Submission #210106

#	Time	Username	Problem	Language	Result	Execution time	Memory
210106		Utaha	Paprike (COI18_paprike)	C++14	100 / 100	86 ms	25080 KiB

paprike

/*input
6 9
5 4 1 3 3 3
3 1
3 5
4 3
4 2
2 6
*/
#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops,no-stack-protector")
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ld,ld> pdd;
#define IOS ios_base::sync_with_stdio(0); cin.tie(0)
#define ALL(a) a.begin(),a.end()
#define SZ(a) ((int)a.size())
#define F first
#define S second
#define REP(i,n) for(int i=0;i<((int)n);i++)
#define eb emplace_back
#define pb push_back
#define MP(a,b) make_pair(a,b)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (int)(lower_bound(c.begin(),c.end(),x)-c.begin())
#define EL cout<<'\n'
#define BS(a,x) binary_search(ALL(a),x)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<'('<<P.F<<','<<P.S<<')';
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,SZ(V)) out<<V[i]<<((i!=SZ(V)-1)?" ":"");
    return out;
}
//}}}
const ll maxn=300005;
const ll maxlg=20;
const ll INF64=1e18;
const int INF=0x3f3f3f3f;
const ll MOD=ll(1e9+7);
const ld PI=acos(-1);
const ld eps=1e-9;
//const ll p=880301;
//const ll P=31;

ll mypow(ll a,ll b){
    ll res=1LL;
    while(b){
        if(b&1) res=res*a%MOD;
        a=a*a%MOD;
        b>>=1;
    }
    return res;
}

ll a[maxn];
vector<int> edge[maxn];
int ans=0;

ll n,k;
ll rem[maxn];

void dfs(int u,int p){
    vector<ll> tmp;
    for(int v:edge[u]) if(v!=p){
        dfs(v,u);
        tmp.pb(rem[v]);
    }
    sort(ALL(tmp));
    ll cur=a[u];

    for(ll i:tmp){
        if(cur+i>k){
            ans++;
        }
        else cur+=i;
    }
    rem[u]=cur;
}

int main(){
    IOS;
    cin>>n>>k;
    REP(i,n) cin>>a[i];
    REP(i,n-1){
        int u,v;
        cin>>u>>v;
        u--;v--;
        edge[u].pb(v);
        edge[v].pb(u);
    }
    dfs(0,-1);
    cout<<ans<<'\n';
    return 0;
}

• Subtask 1: 11 / 11
• Subtask 2: 13 / 13
• Subtask 3: 27 / 27
• Subtask 4: 49 / 49