Submission #577603

#	Time	Username	Problem	Language	Result	Execution time	Memory
577603		jiahng	The Xana coup (BOI21_xanadu)	C++14	100 / 100	126 ms	25036 KiB

xanadu

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define int ll
typedef pair<int32_t, int32_t> pi;
typedef vector <int> vi;
typedef vector <pi> vpi;
typedef pair<pi, ll> pii;
typedef set <ll> si;
typedef long double ld;
#define f first
#define s second
#define mp make_pair
#define FOR(i,s,e) for(int i=s;i<=int(e);++i)
#define DEC(i,s,e) for(int i=s;i>=int(e);--i)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
#define aFOR(i,x) for (auto i: x)
#define mem(x,i) memset(x,i,sizeof x)
#define fast ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define maxn 100010
#define INF (ll)1e9
#define MOD 1000000007
typedef pair <vi, int> pvi;
typedef pair <int,pi> ipi;
typedef vector <pii> vpii;

int N,A[maxn];
vi adj[maxn];
int a,b;
int dp[maxn][2][2];
int dpf(int i,int j,int k,int p){
    // i is the node, j is current state and k is whether node is selected
    if (dp[i][j][k] != -1) return dp[i][j][k];
    vi v;

    int cur = 0;
    aFOR(x,adj[i]) if (x != p){
       v.pb(dpf(x,k^A[x]^1,1,i)+1 - dpf(x,k^A[x],0,i));
       cur += dpf(x,k^A[x],0,i);
    }
    sort(all(v)); int ptr = 0;
    if (j){
        if (v.empty()) return dp[i][j][k] = INF;
        cur += v[ptr]; ptr++;
    }
    dp[i][j][k] = cur;
    while (ptr + 1 < (int)v.size()){
        cur += v[ptr] + v[ptr+1];
        ptr += 2;
        dp[i][j][k] = min(dp[i][j][k], cur);
    }
    assert(dp[i][j][k] >= 0);
    // cout << i << ' ' << j << ' ' << dp[i][j][k] << '\n';        
    return dp[i][j][k];
}
int32_t main(){
	fast;
	cin >> N;
	FOR(i,1,N-1){
        cin >> a >> b;
        adj[a].pb(b); adj[b].pb(a);
    }
    mem(dp,-1);
    FOR(i,1,N) cin >> A[i];	
    int ans = min(dpf(1,A[1],0,-1), dpf(1,!A[1],1,-1) + 1);
    if (ans >= INF) cout << "impossible";
    else cout << ans;
}

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