제출 #813990

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
813990		PixelCat	열쇠 (IOI21_keys)	C++17	9 / 100	3115 ms	1353604 KiB

keys

#ifdef NYAOWO
#include "grader.cpp"
#endif

#include <bits/stdc++.h>
#define For(i, a, b) for(int i = a; i <= b; i++)
#define Forr(i, a, b) for(int i = a; i >= b; i--)
#define F first
#define S second
#define sz(x) ((int)x.size())
#define all(x) x.begin(), x.end()
#define eb emplace_back
// #define int LL
using namespace std;
using i32 = int32_t;
using LL = long long;
using pii = pair<int, int>;

const int MAXN = 300'000;
const int MAXK = 30;  // max types of keys
const int MAXND = MAXN * MAXK;

struct DSU {
    int p[MAXN + 10];
    void init() {
        memset(p, -1, sizeof(p));
    }
    int find(int n) {
        if(p[n] < 0) return n;
        return p[n] = find(p[n]);
    }
    void uni(int a, int b) {
        a = find(a); b = find(b);
        if(a != b) p[b] = a;
    }
} dsu[MAXK + 10];

struct SCC {
    vector<int> adj[MAXND + 10];
    vector<int> rev[MAXND + 10];
    int vis[MAXND + 10];
    vector<int> stk;
    vector<int> scc[MAXND + 10];
    bool broken[MAXND + 10];
    
    int rk2id(int r, int k) {
        return r * MAXK + k;
    }
    pii id2rk(int id) {
        return pii(id / MAXK, id % MAXK);
    }

    void link(int r1, int k1, int r2, int k2) {
        int x = rk2id(r1, k1);
        int y = rk2id(r2, k2);
        adj[x].eb(y);
        rev[y].eb(x);
    }
    void bilink(int r1, int k1, int r2, int k2) {
        link(r1, k1, r2, k2);
        link(r2, k2, r1, k1);
    }

    void dfs1(int n) {
        if(vis[n] < 0) return;
        vis[n] = -1;
        for(auto &i:rev[n]) dfs1(i);
        stk.eb(n);
    }
    void dfs2(int n, int tag) {
        if(vis[n] != -1 && vis[n] != tag) broken[tag] = 1;
        if(vis[n] != -1) return;
        vis[n] = tag;
        for(auto &i:adj[n]) dfs2(i, tag);
    }
    int build_scc(int n) {
        memset(vis, 0, sizeof(vis));
        For(i, 0, n * MAXK - 1) dfs1(i);
        int cur_scc = 0;
        while(sz(stk)) {
            dfs2(stk.back(), cur_scc);
            while(sz(stk) && vis[stk.back()] != -1) stk.pop_back();
            cur_scc++;
        }
        For(i, 0, n * MAXK - 1) {
            scc[vis[i]].eb(i);
        }
        return cur_scc;
    }
} ayaya;

int in_use[MAXN + 10];

vector<i32> find_reachable(vector<i32> R, vector<i32> U, vector<i32> V, vector<i32> C) {
    int n = sz(R);
    int m = sz(C);

    For(i, 0, MAXK - 1) dsu[i].init();
    For(i, 0, m - 1) dsu[C[i]].uni(U[i], V[i]);
    For(k, 0, 29) For(r, 0, n - 1) {
        ayaya.bilink(r, k, dsu[k].find(r), k);
        ayaya.link(r, k, r, R[r]);
    }
    int nscc = ayaya.build_scc(n);

    vector<int> reach(n, -1);
    int mn = n + 48763;
    For(sccid, 0, nscc - 1) if(!ayaya.broken[sccid]) {
        int cnt = 0;
        vector<int> start;
        for(auto &id:ayaya.scc[sccid]) {
            int r, k;
            tie(r, k) = ayaya.id2rk(id);
            if(k == R[r]) start.eb(r);
            cnt += (in_use[r] == 0);
            in_use[r] = 1;
        }
        for(auto &i:start) {
            reach[i] = cnt;
            mn = min(mn, cnt);
        }
        for(auto &id:ayaya.scc[sccid]) {
            in_use[ayaya.id2rk(id).F] = 0;
        }
    }

    vector<i32> ans(n);
    For(i, 0, n - 1) {
        ans[i] = (reach[i] == mn);
    }
    return ans;
}

/*

4 5
0 1 1 2
0 1 0
0 2 0
1 2 1
1 3 0
3 1 2

0 1 1 0


7 10
0 1 1 2 2 1 2
0 1 0
0 2 0
1 2 1
1 3 0
2 3 0
3 4 1
3 5 2
4 5 0
4 6 2
5 6 1

0 1 1 0 1 0 1


3 1
0 0 0
0 1 0

0 0 1

*/

• Subtask 1: 9 / 9
• Subtask 2: 0 / 11
• Subtask 3: 0 / 17
• Subtask 4: 0 / 30
• Subtask 5: 0 / 33