Submission #61485

#	Time	Username	Problem	Language	Result	Execution time	Memory
61485		Benq	Min-max tree (BOI18_minmaxtree)	C++11	100 / 100	637 ms	39652 KiB

minmaxtree

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

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 70010;

vi adj[MX];
set<array<int,3>> ADJ[MX];
priority_queue<int> mx[MX], mn[MX];
int N;
bool done[MX];
map<int,int> m;
vi rm;

void merge(priority_queue<int>& a, priority_queue<int>& b) {
    if (sz(a) < sz(b)) swap(a,b);
    while (sz(b)) a.push(b.top()), b.pop();
}

void del(priority_queue<int>& a) {
    while (sz(a) > 1) {
        int x = a.top(); a.pop();
        if (a.top() == x) a.pop();
        else {
            a.push(x);
            break;
        }
    }
}
void merge(int a, int b) {
    // cout << "TT " << sz(mn[a]) << " " << sz(mn[b]) << "\n";
    merge(mn[a],mn[b]);
    //cout << "TT " << sz(mn[a]) << " " << sz(mn[b]) << "\n";
    del(mn[a]);
    merge(mx[a],mx[b]);
    del(mx[a]);
}

void dfs(int x, int p) {
    for (int i: adj[x]) if (i != p) {
        dfs(i,x);
        // cout << "OH " << i << " " << x << "\n";
        merge(x,i);
    }
    
    int lo = (sz(mn[x]) ? mn[x].top() : -MOD);
    int hi = (sz(mx[x]) ? -mx[x].top() : MOD);
    
    if (p) {
        // cout << "ZZ " << x << " " << p << "\n";
        ADJ[m[lo]].insert({m[hi],x,p});
        ADJ[m[hi]].insert({m[lo],x,p});
    }
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N;
    F0R(i,N-1) {
        // int A = rand() % (i+1)+1, B = i+2;
        int A,B; cin >> A >> B;
        adj[A].pb(B), adj[B].pb(A);
    }
    int K; cin >> K;
    m[MOD] = 0, m[-MOD] = 1;
    rm.pb(MOD), rm.pb(-MOD);
    F0R(i,K) {
        char c; cin >> c;
        int x,y,z; cin >> x >> y >> z;
        if (c == 'M') mx[x].push(-z), mx[y].push(-z);
        else mn[x].push(z), mn[y].push(z);
        m[z] = i+2, rm.pb(z);
    }
    dfs(1,0);
    set<pi> S;
    FOR(i,2,K+2) S.insert({sz(ADJ[i]),i});
    done[0] = done[1] = 1;
    
    while (sz(S)) {
        auto a = *S.begin(); S.erase(a); done[a.s] = 1;
        // cout << "OOOPS " << sz(ADJ[a.s]) << "\n";
        auto it = *ADJ[a.s].begin(); ADJ[a.s].erase(it);
        if (!done[it[0]]) S.erase({sz(ADJ[it[0]]),it[0]});
        ADJ[it[0]].erase({a.s,it[1],it[2]});
        if (!done[it[0]]) {
            S.insert({sz(ADJ[it[0]]),it[0]});
            assert(sz(ADJ[it[0]]));
        }
        cout << it[1] << " " << it[2] << " " << rm[a.s] << "\n";
    }
    
    F0R(i,K+2) for (auto a: ADJ[i]) if (a[0] >= i)
        cout << a[1] << " " << a[2] << " " << rm[i] << "\n";
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 7 / 7
• Subtask 2: 22 / 22
• Subtask 3: 29 / 29
• Subtask 4: 42 / 42