Submission #51377

#	Time	Username	Problem	Language	Result	Execution time	Memory
51377		spencercompton	Popeala (CEOI16_popeala)	C++17	100 / 100	428 ms	17192 KiB

popeala

#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

int main() {
    long t, n, s;
    cin >> n >> t >> s;
    long long cumu[t+1];
    cumu[0] = 0;
    for(long i = 1; i<=t; i++){
        cin >> cumu[i];
        cumu[i]+=cumu[i-1];
    }
    bool got[n][t];
    for(long i = 0; i<n; i++){
        string str;
        cin >> str;
        for(long j = 0; j<t; j++){
            got[i][j] = str[j]=='1';
        }
    }
    vector<vector<long long> > best;
    best.resize(s+1);
    
    for(long a = 0; a<=s; a++){
        for(long b = 0; b<=t; b++){
            best[a].push_back(2000000002L);
        }
    }
    best[0][0] = 0L;
    for(long i = 0; i<s; i++){
        long long last[n];
        for(long j= 0; j<n; j++){
            last[j] = -1L;
        }
        long long inactive[t+1];
        long long mini[n+1];
        for(long j = 0; j<=n; j++){
            mini[j] = 2000000002L;
        }
        for(long j = 0; j<t; j++){
            mini[0] = min(mini[0],best[i][j]);
            inactive[j] = 0;
            for(long k = 0; k<=n; k++){
                mini[k] += (long long)(n-k)*(cumu[j+1]-cumu[j]);
            }
            for(long k = 0; k<n; k++){
                if(got[k][j]){
                    continue;
                }
                for(long l = j; l>=0; l--){
                    if(l>last[k]){
                        inactive[l]++;
                        mini[inactive[l]] = min(mini[inactive[l]],best[i][l]+(cumu[j+1]-cumu[l])*(long long)(n-inactive[l]));
                    }
                    else{
                        break;
                    }
                }
                last[k] = j;
            }
            for(long k = 0; k<=n; k++){
                best[i+1][j+1] = min(best[i+1][j+1],mini[k]);
            }
        }
    }
    for(long i = 1; i<=s; i++){
        cout << best[i][t] << endl;
    }
    return 0;
}

• Subtask 1: 8 / 8
• Subtask 2: 9 / 9
• Subtask 3: 9 / 9
• Subtask 4: 74 / 74