Submission #1065845

#	Time	Username	Problem	Language	Result	Execution time	Memory
1065845		GrindMachine	축구 경기장 (IOI23_soccer)	C++17	64 / 100	4558 ms	102472 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 = 2e3 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "soccer.h"

int mnl[N][N], mxr[N][N], mnu[N][N];

int extendl(int i, int j, int c){
    int mx = -inf1;
    for(int r = i; r <= j; ++r){
        amax(mx,mnl[r][c]);
    }
    return mx;
}

int extendr(int i, int j, int c){
    int mn = inf1;
    for(int r = i; r <= j; ++r){
        amin(mn,mxr[r][c]);
    }
    return mn;
}

int n;
map<array<int,3>,int> mp;

int go(int i, int j, int l){
    if(i > j) return 0;

    l = extendl(i,j,l);
    int r = extendr(i,j,l);
    array<int,3> ar = {i,j,l};
    if(mp.count(ar)) return mp[ar];

    int len = r-l+1;

    // shrink j
    int ans = len+go(i,j-1,l);

    // shrink i till l expansion possible
    if(l-1){
        int p = mnu[j][l-1];
        amax(ans,len*(p-i)+go(p,j,l));
    }

    // shrink i till r expansion possible
    if(r+1 <= n){
        int p = mnu[j][r+1];
        amax(ans,len*(p-i)+go(p,j,l));
    }

    return mp[ar] = ans;
}

int biggest_stadium(int n_, std::vector<std::vector<int>> A)
{
    n = n_;

    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];

    rep1(i,n) mnl[i][0] = 1;
    rep1(i,n){
        rep1(j,n){
            if(!a[i][j]){
                mnl[i][j] = mnl[i][j-1];
            }
            else{
                mnl[i][j] = j+1;
            }
        }
    }

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

    rep1(j,n) mnu[0][j] = 1;
    rep1(i,n){
        rep1(j,n){
            if(!a[i][j]){
                mnu[i][j] = mnu[i-1][j];
            }
            else{
                mnu[i][j] = i+1;
            }
        }
    }

    int ans = 0;
    vector<int> lx(n+5), rx(n+5);

    rep1(i,n){
        rep1(j,n) lx[j] = 1, rx[j] = n;

        {
            stack<int> stk;
            rep1(j,n){
                while(!stk.empty() and mnu[i][j] > mnu[i][stk.top()]){
                    int k = stk.top();
                    rx[k] = j-1;
                    stk.pop();
                }
                stk.push(j);
            }
        }

        {
            stack<int> stk;
            rev(j,n,1){
                while(!stk.empty() and mnu[i][j] > mnu[i][stk.top()]){
                    int k = stk.top();
                    lx[k] = j+1;
                    stk.pop();
                }
                stk.push(j);
            }
        }

        rep1(j,n){
            int l = lx[j];
            amax(ans,go(mnu[i][j],i,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