제출 #1176394

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1176394		nrg_studio	Plahte (COCI17_plahte)	C++20	0 / 160	126 ms	26444 KiB

plahte

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

#define ll long long
#define pb push_back
#define pii pair<ll,ll>
#define f first
#define s second
#define chmin(a, b) a = min(a,b)
#define chmax(a, b) a = max(a,b)
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define all(x) x.begin(),x.end()
#define vec vector

const int MAX_N = 8e4+1, l2d = 18;
pair<pii,int> pts[3*MAX_N];
pair<pii,pii> p2[MAX_N];
int lift[MAX_N][l2d];
vec<int> adj[MAX_N];
set<int> col[MAX_N];
int ans[MAX_N], c[MAX_N];

int jump(int a, int d) {
    for (int i=0;i<l2d;i++) {
        if ((1<<i)&d) {
            a = lift[a][d];
            if (a==-1) {a = 0;}
        }
    } return a;
}

int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    
    int n, m; cin >> n >> m;
    memset(lift,-1,sizeof(lift));
    p2[0] = {{0,0},{2e9,2e9}};

    for (int i=0;i<n;i++) {
        cin >> pts[i].f.f >> pts[i].f.s;
        cin >> pts[i+n+m].f.f >> pts[i+n+m].f.s;
        pts[i].s = i+1; pts[i+n+m].s = i+n+m+1;
        p2[i+1] = {pts[i].f,pts[i+n+m].f};
    } // [1,n]: pts, [n+1,n+m] balls, [n+m+1,2*n+m] pts2
    for (int i=n;i<n+m;i++) {
        cin >> pts[i].f.f >> pts[i].f.s >> c[i-n+1];
        pts[i].s = i+1;
    }
    sort(pts,pts+2*n+m);
    set<pii> active;
    active.insert({0,0});

    for (int i=0;i<2*n+m;i++) {
        int idx = pts[i].s;
        auto[x,y] = pts[i].f;
        
        if (idx<=n+m) {
            int it = prev(active.upper_bound({y,1e9}))->s;
            int lo = 0, hi = n, mid = (lo+hi)/2, ans = -1;
            int goal = (idx<=n ? p2[idx].s.s : y);
            while (lo <= hi) {
                int x2 = jump(it,mid);
                if (p2[x2].s.s>=goal) {hi = mid-1; ans = x2;}
                else {lo = mid+1;}
                mid = (lo+hi)/2;
            }
            
            if (idx<=n) {
                active.insert({y,idx});
                adj[ans].pb(idx);
                lift[idx][0] = ans;
                for (int j=1;j<l2d;j++) {
                    if (lift[idx][j-1]!=-1) {lift[idx][j] = lift[lift[idx][j-1]][j-1];}
                }
            } else {col[ans].insert(c[idx-n]);}
        } else {
            active.erase(active.find({p2[idx-n-m].f.s,idx-n-m}));
        }
    }

    function<void(int,int)> dfs = [&](int cur, int par) {
        for (int x : adj[cur]) {
            if (x!=par) {
                dfs(x,cur);
                if (col[x].size()>col[cur].size()) {swap(col[x],col[cur]);}
                for (int i : col[x]) {col[cur].insert(i);}
            }
        } ans[cur] = col[cur].size();
    };
    dfs(0,0);
    for (int i=1;i<=n;i++) {cout << ans[i] << '\n';}

    /*
    sort points and ranges
    process start range, then point, then end range
    binary search ez to build tree
    get the thing in set and optimal must be an ancestor of it
    nlogn; sparse table get max with binary search so (m+n)logn

    now calc distinct values
    nlog^2m because merging
    */
}

• Subtask 1: 0 / 32
• Subtask 2: 0 / 32
• Subtask 3: 0 / 32
• Subtask 4: 0 / 32
• Subtask 5: 0 / 32