Submission #1065710

#	Time	Username	Problem	Language	Result	Execution time	Memory
1065710		GrindMachine	Soccer Stadium (IOI23_soccer)	C++17	64 / 100	1252 ms	2097152 KiB

soccer

#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;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
https://blog.brucemerry.org.za/2023/11/ioi-2023-day-1.html

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "soccer.h"

int biggest_stadium(int n, std::vector<std::vector<int>> A)
{
    int a[n+5][n+5];
    memset(a,0,sizeof a);
    rep1(i,n) rep1(j,n) a[i][j] = A[i-1][j-1];

    int mx_extend[n+5][n+5];
    memset(mx_extend,0,sizeof mx_extend);
    rep1(i,n) mx_extend[i][n+1] = n;

    rev(j,n,1){
        rep1(i,n){
            if(!a[i][j]){
                mx_extend[i][j] = mx_extend[i][j+1];
            }
            else{
                mx_extend[i][j] = j-1;
            }
        }
    }

    int p[n+5][n+5];
    memset(p,0,sizeof p);
    rep1(i,n){
        rep1(j,n){
            p[i][j] = p[i-1][j]+a[i][j];
        }
    }

    int dp[n+5][n+5][n+5];
    memset(dp,0,sizeof dp);

    int ans = 0;

    rep1(l,n){
        rev(i,n,1){
            int r = inf1;
            for(int j = i; j <= n; ++j){
                amin(r,mx_extend[j][l]);
                if(p[j][l]-p[i-1][l]) conts;
                int len = r-l+1;
                dp[i][j][l] = max(dp[i+1][j][l],dp[i][j-1][l])+len;
                amax(dp[i][j][l],dp[i][j][l-1]);
                amax(ans,dp[i][j][l]);
            }
        }
    }

    return ans;
}

• Subtask 0: 0 / 0
• Subtask 1: 0 / 6
• Subtask 2: 8 / 8
• Subtask 3: 22 / 22
• Subtask 4: 18 / 18
• Subtask 5: 16 / 16
• Subtask 6: 0 / 30