Submission #1028167

#	Time	Username	Problem	Language	Result	Execution time	Memory
1028167		anango	Boarding Passes (BOI22_passes)	C++17	0 / 100	1 ms	348 KiB

passes

#include <bits/stdc++.h>
#define int long long
using namespace std;
int INF = 1LL<<62;

void solve() {

    string S; cin >> S;
    int n = S.size();
    vector<int> ar(n); for (int i=0; i<n; i++) ar[i] = (S[i]-'A');
    int G = *max_element(ar.begin(), ar.end()); G++;
    int pref[n+1][G]; //count of character c, before index i
    int suff[n+1][G]; //count of character c, after or equal to index i
    for (int i=0; i<=n; i++) {
        for (int j=0; j<G; j++) {
            pref[i][j] = suff[i][j] = 0;
        }
    }
    for (int i=0; i<n; i++) {
        for (int j=0; j<=G; j++) {
            if (j==ar[i]) {
                pref[i+1][j] = pref[i][j]+1;
            }
            else {
                pref[i+1][j] = pref[i][j];
            }
        }
    }
    
    for (int i=n-1; i>=0; i--) {
        for (int j=0; j<G; j++) {
            if (j==ar[i]) {
                suff[i][j] = suff[i+1][j]+1;
            }
            else {
                suff[i][j] = suff[i+1][j];
            }
        }
    }
    //everyone before L boards from the front, everyone after or equal to it boards from the back. 
    vector<vector<int>> counts(G,vector<int>(G,0));     //counts[i][j] = sum of subsequences of (i,j) before l, and subsequences of (j,i) after L
    //counts[i][j] = sum of pref[k][j] for k<=L and ar[k]=i, and analogous expression for the suffix
    //first, get initial count values (for everyone boarding from the back)
    
    for (int j=0; j<G; j++) {
        for (int k=0; k<n; k++) {
            int i = ar[k];
            //cout << "adding " << i <<" " << j <<" " << k <<" " << j << " " << suff[k][j] << endl;
            counts[i][j] += suff[k+1][j];
        }
    }
    /*for (int i=0; i<G; i++) {
        for (int j=0; j<G; j++) {
            cout << i <<" " << j <<" " << counts[i][j] << endl;
        }
    }*/
    int ans = 1LL<<62;
    vector<int> dp;
    for (int L=0; L<n; L++) {

        //counts[i][j] = number of swaps required if i goes after j

        dp=vector<int>(1<<G,INF); //dp[mask] = min cost if this mask is at the beginning of the ordering
        dp[0] = 0;
        for (int mask=0; mask<1<<G; mask++) {
            for (int addon=0; addon<G; addon++) {
                if (mask&(1<<addon)) continue;
                int al = 0;
                for (int j=0; j<G; j++) {
                    if (mask&(1<<j)) {
                        al+=counts[addon][j];
                    }
                }
                //cout << mask <<" " << addon <<" " << al <<endl;
                dp[mask | (1<<addon)] = min(dp[mask | (1<<addon)],dp[mask]+al);
            }
        }
        int self=0;
        for (int i=0; i<G; i++) {
            self+=counts[i][i];
        }
        ans=min(ans,2*dp[(1<<G)-1]+self);

        //transfer L to the right
        int cur = ar[L];
        for (int j=0; j<G; j++) {
            int delta = suff[L+1][j];
            int delta2 = pref[L][j];
            counts[cur][j] -= delta-delta2;
        }
        
        /*cout << "TRANSFERRED TO " << L+1 << endl;
        for (int i=0; i<G; i++) {
            for (int j=0; j<G; j++) {
                cout << i <<" " << j <<" " << counts[i][j] << endl;
            }
        }
        cout << endl;*/
    }
    dp=vector<int>(1<<G,INF); //dp[mask] = min cost if this mask is at the beginning of the ordering
    dp[0] = 0;
    for (int mask=0; mask<1<<G; mask++) {
        for (int addon=0; addon<G; addon++) {
            if (mask&(1<<addon)) continue;
            int al = 0;
            for (int j=0; j<G; j++) {
                if (mask&(1<<j)) {
                    al+=counts[addon][j];
                }
            }
            //cout << mask <<" " << addon <<" " << al <<endl;
            dp[mask | (1<<addon)] = min(dp[mask | (1<<addon)],dp[mask]+al);
        }
    }
    int self=0;
    for (int i=0; i<G; i++) {
        self+=counts[i][i];
    }
    ans=min(ans,2*dp[(1<<G)-1]+self);

    long double a2 = ans;
    a2/=(long double)2;
    cout << setprecision(100) << a2 << endl;
}

signed main() {
    int local=1;
    if (local) {
        // for getting input from input.txt
        freopen("input.txt", "r", stdin);
        // for writing output to output.txt
        freopen("output.txt", "w", stdout);
    }
    /*#ifdef ONLINE_JUDGE
    	ios_base::sync_with_stdio(false);
    	cin.tie(NULL);
    #endif*/ //fast IO
    solve();
    
}

Compilation message (stderr)

passes.cpp: In function 'int main()':
passes.cpp:130:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  130 |         freopen("input.txt", "r", stdin);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~
passes.cpp:132:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  132 |         freopen("output.txt", "w", stdout);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 0 / 5
• Subtask 2: 0 / 25
• Subtask 3: 0 / 30
• Subtask 4: 0 / 40