Submission #938430

#	Time	Username	Problem	Language	Result	Execution time	Memory
938430		GrindMachine	Soccer Stadium (IOI23_soccer)	C++17	48 / 100	4595 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

/*

already know some key ideas

*/

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

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

    rep(l1,n){
        rev(r1,n+1,l1+1){
            // l1 --> l1+1
            rep1(l2,n){
                for(int r2 = l2; r2 <= n; ++r2){
                    rep1(l3,l2){
                        for(int r3 = r2; r3 <= n; ++r3){
                            if(p[l1+1][r3]-p[l1+1][l3-1]) conts;
                            amax(dp[l1+1][r1][l3][r3],dp[l1][r1][l2][r2]+r3-l3+1);
                        }
                    }
                }
            }

            // r1 --> r1-1
            rep1(l2,n){
                for(int r2 = l2; r2 <= n; ++r2){
                    rep1(l3,l2){
                        for(int r3 = r2; r3 <= n; ++r3){
                            if(p[r1-1][r3]-p[r1-1][l3-1]) conts;
                            amax(dp[l1][r1-1][l3][r3],dp[l1][r1][l2][r2]+r3-l3+1);
                        }
                    }
                }
            }
        }
    }

    int ans = 0;
    rep(l1,n+1){
        int r1 = l1+1;
        rep1(l2,n){
            for(int r2 = l2; r2 <= n; ++r2){
                amax(ans,dp[l1][r1][l2][r2]);
            }
        }
    }

    return ans;
}

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