Submission #654844

#	Time	Username	Problem	Language	Result	Execution time	Memory
654844		Lobo	Boarding Passes (BOI22_passes)	C++17	100 / 100	473 ms	390464 KiB

passes

#include<bits/stdc++.h>
using namespace std;
const long long inf = (long long) 1e18 + 10;
const int inf1 = (int) 1e9 + 10;
#define int long long
#define dbl long double
#define endl '\n'
#define sc second
#define fr first
#define mp make_pair
#define pb push_back
#define all(x) x.begin(), x.end()

const int maxn1 = 1e5+10;
const int maxn = 1e5+10;
const int maxg = 17;

int n, a[maxn1];
int pf[(maxg)][maxn], dp[(1<<maxg)], prepl[maxg][maxg][maxn], prepr[maxg][maxg][maxn], pfl[maxg][maxn], sfr[maxg][maxn];
vector<int> pos[maxg];

void solve() {
    string s; cin >> s;
    n = s.size();
    vector<char> cc;
    for(auto x : s) cc.pb(x);
    sort(all(cc));
    cc.erase(unique(all(cc)),cc.end());
    for(int i = 1; i <= n; i++) {
        a[i] = lower_bound(all(cc),s[i-1])-cc.begin();
        pos[a[i]].pb(i);
    }
    int g = cc.size();
    if(g == 1) {
        dbl ans = 0;
        for(int i = 1; i <= n; i++) ans+= min((dbl) (i-1)/2,(dbl) (n-i)/2);
        cout.setf(ios::fixed);
        cout.precision(10);
        cout << ans << endl;
        return;
    }

    // for(int mask = 1; mask < (1<<g); mask++) {
    //     pf[mask][0] = 0;
    //     for(int i = 1; i <= n; i++) {
    //         pf[mask][i] = pf[mask][i-1];
    //         if((1<<a[i])&mask) pf[mask][i]++;
    //     }
    // }
    for(int i = 0; i < g; i++) {
        for(int x = 1; x <= n; x++) {
            pf[i][x] = pf[i][x-1];
            if(a[x] == i) pf[i][x]++;
        }
    }

    for(int i = 0; i < g; i++) {
        for(int j = 0; j < g; j++) {
            for(int x = 1; x <= n; x++) {
                prepl[i][j][x] = prepl[i][j][x-1];
                if(pf[i][x]-pf[i][x-1]) prepl[i][j][x]+= pf[j][x-1];
            }
        }

        for(int x = 1; x <= n; x++) {
            pfl[i][x] = pfl[i][x-1];
            if(pf[i][x]-pf[i][x-1]) pfl[i][x]+= ((dbl) pf[i][x-1]);
        }
    }

    for(int i = 0; i < g; i++) {
        for(int j = 0; j < g; j++) {
            for(int x = n; x >= 1; x--) {
                prepr[i][j][x] = prepr[i][j][x+1];
                if(pf[i][x]-pf[i][x-1]) prepr[i][j][x]+= pf[j][n]-pf[j][x];
            }
        }
        for(int x = n; x >= 1; x--) {
            sfr[i][x] = sfr[i][x+1];
            if(pf[i][x]-pf[i][x-1]) sfr[i][x]+= ((dbl) pf[i][n]-pf[i][x]);
        }
    }

    for(int mask = 1; mask < (1<<g); mask++) {

        dp[mask] = inf;
        for(int i = 0; i < g; i++) {
            if(((1<<i)&mask) == 0) continue;
            int mask1 = mask-(1<<i);

            int l = 1;
            int r = pos[i].size();
            int ans = 0;
            // I want to find the greatest i such that I going to 
            // the left if better than i going to the right
            while(l <= r) {
                int mid = (l+r)/2;

                int x = pos[i][mid-1];
                // dbl ansl = pf[mask1][x]+((dbl) j-1)/2;
                int ansl = ((dbl) mid-1); 
                for(int j = 0; j < g; j++) {
                    if(((1<<j)&mask1) == 0) continue;
                    ansl+= 2*(pf[(j)][x]);
                }

                // dbl ansr = pf[mask1][n]-pf[mask1][x]+((dbl) sz-j)/2;
                int ansr = ((dbl) pos[i].size()-mid); 
                for(int j = 0; j < g; j++) {
                    if(((1<<j)&mask1) == 0) continue;
                    ansr+= 2*(pf[(j)][n]-pf[(j)][x]);
                }

                if(ansl <= ansr) {
                    ans = mid;
                    l = mid+1;
                }
                else {
                    r = mid-1;
                }
            }

            int ans1 = dp[mask1];
            int x = 0;
            if(ans != 0) x = pos[i][ans-1];
            if(ans != 0) {
                ans1+= pfl[i][x];
                for(int j = 0; j < g; j++) {
                    if(((1<<j)&mask1) == 0) continue;
                    ans1+= 2*prepl[i][j][pos[i][ans-1]];
                }
            }
            if(ans != pos[i].size()) {
                ans1+= sfr[i][x+1];
                for(int j = 0; j < g; j++) {
                    if(((1<<j)&mask1) == 0) continue;
                    ans1+= 2*prepr[i][j][x+1];
                }
            }
            // for(int j = ans+1; j <= pos[i].size(); j++) {
            //     int x = pos[i][j-1];
            //     ans1+= pf[mask1][n]-pf[mask1][x]+((dbl) sz-j)/2;
            //     // cout << " " << min(pf[mask1][x]+((dbl) j-1)/2, pf[mask1][n]-pf[mask1][x]+((dbl) pos[i].size()-j)/2) << endl;
            // }
            dp[mask] = min(dp[mask],ans1);

        }



        // dp[mask] = inf;
        // for(int i = 0; i < g; i++) {
        //     if(((1<<i)&mask) == 0) continue;
        //     int mask1 = mask-(1<<i);
        //     dbl ans1 = dp[mask1];
        //     int sz = pos[i].size();
        //     for(int j = 1; j <= pos[i].size(); j++) {
        //         int x = pos[i][j-1];
        //         ans1+= min(pf[mask1][x]+((dbl) j-1)/2, pf[mask1][n]-pf[mask1][x]+((dbl) sz-j)/2);
        //         // cout << " " << min(pf[mask1][x]+((dbl) j-1)/2, pf[mask1][n]-pf[mask1][x]+((dbl) pos[i].size()-j)/2) << endl;
        //     }
        //     dp[mask] = min(dp[mask],ans1);
        //     // cout << mask << " " << i << " = " << ans1 << " " << pf[mask1][1] << endl;
        // }
    }
    cout.setf(ios::fixed);
    cout.precision(10);
    cout << (dbl) dp[(1<<g)-1]/2 << endl;
}

int32_t main() {
    ios::sync_with_stdio(false); cin.tie(0);

    // freopen("in.in", "r", stdin);
    // freopen("out.out", "w", stdout);
    int tt = 1;
    // cin >> tt;
    while(tt--) {
        solve();
    }

}

Compilation message (stderr)

passes.cpp: In function 'void solve()':
passes.cpp:133:20: warning: comparison of integer expressions of different signedness: 'long long int' and 'std::vector<long long int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
  133 |             if(ans != pos[i].size()) {
      |                ~~~~^~~~~~~~~~~~~~~~

• Subtask 1: 5 / 5
• Subtask 2: 25 / 25
• Subtask 3: 30 / 30
• Subtask 4: 40 / 40