Submission #1181188

#	Time	Username	Problem	Language	Result	Execution time	Memory
1181188		anteknne	Sandcastle 2 (JOI22_ho_t5)	C++20	10 / 100	5095 ms	1860 KiB

Main

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef long double ld;

#define pb push_back
#define pii pair<int, int>
#define pll pair<ll, ll>
#define st first
#define nd second
#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define debug false

const int MAXW = 50 * 1000;

vector<int> a[MAXW];
vector<pii> przesuniecia = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

int ok (int x1, int y1, int x2, int y2) {
    pii pos = {x1, y1};
    for (int i = x1; i <= x2; i ++) {
        for (int j = y1; j <= y2; j ++) {
            if (a[i][j] > a[pos.st][pos.nd]) {
                pos = {i, j};
            }
        }
    }

    int ile = 1;
    while (true) {
        pii maks = {-1, -1};
        for (auto x : przesuniecia) {
            int nx = pos.st + x.st;
            int ny = pos.nd + x.nd;
            if (nx >= x1 && nx <= x2 && ny >= y1 && ny <= y2) {
                if (a[nx][ny] > a[pos.st][pos.nd]) {
                    continue;
                }
                if (maks.st == -1) {
                    maks = {nx, ny};
                } else  if (a[nx][ny] > a[maks.st][maks.nd]) {
                    maks = {nx, ny};
                }
            }
        }
        if (maks.st == -1) {
            break;
        }
        ile ++;
        pos = maks;
    }
    return (ile == (x2 - x1 + 1) * (y2 - y1 + 1));
}

int main () {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    int n, m, x;
    cin >> n >> m;

    for (int i = 0; i < n; i ++) {
        for (int j = 0; j < m; j ++) {
            cin >> x;
            a[i].pb(x);
        }
    }

    int wyn = 0;
    for (int i = 0; i < n; i ++) {
        for (int j = 0; j < m; j ++) {
            for (int k = i; k < n; k ++) {
                for (int l = j; l < m; l ++) {
                    wyn += ok(i, j, k, l);
                }
            }
        }
    }

    cout << wyn << "\n";



    return 0;
}

• Subtask 1: 0 / 9
• Subtask 2: 10 / 10
• Subtask 3: 0 / 5
• Subtask 4: 0 / 56
• Subtask 5: 0 / 20