백준15564 Äventyr 2

작성일 | In

문제 링크

  • http://icpc.me/15564

문제 출처

  • 제1회 웰노운컵 C2번

사용 알고리즘

  • Centroid Decomposition

풀이

트리와 쿼리 5 (풀이)와 동일한 문제입니다.

단, set을 쓰면 시간 제한이 약간 빡빡할 수 있으니 적당히 커팅해주면 됩니다.

전체 코드

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#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
#define x first
#define y second
#define all(v) v.begin(), v.end()
#define compress(v) sort(all(v)), v.erase(unique(all(v)), v.end())
#define pb push_back
using namespace std;
//using namespace __gnu_pbds; //ordered_set : find_by_order(order), order_of_key(key)
//using namespace __gnu_cxx; //crope : append(str), substr(s, e), at(idx)

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> p;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

const int di[] = {1, 0, -1, 0, 1, 1, -1, -1}, dj[] = {0, 1, 0, -1, 1, -1, 1, -1};
ll gcd(ll x, ll y) { return y ? gcd(y, x%y) : x; }
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }
ll mod(ll a, ll b) { return ((a%b) + b) % b; }
ll ext_gcd(ll a, ll b, ll &x, ll &y) { //ax + by = gcd(a, b)
  ll g = a; x = 1, y = 0;
  if (b) g = ext_gcd(b, a % b, y, x), y -= a / b * x;
  return g;
}
ll inv(ll a, ll m){ //return x when ax mod m = 1, fail -> -1
    ll x, y;
    ll g = ext_gcd(a, m, x, y);
    if(g > 1) return -1;
    return mod(x, m);
}
void finish(){ exit(0); }

int sz[101010], use[101010], dep[101010];
int dp[22][101010];
vector<int> g[101010];
int cdtree[101010];
multiset<int> st[101010];
int color[101010];

void dfs(int now, int prv = -1){
    dp[0][now] = prv;
    dep[now] = dep[prv] + 1;
    for(auto i : g[now]){
        if(i == prv) continue;
        dfs(i, now);
    }
}

int lca(int u, int v){
    if(dep[u] < dep[v]) swap(u, v);
    int diff = dep[u] - dep[v];
    for(int i=0; diff; i++){
        if(diff & 1) u = dp[i][u];
        diff >>= 1;
    }
    if(u == v) return u;
    for(int i=21; i>=0; i--){
        if(dp[i][u] ^ dp[i][v]) u = dp[i][u], v = dp[i][v];
    }
    return dp[0][u];
}

int getDist(int u, int v){
    return dep[u] + dep[v] - 2 * dep[lca(u, v)];
}

int getSize(int now, int prv = -1){
    sz[now] = 1;
    for(auto i : g[now]){
        if(use[i] || prv == i) continue;
        sz[now] += getSize(i, now);
    }
    return sz[now];
}

int getCent(int now, int prv, int cnt){
    for(auto i : g[now]){
        if(use[i] || i == prv) continue;
        if(sz[i] > cnt/2) return getCent(i, now, cnt);
    }
    return now;
}

void cd(int now, int prv = -1){
    int cnt = getSize(now, prv);
    int cent = getCent(now, prv, cnt);
    if(prv == -1) cdtree[cent] = cent;
    else cdtree[cent] = prv;
    use[cent] = 1;
    for(auto i : g[cent]){
        if(!use[i]) cd(i, cent);
    }
}

void update(int v){
    color[v] = !color[v];
    int i = v;
    while(1){
        int dist = getDist(i, v);
        if(color[v]) st[i].insert(dist);
        else{
            st[i].erase(st[i].find(dist));
        }
        if(i == cdtree[i]) break;
        i = cdtree[i];
    }
}

int query(int v){
    int i = v;
    int ret = 1e9+7;
    while(1){
        int dist = getDist(i, v);
        if(!st[i].empty()) ret = min(ret, dist + *st[i].begin());
        if(i == cdtree[i]) break;
        i = cdtree[i];
    }
    if(ret > 1e9) return -1;
    return ret;
}

int main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    int n; cin >> n;
    int q; cin >> q;
    for(int i=2; i<=n; i++){
        int s = i, e; cin >> e;
        g[s].push_back(e); g[e].push_back(s);
    }
    dfs(1);
    for(int j=1; j<22; j++){
        for(int i=1; i<=n; i++){
            dp[j][i] = dp[j-1][dp[j-1][i]];
        }
    }
    cd(1);
    bool flag = 0;
    while(q--){
        int op, v; cin >> op >> v;
        if(op == 1){
            update(v);
            flag = 1;
        }else{
            if(!flag) cout << "-1\n";
            else cout << query(v) << "\n";
        }
    }
}