs = V<string>;

#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x) 
#define len(x) (int)((x).size())
#define rsz resize
#define ins insert
#define ft front()
#define bk back()
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define ai2 array<int,2>
#define ai3 array<int,3>
#define ai4 array<int,4>
#define ai5 array<int,5>
template<class T> int lwb(const V<T>& a, const T& b) { return lb(all(a),b)-begin(a); }
template<class T> int upb(const V<T>& a, const T& b) { return ub(all(a),b)-begin(a); }
template<class T> bool ckmin(T& a, const T& b) { return a > b ? a=b, true : false; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a=b, true : false; }
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define clz __builtin_clzll
constexpr int p2(int x) { return (int)1 << x; }
constexpr int bits(int x) { return x == 0 ? 0 : 63-clz(x); } // floor(log2(x)) 
template<class T>void UNIQUE(V<T>& v) { sort(all(v)); v.erase(unique(all(v)),end(v)); }
template<class T,class U>void erase(T& t, const U& u) { auto it = t.find(u); assert(it != end(t)); t.erase(it); }
template<class F> struct y_combinator_result {
    F f;
    template<class T> explicit y_combinator_result(T &&f): f(std::forward<T>(f)) {}
    template<class ...Args> decltype(auto) operator()(Args &&...args) { return f(std::ref(*this), std::forward<Args>(args)...); }
};
template<class Fun> decltype(auto) yy(Fun &&fun) { return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun)); }

// In a tree, you are given that on every path from the root to a leaf there exists a red edge.
// Prove that you can choose a subset of red edges that will cover each leaf exactly once.
// Greedy: Sort the edges in inc order of depth, take it if the leaves were not already covered.
// Remember that edges cover segments of leaves.
// AFC some leaf was not covered. Then all the edges (segments) which cover it intersect another 
// edge (segment) that was used. But then this segment must completely contain the other segment
// (can't be the other way around or else this leaf would be covered) and we would have taken
// this segment first.

const int N = 300, D = 300;
int dp[N][D + 1];
V<ai2> sons[N];

void solve() {
    int n, m; cin >> n >> m;
    vi inp_d(n + m);
    for (int i = 1; i < n + m; i++) {
        int p, w; cin >> p >> w; p--;
        sons[p].pb({ i, w });
        inp_d[i] = inp_d[p] + w;
    }

    for (int i = 0; i < n + m; i++) {
        for (int j = 0; j <= D; j++) {
            dp[i][j] = inf;
        }
    }
    for (int u = n; u < n + m; u++) dp[u][inp_d[u]] = 0;

    yy([&](auto rec, int u) -> void {
        if (u >= n) return;
        for (auto [v, w] : sons[u]) rec(v);
        for (int x = 0; x <= D; x++) {
            dp[u][x] = 0;
            for (auto [v, w] : sons[u]) {
                int mn = inf;
                for (int y = 0; y <= min(D, x + w); y++) {
                    ckmin(mn, dp[v][y] + abs(y - x));
                }
                if (mn == inf) dp[u][x] = inf;
                else dp[u][x] += mn;
            }
        }
    })(0);

    int ans = inf;
    for (int d = 0; d <= D; d++) ckmin(ans, dp[0][d]);
    cout << ans << '\n';
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    solve();
    return 0;
}

• Subtask 1: 0 / 7
• Subtask 2: 19 / 19
• Subtask 3: 0 / 29
• Subtask 4: 0 / 45