Submission #535787

#	Time	Username	Problem	Language	Result	Execution time	Memory
535787		browntoad	Unique Cities (JOI19_ho_t5)	C++14	4 / 100	122 ms	596 KiB

joi2019_ho_t5

#include <bits/stdc++.h>
#pragma GCC optimize ("Ofast", "unroll-loops")
using namespace std;
#define ll long long
#define int ll
#define FOR(i,a,b) for (int i = (a); i<(b); i++)
#define REP(i,n) FOR(i,0,n)
#define REP1(i,n) FOR(i,1,n+1)
#define RREP(i,n) for (int i=(n)-1; i>=0; i--)
#define f first
#define s second
#define pb push_back
#define ALL(x) x.begin(),x.end()
#define SZ(x) (int)(x.size())
#define SQ(x) (x)*(x)
#define pii pair<int, int>
#define pdd pair<double ,double>
#define pcc pair<char, char> 
#define endl '\n'
//#define TOAD
#ifdef TOAD
#define bug(x) cerr<<__LINE__<<": "<<#x<<" is "<<x<<endl
#define IOS()
#else
#define bug(...)
#define IOS() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#endif
 
const ll inf = 1ll<<60;
const int iinf=2147483647;
const ll mod = 1e9+7;
const ll maxn=2005;
const double PI=acos(-1);

ll pw(ll x, ll p, ll m=mod){
    ll ret=1;
    while (p>0){
        if (p&1){
            ret*=x;
            ret%=m;
        }
        x*=x;
        x%=m;
        p>>=1;
    }
    return ret;
}
 
ll inv(ll a, ll m=mod){
    return pw(a,m-2);
}
 
//=======================================================================================
vector<int> graph[maxn];
vector<int> typ(maxn);
vector<int> tot(maxn);
vector<int> c(maxn);
void dfs(int x, int prev, int lay){
    if (lay!=0){
        if (typ[lay]!=0){
            typ[lay]=-1;
        }
        else {
            typ[lay]=c[x];
        }
    }
    REP(i,SZ(graph[x])){
        if (graph[x][i]==prev) continue;
        dfs(graph[x][i], x, lay+1);
    }
}
signed main (){
    IOS();
    int n; cin>>n;
    int m; cin>>m;
    REP(i,n-1){
        int x, y; cin>>x>>y;
        graph[x].pb(y);
        graph[y].pb(x);
    }
    REP1(i, n) cin>>c[i];
    FOR(i,1,n+1){
        REP(i, n+1){
            typ[i]=0;
            tot[i]=0;
        }
        dfs(i, -1, 0);
        int ans=0;
        REP1(j, n){
            if (typ[j]!=-1){
                tot[typ[j]]=1;
            }
        }
        REP1(j, m){
            ans+=tot[j];
        }
        cout<<ans<<endl;
    }
}
/*
2 3
2 1
2 1
2
2
2
*/

• Subtask 1: 4 / 4
• Subtask 2: 0 / 32
• Subtask 3: 0 / 32
• Subtask 4: 0 / 32