제출 #938398

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
938398		GrindMachine	축구 경기장 (IOI23_soccer)	C++17	0 / 100	2035 ms	2097156 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 dp1[n][n][n];
    memset(dp1,0,sizeof dp1);
    rep(i,n){
        rev(l,n-1,0){
            for(int r = l; r < n; ++r){
                if(a[i][r]) break;
                if(l+1 < n){
                    amax(dp1[i][l][r],dp1[i][l+1][r]);
                }
                if(r-1 >= 0){
                    amax(dp1[i][l][r],dp1[i][l][r-1]);
                }
                if(i){
                    amax(dp1[i][l][r],dp1[i-1][l][r]);
                }
            }
        }

        rep(l,n){
            for(int r = l; r < n; ++r){
                if(a[i][r]) break;
                dp1[i][l][r] += r-l+1;
            }
        }
    }

    int dp2[n][n], dp3[n][n];
    memset(dp2,0,sizeof dp2);
    memset(dp3,0,sizeof dp3);
    int ans = 0;

    rev(i,n-1,0){
        rev(l,n-1,0){
            for(int r = l; r < n; ++r){
                if(a[i][r]) break;

                if(l+1 < n){
                    amax(dp3[l][r],dp3[l+1][r]);
                }
                if(r-1 >= 0){
                    amax(dp3[l][r],dp3[l][r-1]);
                }
                if(i+1 < n){
                    amax(dp3[l][r],dp2[l][r]);
                }
            }
        }

        rep(l,n){
            for(int r = l; r < n; ++r){
                if(a[i][r]) break;
                int val = dp1[i][l][r]+dp3[l][r];
                amax(ans,val);
                dp3[l][r] += r-l+1;
            }
        }

        rep(l,n){
            for(int r = l; r < n; ++r){
                dp2[l][r] = dp3[l][r];
                dp3[l][r] = 0;
            }
        }
    }

    return ans;
}

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