Submission #1092694

#	Time	Username	Problem	Language	Result	Execution time	Memory
1092694		ortsac	Let's Win the Election (JOI22_ho_t3)	C++17	100 / 100	822 ms	2644 KiB

Main

#include <bits/stdc++.h>

using namespace std;

#define int long long
#define db double

int INF = 0x3f3f3f3f3f3f3f3f;

struct Val {
    int a, b, idx;
    Val(int x = 0, int y = 0, int z = 0) : a(x), b(y), idx(z) {}
};

bool comp1(const Val& a, const Val& b) {
    return (make_pair(a.b, -a.a) < make_pair(b.b, -b.a));
}

bool comp2(const Val& a, const Val& b) {
    return (make_pair(a.a, a.idx) < make_pair(b.a, b.idx));
}

const int MAXN = 505;
db dp[MAXN][MAXN];

int32_t main() {
    int n, k;
    cin >> n >> k;
    vector<Val> v(n);
    for (int i = 0; i < n; i++) {
        cin >> v[i].a >> v[i].b;
        if (v[i].b == -1) v[i].b = INF;
        v[i].idx = (i + 1);
    }
    sort(v.begin(), v.end(), comp1);
    v.insert(v.begin(), 0);
    cout << fixed << setprecision(2);
    db ans = INF;
    for (int p = 0; p <= k; p++) {
        for (int i = 0; i <= k; i++) {
            for (int j = 0; j <= k; j++) dp[i][j] = INF;
        }
        dp[0][0] = 0;
        for (int i = 1; i <= k; i++) {
            for (int j = 0; j <= i; j++) {
                // case 1, we get i to be a companion
                if (j > 0) {
                    db case1 = dp[i - 1][j - 1] + ((db)v[i].b)/((db)j);
                    dp[i][j] = min(dp[i][j], case1);
                }
                db case2 = dp[i - 1][j] + ((db)v[i].a)/((db)(p + 1));
                dp[i][j] = min(dp[i][j], case2);
            }
        }
        int need = (k - p);
        for (int i = 0; i <= need; i++) {
            int sumI = 0;
            vector<int> rem;
            for (int j = (p + i + 1); j <= n; j++) {
                rem.push_back(v[j].a);
            }
            sort(rem.begin(), rem.end());
            for (int j = 0; j < (need - i); j++) {
                sumI += rem[j];
            }
            db sum = ((db)sumI)/((db)p+1);
            db curr = (sum + dp[p + i][p]);
            if (i != need) {
                sum -= rem.back();
                rem.pop_back();
            }
            if (curr < ans) {
                //cout << sum << " " << i << " " << p << " " << dp[p + 1][p] << "\n";
            }
            ans = min(ans, curr);
        }
    }
    cout << ans << "\n";
}

• 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