제출 #1266216

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1266216		syanvu	Let's Win the Election (JOI22_ho_t3)	C++20	100 / 100	411 ms	2452 KiB

Main

#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp> 
// #include <ext/pb_ds/tree_policy.hpp>

#define SS ios_base::sync_with_stdio(0);cin.tie(nullptr);cout.tie(nullptr);
#pragma optimize("g", on)
#pragma GCC optimize("03")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,avx2,mmx,fma,avx,tune=native")
#define int long long
#define all(x) x.begin(),x.end()
#define F first
#define S second


using namespace std;
// using namespace __gnu_pbds; 
// #define ordered_set tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update>

const int LG = 20,N = 2e5+1,inf = 1e18,MOD = 998244353;
const double eps = 1e-9;
int T;

int n,k;
int a[N],b[N];

double get(int x){
    vector<vector<double>> dp(n+1,vector<double>(n+1,inf));
    dp[0][0]=0;
    for(int i=1;i<=n;i++){
        for(int j=0;j<=i;j++){
            if(j) dp[i][j]=min(dp[i][j],dp[i-1][j-1]+double(a[i])/double(x+1));

            if(i-j<=x) dp[i][j]=min(dp[i][j],dp[i-1][j]+double(b[i])/double(i-j));
            else dp[i][j]=min(dp[i][j],dp[i-1][j]);
        }
    }
    return dp[n][k-x];
}

void solve() {
    cin>>n>>k;
    pair<int,int> p[n+1];
    for(int i=1;i<=n;i++){
        cin>>a[i]>>b[i];
        b[i]=(b[i]==-1 ? inf : b[i]);
        p[i]={b[i],a[i]};
    }
    sort(p+1,p+n+1);
    for(int i=1;i<=n;i++){
        a[i]=p[i].S;
        b[i]=p[i].F;
    }

    double ans=inf;
    for(int x=0;x<=k;x++){
        ans=min(ans,get(x));
    }

    cout<<fixed<<setprecision(6)<<ans;
    return;
}
signed main() {
    //   freopen("deleg.in","r",stdin);    
    // freopen("deleg.out","w",stdout);
    SS
    int t = 1;
    if (T) {
        cin >> t;
    }
    while (t--) {
        solve();
    }
}
/*
dp[i][j]=dp[i-1][j-1]+a[i]/(x+1)
i need to take i as cooperator only on prefix and do not skip any free numbers on prefix
i can take number on prefix if already numbers of non taken elements on prefix is less or equal tham x
so i can take number if i-j<=x
dp[i][j]=dp[i-1][j]+b[i]/(i-j);
*/

• 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