제출 #502315

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
502315		ljuba	The Xana coup (BOI21_xanadu)	C++17	100 / 100	62 ms	18404 KiB

xanadu

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
 
typedef vector<int> vi;
typedef vector<ll> vll;
 
typedef vector<vi> vvi;
typedef vector<vll> vvll;
 
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
 
typedef vector<pii> vpii;
typedef vector<pll> vpll;
 
typedef vector<vpii> vvpii;
typedef vector<vpll> vvpll;
 
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int)(x).size()
#define fi first
#define se second
 
template<class T> bool ckmin(T &a, const T &b) {return a > b ? a = b, 1 : 0;}
template<class T> bool ckmax(T &a, const T &b) {return a < b ? a = b, 1 : 0;}
 
namespace debug {
    void __print(int x) {cerr << x;}
    void __print(long long x) {cerr << x;}
    void __print(double x) {cerr << x;}
    void __print(long double x) {cerr << x;}
    void __print(char x) {cerr << '\'' << x << '\'';}
    void __print(const string &x) {cerr << '\"' << x << '\"';}
    void __print(const char *x) {cerr << '\"' << x << '\"';}
    void __print(bool x) {cerr << (x ? "true" : "false");}
 
    template<typename T, typename V>
    void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
    template<typename T>
    void __print(const T &x) {int f = 0; cerr << '{'; for(auto z : x) cerr << (f++ ? "," : ""), __print(z); cerr << "}";}
    void _print() {cerr << "]\n";}
    template <typename T, typename... V>
    void _print(T t, V... v) {__print(t); if(sizeof...(v)) cerr << ", "; _print(v...);}
 
#ifdef ljuba
#define dbg(x...) cerr << "\e[91m" << "LINE(" << __LINE__ << ") -> " << "[" << #x << "] = ["; _print(x)
#else
#define dbg(x...)
#endif
}
 
using namespace debug;
 
const char nl = '\n';
 
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
 
/*
погледај сампле тестове, пре него што имплементираш (можда имаш погрешну идеју) 
уместо да разматраш неке глупе случајеве на папиру, размишљај заправо о задатку
*/

const int mxN = 1e5 + 12;
vi adj[mxN];
ll dp[mxN][2][2];
int a[mxN];

const ll INF = 1e9 + 12;

void dfs(int s, int p = -1) {
    for(auto e : adj[s]) {
        if(e == p) continue;
        dfs(e, s);
    }

    for(int trenutni = 0; trenutni < 2; ++trenutni) {
        for(int roditelj = 0; roditelj < 2; ++roditelj) {
            ll parno = 0, neparno = INF;

            for(auto e : adj[s]) {
                if(e == p) continue;
                ll deteParno = dp[e][0][trenutni];
                ll deteNeparno = dp[e][1][trenutni];
                ll noviParno = min(parno+deteParno, neparno+deteNeparno);
                ll noviNeparno = min(neparno+deteParno, parno+deteNeparno);
                parno = noviParno;
                neparno = noviNeparno;
            }

            dp[s][trenutni][roditelj] = trenutni;
            if(!(a[s]^trenutni^roditelj)) {
                dp[s][trenutni][roditelj] += parno;
            } else {
                dp[s][trenutni][roditelj] += neparno;
            }
        }
    }
}

void solve() {
    int n;
    cin >> n;
    for(int i = 1; i < n; ++i) {
        int u, v;
        cin >> u >> v;
        --u, --v;
        adj[u].pb(v);
        adj[v].pb(u);
    }

    for(int i = 0; i < n; ++i) cin >> a[i];
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < 2; ++j) {
            for(int k = 0; k < 2; ++k) {
                dp[i][j][k] = INF;
            }
        }
    }

    dfs(0);

    //for(int i = 0; i < n; ++i) {
        //for(int j = 0; j < 2; ++j) {
            //for(int k = 0; k < 2; ++k) {
                //dbg(i, j, k, dp[i][j][k]);
            //}
        //}
    //}
    ll ans = min(dp[0][0][0], dp[0][1][0]);

    cout << ((ans >= INF) ? "impossible" : to_string(ans)) << nl;
}
 
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    int testCases = 1;
    //cin >> testCases;
    while(testCases--)
        solve();
}

• Subtask 1: 5 / 5
• Subtask 2: 15 / 15
• Subtask 3: 10 / 10
• Subtask 4: 40 / 40
• Subtask 5: 30 / 30