제출 #749067

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
749067		GrindMachine	The Xana coup (BOI21_xanadu)	C++17	100 / 100	95 ms	23484 KiB

xanadu

// Om Namah Shivaya

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*



*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

vector<ll> adj[N];
vector<ll> a(N);
ll dp[N][2][2];

void dfs(ll u, ll p){
    ll dp1[2][2], dp2[2][2];
    memset(dp1,0x3f,sizeof dp1);
    memset(dp2,0x3f,sizeof dp2);

    dp1[0][a[u]] = 0;
    dp1[1][a[u]^1] = 1;

    trav(v, adj[u]){
        if(v == p) conts;
        dfs(v, u);

        memset(dp2,0x3f,sizeof dp2);

        rep(op1,2){
            rep(c1,2){
                rep(op2,2){
                    rep(c2,2){
                        if(op1^c2) conts;
                        amin(dp2[op1][c1^op2], dp1[op1][c1] + dp[v][op2][c2]);
                    }
                }
            }
        }

        rep(j,2){
            rep(k,2){
                dp1[j][k] = dp2[j][k];
            }
        }
    }

    rep(j,2){
        rep(k,2){
            dp[u][j][k] = dp1[j][k];
        }
    }
}

void solve(int test_case)
{
    ll n; cin >> n;
    rep1(i,n-1){
        ll u,v; cin >> u >> v;
        adj[u].pb(v), adj[v].pb(u);
    }
    rep1(i,n) cin >> a[i];

    memset(dp,0x3f,sizeof dp);
    dfs(1,-1);

    ll ans = inf2;
    rep(j,2){
        amin(ans, dp[1][j][0]);
    }

    if(ans > n) {
        cout << "impossible" << endl;
    }
    else{
        cout << ans << endl;
    }
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

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