Submission #546999

#	Time	Username	Problem	Language	Result	Execution time	Memory
546999		ngrace	Copy and Paste 3 (JOI22_copypaste3)	C++14	25 / 100	3073 ms	40196 KiB

copypaste3

#include <vector>
#include <iostream>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <string>
using namespace std;
#define v vector
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second


int main(){
    int N;
    cin>>N;
    string S;
    cin>>S;
    ll A,B,C;
    cin>>A>>B>>C;

    map<string, int> subToInd;
    map<int, set<int>> indToStarts;
    int subInd=1;
    for(int i=0; i<N; i++){
        for(int s=1; s<=N-i; s++){
            string sub;
            for(int j=i; j<i+s; j++) sub+=S[j];
            if(subToInd[sub]==0) subToInd[sub]=subInd++;
            indToStarts[subToInd[sub]].insert(i);
        }
    }

    v<v<int>> nextOccur(N,v<int>(N+1,-1));
    for(int i=N-1; i>=0; i--){
        for(int s=1; s<=N-i; s++){
            string sub;
            for(int j=i; j<i+s; j++) sub+=S[j];
            int ind=subToInd[sub];
            auto next = indToStarts[ind].lower_bound(i+s);
            if(next!=indToStarts[ind].end()) nextOccur[i][s]=(*next);
        }
    }

    v<v<ll>> dp(N, v<ll>(N,1e17));
    for(int s=1; s<=N; s++){
        for(int l=0; l<N; l++){
            int r=l+s-1;
            if(r>=N) break;
            if(l==r){
                dp[l][r]=A;
                continue;
            }
            dp[l][r] = min(dp[l][r], A+dp[l+1][r]);
            dp[l][r] = min(dp[l][r], dp[l][r-1]+A);
            //only need to account for copying sub strings starting at l, any others are accounted for above
            for(int k=1; k<s; k++){
                ll num=1, nex=nextOccur[l][k];
                while(nex!=-1 && nex+k-1<=r){
                    num++;
                    nex=nextOccur[nex][k];
                }
                dp[l][r] = min(dp[l][r], A*(s-num*k)+num*C+dp[l][l+k-1]+B);
            }
        }
    }

    /**for(int l=0; l<N; l++){
        for(int r=l; r<N; r++){
            cout<<l<<" "<<r<<" "<<dp[l][r]<<endl;
        }
    }**/

    cout<<dp[0][N-1]<<endl;
}

• Subtask 1: 1 / 1
• Subtask 2: 0 / 5
• Subtask 3: 14 / 14
• Subtask 4: 10 / 10
• Subtask 5: 0 / 32
• Subtask 6: 0 / 38