Submission #526833

#TimeUsernameProblemLanguageResultExecution timeMemory
526833abc864197532Let's Win the Election (JOI22_ho_t3)C++17
100 / 100
299 ms328 KiB
#include <bits/stdc++.h> using namespace std; #define lli long long int #define pb push_back #define eb emplace_back #define pii pair<int,int> #define X first #define Y second #define all(x) x.begin(), x.end() #define rall(x) x.rbegin(), x.rend() void abc() {cout << endl;} template <typename T, typename ...U> void abc(T i, U ...j) { cout << i << ' ', abc(j...); } template <typename T> void printv(T l, T r) { for (; l != r; ++l) cout << *l << " \n"[l + 1 == r]; } #ifdef Doludu #define test(x...) abc("[" + string(#x) + "]", x) #define owo freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout) #else #define test(x...) void(0) #define owo ios::sync_with_stdio(false), cin.tie(0); #endif const int N = 200000; int main () { owo; int n, k; cin >> n >> k; vector <pii> state(n); for (int i = 0; i < n; ++i) { cin >> state[i].X >> state[i].Y; if (state[i].Y == -1) state[i].Y = 1 << 30; } /* let number of choose B be x B[i] / 1 + B[i] / 2 + ... + B[i] / x + A[i] / (x + 1) + A[i] / (x + 1) + ... dp[i][j][k] -> consider first i state, choose j for B, k for C the min answer */ sort(all(state), [&](pii i, pii j) { if (i.Y == j.Y) return i.X < j.X; return i.Y < j.Y; }); double ans = 1e9; for (int x = 0; x <= k; ++x) { vector <double> dp(n + 1, 1e9); dp[0] = 0; for (int i = 0; i < n; ++i) { vector <double> nxt(n + 1, 1e9); for (int j = 0; j <= i + 1; ++j) { if (j) nxt[j] = min(nxt[j], dp[j - 1] + double(state[i].X) / (x + 1)); if (i + 1 - j <= x) nxt[j] = min(nxt[j], dp[j] + double(state[i].Y) / (i + 1 - j)); else nxt[j] = min(nxt[j], dp[j]); } dp = nxt; } ans = min(ans, dp[k - x]); } cout << fixed << setprecision(10) << ans << endl; }
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