Submission #1158985

#	Time	Username	Problem	Language	Result	Execution time	Memory
1158985		pan	Zemljište (COCI22_zemljiste)	C++20	0 / 70	0 ms	320 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;

ll dp[505][505];

// Returns the sum of the submatrix with top-left (x1, y1) and bottom-right (x2, y2)
inline ll cal(ll x1, ll y1, ll x2, ll y2) {
    return dp[x2][y2] - dp[x1-1][y2] - dp[x2][y1-1] + dp[x1-1][y1-1];
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    
    ll r, c, a, b;
    cin >> r >> c >> a >> b;
    
    // Build the prefix sum array
    for (ll i = 1; i <= r; ++i) {
        for (ll j = 1; j <= c; ++j) {
            cin >> dp[i][j];
            dp[i][j] += dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1];
        }
    }
    
    ll ans = LLONG_MAX;
    // Iterate over all choices for bottom-right (i, j) and left column k
    for (ll i = 1; i <= r; ++i) {
        for (ll j = 1; j <= c; ++j) {
            for (ll k = 1; k <= j; ++k) {
                // Instead of binary searching for the optimal top row,
                // we use the observation that the best cost is determined
                // by the maximum (using top row 1) and minimum (using top row i) sums.
                ll S_top = cal(1, k, i, j);
                ll S_bot = cal(i, k, i, j);
                ll curCost;
                if(S_top < a) 
                    curCost = (a - S_top) + (b - S_top);
                else if(S_bot > b)
                    curCost = (S_bot - a) + (S_bot - b);
                else 
                    curCost = b - a; // there is some x with F(x) in [a, b]
                ans = min(ans, curCost);
            }
        }
    }
    
    cout << ans << endl;
    return 0;
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 20
• Subtask 3: 0 / 40