Submission #658296

#	Time	Username	Problem	Language	Result	Execution time	Memory
658296		Kahou	Spring cleaning (CEOI20_cleaning)	C++14	37 / 100	201 ms	26712 KiB

cleaning

#include<bits/stdc++.h>
using namespace std;
#define F first
#define S second
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int N = 1e5 + 50;
int n, q, h[N], par[20][N], st[N], ft[N], t, cnt[N], sub[N];
int oa, ans, lf[N];
int p[N];
bool mark[N];
vector<int> adj[N], adj2[N];
vector<int> vc;

void dfs(int u, int p) {
        st[u] = ++t;
        par[0][u] = p;
        h[u] = h[p]+1;
        for (int v:adj[u]) {
                if (v==p) continue;
                dfs(v, u);
                sub[u] += sub[v];
        }
        ft[u] = t;
}
void dfs2(int u, int p) {
        for (int v:adj[u]) {
                if (v == p) continue;
                cnt[v] = cnt[u] + 1-(sub[v]&1);
                oa += 1-(sub[v]&1);
                dfs2(v, u);
        }
}
int Par(int u, int x) {
        for (int k = 0; k < 20; k++) {
                if (x>>k&1) u = par[k][u];
        }
        return u;
}
int LCA(int u, int v) {
        if (h[u] < h[v]) swap(u, v);
        u = Par(u, h[u]-h[v]);
        for (int k = 19; k >= 0; k--) {
                if (par[k][u] != par[k][v]) {
                        u = par[k][u];
                        v = par[k][v];
                }
        }
        if (u == v) return u;
        return par[0][u];
}

int f(int u, int v) {
        return (h[v]-h[u])-2*(cnt[v]-cnt[u]);
}

void sfd(int u, int p) {
        for (int v:adj2[u]) {
                if (v == p) continue;
                sfd(v, u);
                lf[u] += lf[v];
                if (lf[v]&1) ans += f(u, v);
        }
}

bool cmp(int u, int v) {
        return st[u] < st[v];
}

void solve() {
        cin >> n >> q;
        for (int i = 1; i <= n-1; i++) {
                int u, v;
                cin >> u >> v;
                adj[u].push_back(v);
                adj[v].push_back(u);
        }
        int r = 0;
        for (int u = 1; u <= n; u++) {
                if(adj[u].size() > 1) {
                        if (!r) r = u;
                }
                else {
                        mark[u] = 1;
                        sub[u] = 1;
                }
        }
        dfs(r, 0);
        dfs2(r, 0);
        for (int k = 1; k < 20; k++) {
                for (int u = 1; u <= n; u++) {
                        par[k][u] = par[k-1][par[k-1][u]];
                }
        }

        for (int u = 1; u <= n; u++) {
                lf[u] = mark[u];
        }
        ans = oa;
        while (q--) {
                int d;
                cin >> d;
                for (int i = 1; i <= d; i++) {
                        int u;
                        cin >> u;
                        lf[u]++;
                        vc.push_back(u);
                }
                sort(vc.begin(), vc.end(), cmp);
                vc.resize(unique(vc.begin(), vc.end()) - vc.begin());
                int tmp = vc.size();
                for (int i = 1; i < tmp; i++) {
                        int u = vc[i];
                        int v = vc[i-1];
                        vc.push_back(LCA(u, v));
                }
                sort(vc.begin(), vc.end(), cmp);
                vc.resize(unique(vc.begin(), vc.end()) - vc.begin());
                for (int i = 1; i < (int)vc.size(); i++) {
                        int u = vc[i];
                        p[u] = vc[i-1];
                        while (st[u] > ft[p[u]]) p[u] = p[p[u]];
                        adj2[u].push_back(p[u]);
                        adj2[p[u]].push_back(u);
                }
                sfd(vc[0], 0);
                cout << ((sub[r] + lf[vc[0]])&1 ? -1:n+d-1+ans) << endl;
                for (int v:vc) {
                        lf[v] = mark[v];
                        adj2[v].clear();
                }
                vc.clear();
                ans = oa;
        }
}

int main() {
        ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
        solve();
        return 0;
}

• Subtask 1: 0 / 0
• Subtask 2: 9 / 9
• Subtask 3: 9 / 9
• Subtask 4: 0 / 16
• Subtask 5: 19 / 19
• Subtask 6: 0 / 17
• Subtask 7: 0 / 30