제출 #471040

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
471040		Carmel_Ab1	Birthday gift (IZhO18_treearray)	C++17	100 / 100	1383 ms	91864 KiB

treearray

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

//using namespace __gnu_pbds;
using namespace std;

typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<pl> vpl;
typedef vector<ld> vld;
typedef pair<ld, ld> pld;

//typedef tree<ll, null_type, less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
template<typename T>
ostream &operator<<(ostream &os, vector<T> &a) {
    os << "[";
    for (int i = 0; i < ll(a.size()); i++) { os << a[i] << ((i != ll(a.size() - 1) ? " " : "")); }
    os << "]\n";
    return os;
}

#define all(x) x.begin(),x.end()
#define YES out("YES")
#define NO out("NO")
#define out(x){cout << x << "\n"; return;}
#define GLHF ios_base::sync_with_stdio(false); cin.tie(NULL)
#define print(x){for(auto ait:x) cout << ait << " "; cout << "\n";}
#define pb push_back
#define umap unordered_map

template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
    is >> p.first >> p.second;
    return is;
}

template<typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
    os << "" << p.first << " " << p.second << "";
    return os;
}

void usaco(string taskname) {
    string fin = taskname + ".in";
    string fout = taskname + ".out";
    const char *FIN = fin.c_str();
    const char *FOUT = fout.c_str();
    freopen(FIN, "r", stdin);
    freopen(FOUT, "w", stdout);
}

template<typename T>
void read(vector<T> &v) {
    int n = v.size();
    for (int i = 0; i < n; i++)
        cin >> v[i];
}

template<typename T>
vector<T> UNQ(vector<T> a) {
    vector<T> ans;
    for (T t: a)
        if (ans.empty() || t != ans.back())
            ans.push_back(t);
    return ans;
}

ll ceil(ll a, ll b) {
    ll ans = a / b;
    if (a % b)ans++;
    return ans;
}

ld log(ld a, ld b) { return log(b) / log(a); }

ll power(ll base, ll exp, ll M = 1e18) {//(base^exp)%M
    ll ans = 1;
    while (exp) {
        if (exp % 2 == 1)ans = ((ans % M) * (base % M)) % M;
        base = ((base % M) * (base % M)) % M;
        exp /= 2;
    }
    return ans;
}

string to_base(int n, int new_base) {
    string s;
    int nn = n;
    while (nn) {
        s += to_string(nn % new_base);
        nn /= new_base;
    }
    reverse(all(s));
    return s;
}

ll gcd(ll a, ll b) {
    if (a < b)swap(a, b);
    if (b == 0)return a;
    return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
    ll x = (a / gcd(a, b));
    return b * x;
}

vl generate_array(ll n, ll mn = 1, ll mx = 1e18 + 1) {
    if (mx == 1e18 + 1)
        mx = n;
    vl ans(n);
    for (int i = 0; i < n; i++)
        ans[i] = (mn + rand() % (mx - mn + 1));
    return ans;
}

string substr(string s, int l, int r) {
    string ans;
    for (int i = l; i <= r; i++)
        ans += s[i];
    return ans;
}


void solve();

int main() {
    GLHF;
    int t = 1;
    //cin >> t;
    while (t--)
        solve();
}

int n,m,q;
vvi g;
vvi up;
vi dep;
void dfs1(int src,int par){
    if(par!=-1)
        dep[src]=dep[par]+1;
    up[src][0]=par;
    for(int nbr:g[src])
        if(nbr!=par)
            dfs1(nbr,src);
}
int get(int x,int k){
    for(int i=0;x!=-1 && i<20 && k ; i++){
        if(k&(1<<i)){
            x=up[x][i];
            k-=(1<<i);
        }
    }
    return x;
}
int lca(int x,int y){

    assert(x || y);
    if(!x)return y;
    else if(!y)return x;

    if(dep[x]>dep[y])
        swap(x,y);
    //x<=y
    y=get(y,dep[y]-dep[x]);
    for(int i=19;x!=y && 0<=i; i--)
        if(up[x][i]!=up[y][i]){
            x=up[x][i];
            y=up[y][i];
        }
    if(x==y)return x;
    return up[x][0];
}


void solve() {
    cin >> n >> m>> q;
    g.resize(n+1);
    for(int i=0; i<n-1; i++){
        int u,v;
        cin >> u >> v;
        g[u].pb(v);
        g[v].pb(u);
    }
    dep.resize(n+1);
    up.resize(n+1,vi(20,-1));
    dfs1(1,-1);

    for(int j=1; j<20; j++)
        for(int i=1; i<=n; i++)
            if(up[i][j-1]!=-1)
                up[i][j]=up[up[i][j-1]][j-1];
    vi a(m+1);
    for(int i=1; i<=m; i++)
        cin >> a[i];

    vector<set<int>>who1(n+1),who2(n+1);
    for(int i=1; i<=m; i++)
        who1[a[i]].insert(i);
    for(int i=1; i<m; i++)
        who2[lca(a[i],a[i+1])].insert(i);

    while(q--){
        int op;
        cin >> op;
        if(op==1){
            int i,v;

            cin >> i >> v;

            who1[a[i]].erase(i);
            who1[v].insert(i);

            if(i!=1)
                who2[lca(a[i],a[i-1])].erase(i-1);
            if(i!=m)
                who2[lca(a[i],a[i+1])].erase(i);

            if(i!=1)
                who2[lca(a[i-1],v)].insert(i-1);
            if(i!=m)
                who2[lca(v,a[i+1])].insert(i);

            a[i]=v;
            continue;
        }
        int l,r,v;
        cin >> l >> r >> v;

        pi ans={-1,-1};
        if(who1[v].lower_bound(l)!=who1[v].end() && *who1[v].lower_bound(l)<=r)
            ans={*who1[v].lower_bound(l),*who1[v].lower_bound(l)};
        else if(who2[v].lower_bound(l)!=who2[v].end() && *who2[v].lower_bound(l)<r)
            ans={*who2[v].lower_bound(l),*who2[v].lower_bound(l)+1};
        cout << ans << "\n";
    }
}
/*
5 4 4
1 2
3 1
3 4
5 3
4 5 2 3
2 1 3 1
1 3 5
2 3 4 5
2 1 3 1
 */

컴파일 시 표준 에러 (stderr) 메시지

treearray.cpp: In function 'void usaco(std::string)':
treearray.cpp:56:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   56 |     freopen(FIN, "r", stdin);
      |     ~~~~~~~^~~~~~~~~~~~~~~~~
treearray.cpp:57:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   57 |     freopen(FOUT, "w", stdout);
      |     ~~~~~~~^~~~~~~~~~~~~~~~~~~

• Subtask 1: 12 / 12
• Subtask 2: 18 / 18
• Subtask 3: 26 / 26
• Subtask 4: 44 / 44