Submission #738151

#	Time	Username	Problem	Language	Result	Execution time	Memory
738151		maomao90	Let's Win the Election (JOI22_ho_t3)	C++17	61 / 100	2556 ms	4296 KiB

Main

// Hallelujah, praise the one who set me free
// Hallelujah, death has lost its grip on me
// You have broken every chain, There's salvation in your name
// Jesus Christ, my living hope
#include <bits/stdc++.h> 
using namespace std;

#define REP(i, s, e) for (int i = (s); i < (e); i++)
#define RREP(i, s, e) for (int i = (s); i >= (e); i--)
template <class T>
inline bool mnto(T& a, T b) {return a > b ? a = b, 1 : 0;}
template <class T>
inline bool mxto(T& a, T b) {return a < b ? a = b, 1: 0;}
typedef long long ll;
typedef long double ld;
#define FI first
#define SE second
typedef pair<int, int> ii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> iii;
#define ALL(_a) _a.begin(), _a.end()
#define SZ(_a) (int) _a.size()
#define pb push_back
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<ii> vii;
typedef vector<iii> viii;

#ifndef DEBUG
#define cerr if (0) cerr
#endif

const int INF = 1000000005;
const ll LINF = 1000000000000000005ll;
const int MAXN = 505;

int n, k, m;
int a[MAXN], b[MAXN];
int aid[MAXN], bid[MAXN];
bool done[MAXN];
ld dp[MAXN][MAXN];
ld ans;

int main() {
#ifndef DEBUG
    ios::sync_with_stdio(0), cin.tie(0);
#endif
    cin >> n >> k;
    REP (i, 1, n + 1) {
        cin >> a[i] >> b[i];
        if (b[i] != -1) {
            m++;
        }
    }
    iota(aid + 1, aid + n + 1, 1);
    iota(bid + 1, bid + n + 1, 1);
    sort(aid + 1, aid + n + 1, [&] (int l, int r) {
            return a[l] < a[r];
            });
    sort(bid + 1, bid + n + 1, [&] (int l, int r) {
            return b[l] < b[r];
            });
    ans = INF;
    REP (i, 0, min(k, m) + 1) {
        REP (l, 0, n + 1) {
            REP (o, 0, n + 1) {
                dp[l][o] = INF;
            }
        }
        dp[0][0] = 0;
        REP (j, 1, n + 1) {
            RREP (l, n, 0) {
                RREP (o, n, 0) {
                    if (o) {
                        mnto(dp[l][o],
                                dp[l][o - 1] + a[bid[j]] / (ld) (i + 1));
                    }
                    if (l && b[bid[j]] != -1) {
                        mnto(dp[l][o],
                                dp[l - 1][o] + b[bid[j]] / (ld) l);
                    }
                    //cerr << j << ' ' << l << ' ' << o << ": " << dp[j][l][o] << '\n';
                }
            }
        }
        //cerr << i << ": " << dp[n][i][k - i] << '\n';
        mnto(ans, dp[i][k - i]);
    }
    cout << setprecision(10) << ans << '\n';
    return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 0 / 5
• Subtask 3: 11 / 11
• Subtask 4: 12 / 12
• Subtask 5: 33 / 33
• Subtask 6: 0 / 11
• Subtask 7: 0 / 23