Submission #738133

#TimeUsernameProblemLanguageResultExecution timeMemory
738133maomao90Let's Win the Election (JOI22_ho_t3)C++17
0 / 100
1024 ms1048576 KiB
// 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][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) { if (b[l] == -1) return false; 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[0][l][o] = INF; } } dp[0][0][0] = 0; REP (j, 1, n + 1) { REP (l, 0, n + 1) { REP (o, 0, n + 1) { dp[j][l][o] = dp[j - 1][l][o]; if (o) { mnto(dp[j][l][o], dp[j - 1][l][o - 1] + a[j] / (ld) (i + 1)); } if (l && b[j] != -1) { mnto(dp[j][l][o], dp[j - 1][l - 1][o] + b[j] / (ld) l); } //cerr << j << ' ' << l << ' ' << o << ": " << dp[j][l][o] << '\n'; } } } cerr << i << ": " << dp[n][i][k - i] << '\n'; mnto(ans, dp[n][i][k - i]); } cout << setprecision(10) << ans << '\n'; return 0; }
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