Submission #1040697

#	Time	Username	Problem	Language	Result	Execution time	Memory
1040697		Plurm	Inside information (BOI21_servers)	C++11	100 / 100	3412 ms	232788 KiB

servers

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

vector<int> g[1 << 19];
vector<int> rg[1 << 19];
int st[1 << 19];
int dep[1 << 19];
int ddep[1 << 19];
int t;
int ett[1048576];
int rev[1048576];
void dfs(int u, int p = 0, int d = 0){
    t++;
    ddep[u] = d;
    ett[t] = dep[u] = dep[p]+1;
    st[u] = t;
    rev[t] = u;
    for(int v : g[u]){
        if(v == p) continue;
        dfs(v, u, d+1);
        t++;
        ett[t] = dep[u];
        rev[t] = u;
    }
    for(int v : rg[u]){
        if(v == p) continue;
        dfs(v, u, d-1);
        t++;
        ett[t] = dep[u];
        rev[t] = u;
    }
}
pair<int, int> sparse[21][1048576];
void build(){
    for(int i = 0; i < 1048576; i++) sparse[0][i] = {ett[i], rev[i]};
    for(int i = 1; i < 21; i++){
        for(int j = 0; j + (1 << i) < 1048576; j++){
            sparse[i][j] = min(sparse[i-1][j], sparse[i-1][j+(1<<(i-1))]);
        }
    }
}
inline bool search(const int& a, const int& d){
    int l, r;
    tie(l, r) = minmax(st[a], st[d]);
    int lg = 31 - __builtin_clz(r-l+1);
    return dep[a] + dep[d] - 2*dep[min(sparse[lg][l], sparse[lg][r-(1<<lg)+1]).second] == ddep[d] - ddep[a];
}
tuple<int,int,int> buffer[1 << 19];
int bsz;
const int THRESH = 1024;
int qs[1 << 19];
int evp[1 << 19];
int update(int u, int p = 0, int d = 0){
    for(int v : rg[u]){
        if(v == p) continue;
        d += update(v, u, 0);
    }
    d += evp[u];
    qs[u] += d;
    for(int v : g[u]){
        if(v == p) continue;
        update(v, u, d);
    }
    return d;
}
int count(int d){
    if(bsz > THRESH){
        for(int i = 0; i < bsz; i++){
            evp[get<0>(buffer[i])]--;
            evp[get<1>(buffer[i])]--;
            evp[get<2>(buffer[i])] += 2;
        }
        update(1);
        for(int i = 0; i < bsz; i++){
            evp[get<0>(buffer[i])]++;
            evp[get<1>(buffer[i])]++;
            evp[get<2>(buffer[i])] -= 2;
        }
        bsz = 0;
    }
    int ret = qs[d];
    for(int i = 0; i < bsz; i++){
        if(search(get<0>(buffer[i]), d) || search(get<1>(buffer[i]), d)){
            ret++;
        }else if(search(get<2>(buffer[i]), d)){
            ret += 2;
        }
    }
    return ret;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    k += n-1;
    vector<tuple<int,int,int>> cmds;
    vector<int> ids(n+1);
    for(int i = 1; i <= n; i++){
        ids[i] = i;
    }
    int last = n;
    for(int i = 0; i < k; i++){
        string cmd;
        cin >> cmd;
        int a, b, d;
        switch(cmd[0]){
            case 'S':
                cin >> a >> b;
                last++;
                g[last].push_back(ids[a]);
                rg[ids[a]].push_back(last);
                g[last].push_back(ids[b]);
                rg[ids[b]].push_back(last);
                ids[a] = ids[b] = last;
                cmds.push_back({0, a, b});
                break;
            case 'Q':
                cin >> a >> d;
                cmds.push_back({1, a, d});
                break;
            case 'C':
                cin >> d;
                cmds.push_back({2, d, 0});
                break;
        }
    }
    dfs(1);
    build();
    for(int i = 1; i <= n; i++){
        ids[i] = i;
        qs[i]++;
    }
    last = n;
    for(int i = 0; i < k; i++){
        int a, b, d;
        switch(get<0>(cmds[i])){
            case 0:
                a = get<1>(cmds[i]);
                b = get<2>(cmds[i]);
                last++;
                buffer[bsz++] = {ids[a], ids[b], last};
                ids[a] = ids[b] = last;
                break;
            case 1:
                a = get<1>(cmds[i]);
                d = get<2>(cmds[i]);
                if(search(ids[a], d)) cout << "yes\n";
                else cout << "no\n";
                break;
            case 2:
                d = get<1>(cmds[i]);
                cout << count(d) << "\n";
                break;
        }
    }
    return 0;
}

• Subtask 1: 2.5 / 2.5
• Subtask 2: 2.5 / 2.5
• Subtask 3: 2.5 / 2.5
• Subtask 4: 2.5 / 2.5
• Subtask 5: 5 / 5
• Subtask 6: 5 / 5
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 12.5 / 12.5
• Subtask 10: 12.5 / 12.5
• Subtask 11: 17.5 / 17.5
• Subtask 12: 17.5 / 17.5