Submission #220797

#	Time	Username	Problem	Language	Result	Execution time	Memory
220797		qkxwsm	Regions (IOI09_regions)	C++14	0 / 100	112 ms	45820 KiB

regions

#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
    if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
    if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

const int MAXN = 200013;

int N, M, Q, T, K;
int parent[MAXN], arr[MAXN];
vi edge[MAXN];
int st[MAXN], ft[MAXN], loc[MAXN];
int ord[MAXN];
vi inds[MAXN], del[MAXN];
map<int, int> ans[MAXN];

void dfs(int u)
{
    st[u] = T;
    ft[u] = T;
    ord[T] = u;
    T++;
    inds[arr[u]].PB(st[u]);
    // cerr << "inds " << arr[u] << " PB " << st[u] << endl;
    for (int v : edge[u])
    {
        dfs(v);
        ft[u] = ft[v];
    }
    del[arr[u]].PB(ft[u]);
}
int order_of_key(vi &x, int n) //counts how many are strictly less than n.
{
    return LB(ALL(x), n) - x.begin();
}

int32_t main()
{
    cout << fixed << setprecision(12);
    cerr << fixed << setprecision(4);
    ios_base::sync_with_stdio(false); cin.tie(0);
    cin >> N >> M >> Q;
    FOR(i, 0, N)
    {
        if (i)
        {
            cin >> parent[i];
            parent[i]--;
            edge[parent[i]].PB(i);
        }
        cin >> arr[i]; arr[i]--;
    }
    dfs(0);
    while(Q--)
    {
        int x, y;
        cin >> x >> y; x--; y--;
        if (!ans[x].count(y))
        {
            //well shit.
            int res = 0;
            if (SZ(inds[x]) > SZ(inds[y]))
            {
                //for each node in y, count how many are in its parenthood.
                for (int c : inds[y])
                {
                    res += order_of_key(inds[x], c + 1) - order_of_key(del[x], c);
                }
            }
            else
            {
                //for each node in x, count how many y's are in its subtree.
                for (int c : inds[x])
                {
                    int u = ord[c];
                    // cerr << "u = " << u << endl;
                    res += order_of_key(inds[y], ft[u] + 1) - order_of_key(inds[y], st[u]);
                    // cerr << "res from " << u << " = " << ft[u] + 1 << ' ' << st[u] << endl;
                }
            }
            ans[x][y] = res;
        }
        cout << ans[x][y] << '\n';
        fflush(stdout);
        //how many are in subtree of some node of x.
        //x is smaller.
    }
}

• Subtask 1: 0 / 15
• Subtask 2: 0 / 85