Cod sursa(job #1068519)

Utilizator AlexandruValeanuAlexandru Valeanu AlexandruValeanu Data 28 decembrie 2013 14:03:05
Problema Balans Scor 35
Compilator cpp Status done
Runda Arhiva de probleme Marime 2.15 kb
#include <iostream>
#include <fstream>
#include <deque>
#include <iomanip>

using namespace std;

const int Nmax = 302;

int N, M, R, C;

deque <int> deck;

long long A[Nmax][Nmax];
long long sum[Nmax][Nmax];
long long sume_coloana[Nmax];
long long sumCol[Nmax];

int valid( long long cost )
{
    long long maxim = -10000000;


    for ( int i = 1; i <= 2 * N; ++i )
    {
        for ( int j = i + R - 1; j - i < N && j <= 2 * N; ++j )
        {
            for ( int k = 1; k <= 2 * M; ++k )
            {
                sume_coloana[k] = sum[j][k] - sum[i - 1][k] - cost * ( j - i + 1 );
                sumCol[k] = sumCol[k - 1] + sume_coloana[k];
            }

            /// intre C-M
            deck.clear();

            for ( int k = C; k <= 2 * M; ++k )
            {
                while ( deck.size() && sumCol[ deck.back() ] > sumCol[k - C] ) deck.pop_back();
                while ( deck.size() && deck.front() < k - M ) deck.pop_front();

                deck.push_back( k - C );

                maxim = max( maxim, sumCol[k] - sumCol[ deck.front() ] );
            }
        }
    }

    return ( maxim >= 0 );
}

long long cautare_binara()
{
    long long st = 0;
    long long dr = 100001 * 1000;
    long long sol = 0;
    long long m;

    while ( st <= dr )
    {
        m = ( st + dr ) / 2;

        if ( valid( m ) )
        {
            st = m + 1;
            sol = m;
        }
        else
        {
            dr = m - 1;
        }
    }

    return sol;
}

int main()
{
    ifstream f("balans.in");
    ofstream g("balans.out");

    f >> N >> M >> R >> C;

    for ( int i = 1; i <= N; ++i )
            for ( int j = 1; j <= M; ++j )
            {
                f >> A[i][j];

                A[i][j] *= 1000;
                A[i + N][j] = A[i][j + M] = A[i + N][j + M] = A[i][j];
            }

    for ( int i = 1; i <= 2 * N; ++i )
    {
        for ( int j = 1; j <= 2 * N; ++j )
        {
            sum[i][j] = sum[i - 1][j] + A[i][j];
        }
    }

    g << fixed << setprecision( 3 );
    g << (double)cautare_binara() / 1000;

    return 0;
}