Submission #1238293

#	Time	Username	Problem	Language	Result	Execution time	Memory
1238293		ema_nicole	Quality Of Living (IOI10_quality)	C++17	20 / 100	6 ms	1352 KiB

quality

#pragma GCC optimize ("Ofast", "unroll-loops")
#include <iostream>
#include <cmath>

using namespace std;
const int NMAX = 3000;

int maxx, ans, mid, LOG;
int aib[NMAX * NMAX + 2], frecv[NMAX * NMAX + 2];

void update2(int pos, int val) {
    while(pos <= maxx) {
        frecv[pos] += val;
        pos += pos & (-pos);
    }
}
int query(int pos) {
    int sum = 0;
    while(pos > 0) {
        sum += frecv[pos];
        pos -= pos & (-pos);
    }
    return sum;
}

void update(int pos, int val) {
    while(pos <= maxx) {
        aib[pos] += val;
        pos += pos & (-pos);
    }
}

int check() { ///de gasit mediana
    int pos = 0;
    int cnt = 0; ///suma curenta
    for(int bit = LOG; bit >= 0; bit--) {
        if(pos + (1 << bit) <= maxx && cnt + aib[pos + (1 << bit)] < mid) {
            pos += (1 << bit);
            cnt += aib[pos];
        }
    }
    return pos + 1;
}

int v[NMAX + 2][NMAX + 2];
int rectangle(int n, int m, int r, int c, int q[NMAX + 1][NMAX + 1]) {
    if(c < r) { ///swap
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                v[j][i] = q[i - 1][j - 1];
        swap(n, m);
        swap(r, c); ///vrem r < c pt cat mai putine switch-uri
        ///TBD DC NU ERA N < M!!!
    }
    else {
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                v[i][j] = q[i - 1][j - 1];
    }
    maxx = n * m;
    ans = n * m + 1;
    mid = (r * c + 1) / 2;
    LOG = log2(n * m);
    ///INIT PT AIB
    for(int i = 1; i <= r; i++) {
        for(int j = 1; j <= c; j++) {
            update(v[i][j], 1);
        }
    }
    for(int i = 1; i <= n * m; i++)
        update2(i, 1);

    int ultlin = n - r + 1, ultcol;
    bool sens = 0; ///0 --> dr, 1 --> st
    if(ultlin % 2 == 0)
        ultcol = 1;
    else
        ultcol = m - c + 1;

    int ramase = 0;
    int i = 1, j = 1; ///coltul din st sus
    while(1) {
        //cout << i << " " << j << "  " << check() << '\n';
        int dif = ans;
        ans = min(ans, check()); ///ver unde am ajuns
        if(dif > ans) ///s-a schimbat --> updatam
            ramase = query(ans - 1) - query(mid - 1);
        if(!ramase) ///nu mai putem imbunatati
            return ans;

        if(i == ultlin && j == ultcol) ///am term, done
            break;
        if(!sens) { ///o luam in dr
            if(j + c - 1 == m) { ///am ajuns la fin
                for(int col = j; col <= m; col++) {
                    update(v[i][col], -1); ///scoatem vechi
                    update2(v[i][col], -1);
                    if(mid <= v[i][col] && v[i][col] < ans) {
                        ramase--;
                        if(!ramase)
                            return ans;
                    }
                    update(v[i + r][col], 1); ///luam nou
                }
                i++;
                sens = 1;
            }
            else { ///scoatem col veche
                for(int lin = i; lin <= i + r - 1; lin++) {
                    update(v[lin][j], -1);
                    update2(v[lin][j], -1);
                    update(v[lin][j + c], 1);
                }
                if(mid <= v[i][j] && v[i][j] < ans) {
                    ramase--;
                    if(!ramase)
                        return ans;
                }
                j++;
            }
        }
        else { ///o luam in st
            if(j == 1) { ///fin
                for(int col = 1; col <= c; col++) {
                    update(v[i][col], -1);
                    update2(v[i][col], -1);
                    if(mid <= v[i][col] && v[i][col] < ans) {
                        ramase--;
                        if(!ramase)
                            return ans;
                    }
                    update(v[i + r][col], 1);
                }
                i++;
                sens = 0;
            }
            else { ///ne shiftam la dr
                for(int lin = i; lin <= i + r - 1; lin++) {
                    update(v[lin][j + c - 1], -1);
                    update2(v[lin][j + c - 1], -1);
                    update(v[lin][j - 1], 1);
                }
                if(mid <= v[i][j + c - 1] && v[i][j + c - 1] < ans) {
                    ramase--;
                    if(!ramase)
                        return ans;
                }
                j--;
            }
        }
    }
    return ans;
}

• Subtask 1: 20 / 20
• Subtask 2: 0 / 20
• Subtask 3: 0 / 20
• Subtask 4: 0 / 20
• Subtask 5: 0 / 20