제출 #1305596

#제출 시각아이디문제언어결과실행 시간메모리
1305596darkdevilvaqifStranded Far From Home (BOI22_island)C++20
100 / 100
156 ms43492 KiB
/*
  _      __        __         __        ____                    ___    _                   __
 | | /| / / ___ _ / /_ ____  / /       / __ \  ___  ___        / _ \  (_) ___  ____ ___   / /
 | |/ |/ / / _ `// __// __/ / _ \     / /_/ / / _ \/ -_)      / ___/ / / / -_)/ __// -_) /_/ 
 |__/|__/  \_,_/ \__/ \__/ /_//_/     \____/ /_//_/\__/      /_/    /_/  \__/ \__/ \__/ (_)  
                                                                                             
*/

#pragma GCC optimize("O3")
#pragma GCC optimize ("unroll-loops")
// #pragma GCC target("avx2")
#include <bits/stdc++.h>
#define ll long long
#define ull unsigned ll
#define ld long double
#define all(v, l) v.begin() + l, v.end()
#define rall(v, l) v.rbegin(), v.rend() - l
#define pb push_back
#define rsz resize
#define fi first
#define se second
#define LMAX LLONG_MAX
#define LMIN LLONG_MIN
#define IMAX INT_MAX
#define IMIN INT_MIN
#define endl "\n"
#define newline cout << endl;
using namespace std;
 
// constants

// functions
ll GCD(ll A, ll B);
ll LCM(ll A, ll B);
ll power(ll base, ll exponent);

// structs
struct graph
{
    vector <ll> lvl;
    vector <int> par;
    vector <vector <int> > g;
    
    void prep(int n)
    {
        par.rsz(n);
        g.rsz(n);
        lvl.rsz(n);
    }
    
    void add(int u, int v)
    {
        g[u].pb(v);
        g[v].pb(u);
    }
    
    int get(int v)
    {
        if (par[v] == v)
        {
            return v;
        }
        return par[v] = get(par[v]);
    }
    
    void merge(int u, int v)
    {
        lvl[u] += lvl[v];
        par[v] = u;
    }
    
    void cleanse()
    {
        iota(all(par, 0), 0);
    }
};

// globals

// variables
int n, m;

// iterators
int i;

// notes
/*
One piece on top!

 -stuff you should look for-
* int overflow, array bounds
* special cases (n=1?)
* do something instead of nothing and stay organized
* WRITE STUFF DOWN
* DON'T GET STUCK ON ONE APPROACH

continue - skip the rest in the loop
*/
 
void solve()
{
    graph G;
    cin >> n >> m;
    G.prep(n + 1);
    vector <pair <int, int> > v(n + 1);
    for (i = 1; i <= n; i++)
    {
        cin >> v[i].fi;
        G.lvl[i] = 1LL * v[i].fi;
        v[i].se = i;
    }
    for (i = 1; i <= m; i++)
    {
        int a, b;
        cin >> a >> b;
        
        G.add(a, b);
    }
    
    G.cleanse();
    sort(all(v, 1));
    
    vector <bool> ok(n + 1, false);
    vector <vector <int> > nw(n + 1);
    for (auto [d, a] : v)
    {
        ok[a] = true;
        for (auto b : G.g[a])
        {
            if (!ok[b])
            {
                continue;
            }
            
            b = G.get(b);
            if (b == a)
            {
                continue;
            }
            
            nw[a].pb(b);
            G.merge(a, b);
        }
    }
    
    int c = v[n].se;
    vector <int> win(n + 1, 0);
    for (i = 1; i <= n; i++)
    {
        swap(v[i].fi, v[i].se);
    }
    sort(all(v, 1));
    function<void(int, int)> DFS = [&](int k, int b)
    {
        win[k] = b;
        for (auto u : nw[k])
        {
            DFS(u, (b && (G.lvl[u] >= 1LL * v[k].se)));
        }
    };
    
    DFS(c, true);
    
    for (i = 1; i <= n; i++)
    {
        cout << win[i];
    }
}
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
        newline
    }
}
 
int GCD(int A, int B)
{
    if (!A)
    {
        return B;
    }
    return GCD(B % A, A);
}

int LCM(int A, int B)
{
    return A * B / GCD(A, B);
}

int power(int base, int exponent)
{
    int res = 1;
    while (exponent) 
    {
        if (exponent & 1)
        { 
            res *= base;
        }
        base *= base;
        exponent >>= 1;    
    }
    return res;
}

/*
$$$$$$$$\ $$$$$$$$\ 
$$  _____|\____$$  |
$$ |          $$  / 
$$$$$\       $$  /  
$$  __|     $$  /   
$$ |       $$  /    
$$$$$$$$\ $$$$$$$$\ 
\________|\________|
*/
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