Submission #907857

#	Time	Username	Problem	Language	Result	Execution time	Memory
907857		vjudge1	Let's Win the Election (JOI22_ho_t3)	C++17	100 / 100	361 ms	4696 KiB

Main

#pragma GCC target ("avx2")
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")

#include<bits/stdc++.h>
#include<math.h>
using namespace std;

typedef int ll;
typedef double ld;
typedef pair<ll, ll> pl;
typedef vector<ll> vl;
#define FD(i, r, l) for(ll i = r; i > (l); --i)

#define K first
#define V second
#define G(x) ll x; cin >> x;
#define GD(x) ld x; cin >> x;
#define GS(s) string s; cin >> s;
#define EX(x) { cout << x << '\n'; exit(0); }
#define A(a) (a).begin(), (a).end()
#define F(i, l, r) for (ll i = l; i < (r); ++i)

#define NN 510

vector<pl> votes;
vector<pl> votes2;
ll n, k;
ll take;

// The final observation is this: 
// We can't actually "skip" votes in the optimal construction.
// Since there's absolutely no world in which we skip b[i] but take b[i+1]. 
// If we don't make b[i] a comrade, we MUST make him a voter. 

ld dp[NN][NN]; 
ld dp2[NN][NN];

ld query(ll i, ll need) {
    if (need <= 0) return 0;
    return dp2[i][need];
}

ld rec(ll i, ll bs) {
    if (bs == take) return query(i, k - i) / (1 + take);
    if (i == n) return 1e9;
    auto &DP = dp[i][bs];
    if (DP != DP) {
        DP = 1e9;
        DP = min(DP, rec(i+1, bs) + ld(votes[i].V)/(1 + take)); 
        DP = min(DP, rec(i+1, bs+1) + ld(votes[i].K)/(1 + bs)); 
    }
    return DP;
}

int main(){
//    freopen("a.in", "r", stdin);
//    freopen("a.out", "w", stdout);

    ios_base::sync_with_stdio(false); cin.tie(0);
    cout << fixed << setprecision(20);
    cin >> n >> k;
    votes.resize(n);
    for (auto &[b, a]: votes) {
        cin >> a >> b;
        if (!~b) b = 1e9;
    }
    sort(A(votes));
    F(i, 0, NN) F(j, 0, NN) dp2[i][j] = 1e9;
    F(suff, 0, n+1) {
        votes2 = votes;
        sort(suff + A(votes2), [&](const pl &a, const pl &b) {
            return a.V < b.V;
        });
        ld sm = 0;
        ll take = 1;
        for (auto it = votes2.begin() + suff; it != votes2.end(); it++) {
            sm += it->V;
            dp2[suff][take++] = sm;
        }
    }

    ld ans = 1e9; 

    for (take = 0; take <= n; take++) {
        memset(dp, -1, sizeof dp);
        ans = min(ans, rec(0, 0));
    }

    cout << ans << endl;

}

• Subtask 1: 5 / 5
• Subtask 2: 5 / 5
• Subtask 3: 11 / 11
• Subtask 4: 12 / 12
• Subtask 5: 33 / 33
• Subtask 6: 11 / 11
• Subtask 7: 23 / 23