Submission #403510

#	Time	Username	Problem	Language	Result	Execution time	Memory
403510		lyc	족보 (KOI18_family)	C++14	0 / 100	24 ms	47340 KiB

family

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

#define TRACE(x) cerr << #x << " :: " << x << endl
#define _ << " " <<
#define SZ(x) (int)(x).size()
#define ALL(x) (x).begin(),(x).end()
#define FOR(i,a,b) for(int i=(a);i<=(b);++i)
#define RFOR(i,a,b) for (int i=(a);i>=(b);--i)

const int mxN = 3e5+5;
const int mxK = 3e5+5;

int K;

struct Tree {
    int N;
    vector<int> g[mxN];
    int pa[mxN][14];
    int dep[mxN];
    int lcd[mxK];

    void read() {
        FOR(i,1,N){
            int P; cin >> P;
            g[P].push_back(i);
        }
    }

    void dfs(int u) {
        FOR(k,1,13) pa[u][k] = (pa[u][k-1] == -1 ? -1 : pa[pa[u][k-1]][k-1]);
        for (int& v : g[u]) {
            pa[v][0] = u;
            dep[v] = dep[u]+1;
            dfs(v);
        }
    }

    int lca(int a, int b) {
        if (dep[a] < dep[b]) swap(a,b);
        RFOR(k,13,0) if (pa[a][k] != -1 && dep[pa[a][k]] >= dep[b]) {
            a = pa[a][k];
        }
        if (a == b) return a;
        RFOR(k,13,0) if (pa[a][k] != pa[b][k]) {
            a = pa[a][k], b = pa[b][k];
        }
        return pa[a][0];
    }

    void run() {
        memset(pa,-1,sizeof pa);
        dep[0] = 0;
        dfs(0);

        FOR(i,1,K-1){
            lcd[i] = dep[lca(i,i+1)];
        }
    }
} A, B;

const int inf = 1e9+5;

int maxV[mxN], minY[mxN], stk[mxN], tp, ls[mxN], rs[mxN];

struct node {
    int s, e, m, v;
    node *l, *r;
    node (int s, int e): s(s), e(e), m((s+e)/2), v(inf) {
        if (s != e) {
            l = new node(s,m);
            r = new node(m+1,e);
        }
    }

    void update(int a, int b) {
        if (s == e) { v = b; return; }
        if (a <= m) l->update(a,b);
        else r->update(a,b);
        v = min(l->v,r->v);
    }

    int qmin(int a, int b) {
        if (s == a && e == b) return v;
        if (b <= m) return l->qmin(a,b);
        if (a > m) return r->qmin(a,b);
        return min(l->qmin(a,m), r->qmin(m+1,b));
    }
} *root, *root2, *root3;

int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

    cin >> A.N >> B.N >> K;
    A.read();
    B.read();

    A.run();
    B.run();

    tp = 0;
    root = new node(1,K);
    FOR(i,1,K){
        while (tp > 0 && A.lcd[stk[tp-1]] >= A.lcd[i]) {
            rs[stk[--tp]] = i-1;
        }
        ls[i] = (tp > 0 ? stk[tp-1]+1 : 1);
        stk[tp++] = i;

        root->update(i, B.lcd[i]);
    }
    while (tp > 0) {
        rs[stk[--tp]] = A.N;
    }



    FOR(i,1,A.N){
        maxV[i] = -inf;
        minY[i] = inf;
    }

    FOR(i,1,K-1){
        maxV[A.lcd[i]] = max(maxV[A.lcd[i]], B.lcd[i]);
        minY[A.lcd[i]] = min(minY[A.lcd[i]], root->qmin(ls[i], rs[i]));
    }



    root2 = new node(1,A.N);
    root3 = new node(1,A.N);
    FOR(i,1,A.N){
        root2->update(i, -maxV[i]);
        root3->update(i, minY[i]);
    }



    bool die = 0;
    FOR(i,2,A.N){
        die |= -root2->qmin(1,i-1) > root3->qmin(i,A.N);
    }

    cout << (die ? "NO" : "YES") << '\n';
}

• Subtask 1: 0 / 17
• Subtask 2: 0 / 27
• Subtask 3: 0 / 15
• Subtask 4: 0 / 41