Submission #1021355

#TimeUsernameProblemLanguageResultExecution timeMemory
1021355vjudge1Split the Attractions (IOI19_split)C++17
18 / 100
62 ms19828 KiB
#include <bits/stdc++.h>
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx2")
// #pragma GCC optimization ("O3")
// #pragma GCC optimization ("unroll-loops")
// #pragma GCC target("popcnt")
#define endl '\n'
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fo(i,n) for(auto i =0 ; i < n;i++)
#define fore(i,l,r) for(auto i = l; i < r;i++)
#define forex(i,r,l) for(auto i = r; i >= l; i--)
#define ffo(i,n) forex(i,n-1,0)
#define all(x) x.begin(),x.end()
#define lsb(x) x&(-x)
#define sz(x) (int)x.size()
#define gcd(a,b) __gcd(a,b)
#define vii vector<ii>
#define pq_min(a) priority_queue<a, vector<a>, greater<a>>
#define fls cout.flush()
using namespace std;
using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using ii = pair<int,int>;
using mii = map<int,int>;
using lld = long double;
void valid(ll in){cout<<((in)?"YES\n":"NO\n");}
const int N = 2e5 + 7;
vi graph[N];
void dfs(int nodo, vi &ans, vi &vis, int a, int b, int c){
    vis[nodo]=1;
    if(a>0)ans[nodo]=1,a--;
    else if(b>0)ans[nodo]=2,b--;
    else ans[nodo]=3,c--;
    for(int v :graph[nodo]){
        if(vis[v])continue;
        dfs(v, ans, vis, a, b ,c);
    }
}
vi find_split(int n,int a,int b,int c,vi A,vi B){
    int m = sz(A);
    vi cn(n, 0);
    fo(i,m){
        graph[A[i]].pb(B[i]);
        graph[B[i]].pb(A[i]);
        cn[A[i]]++;
        cn[B[i]]++;
    }
    vi ans(n,0);
    if(a==1){
        queue<int> q;
        q.push(0);
        ans[0]=2;b--;
        while(!q.empty()){
            int nodo = q.front();q.pop();
            for(int v : graph[nodo]){
                if(ans[v]==0&&b>0){
                    ans[v]=2;
                    q.push(v);
                    b--;
                }
            }
        }
        fo(i,n){
            if(ans[i]==0&&a==1){
                a--;
                ans[i]=1;
            }else if(ans[i]==0){
                ans[i]=3;
            }
        }
        return ans;
    }
    int ini = 0;
    fo(i,n)
        if(cn[ini]>cn[i])
            ini=i;
    queue<int> q;
    vi v(n, 0);
    dfs(ini,ans,v,a,b,c);
    fo(i,n)graph[i].clear();
    return ans;
}
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