Submission #1188218

#	Time	Username	Problem	Language	Result	Execution time	Memory
1188218		duyngadocton	September (APIO24_september)	C++20	0 / 100	16 ms	2884 KiB

september

#include "september.h"
#include<bits/stdc++.h>

using namespace std;

#define fi first
#define se second
#define ii pair<int,int>
#define iii pair<ii,int>
#define pb push_back
#define eb emplace_back
#define ll long long

const int MAX = (int) 1e5;
const int maxp = (int) 7e6;

int n, m;
vector<int> adj[MAX + 5];
int info[6][MAX + 5];
int e[maxp + 5], ans;
int a[6][MAX + 5];
int smax[6][MAX + 5];
int id;
int start[MAX + 5], arr[MAX + 5];
int f[25][MAX + 5];
int cur[MAX + 5];
int stop[MAX + 5];
int num, ok;

int get(int l, int r) {
    int k = log2(r - l + 1);
    return max(f[k][l], f[k][r - (1 << k) + 1]);
}

void dfs(int u, int dad) {
    start[u] = ++id;
    for(int v : adj[u]) {
        dfs(v, u);
    }
    stop[u] = id;
}

void add(int x) {
    cur[x] ++;
    if(cur[x] == m && cur[x] != 1) --num;
    if(num == 0) ok = 1;
    if(cur[x] == 1 && cur[x] != m) ++num, ok = 0;
}

void calc() {
    dfs(1, 0);
    for(int p = 1; p <= m; ++p) {
        for(int i = 1; i < n; ++i) {
            int x = info[p][i];
            arr[start[x]] = i;
        }
        int maxk = log2(n) + 1;
        for(int i = 1; i <= n; ++i) f[0][i] = arr[i];
        for(int k = 1; k <= maxk; ++k)
            for(int i = 1; i + (1 << k) - 1 <= n; ++i) f[k][i] = max(f[k - 1][i], f[k - 1][i + (1 << (k - 1))]);
        for(int i = 1; i < n; ++i) {
            int x = info[p][i];
            a[p][i] = get(start[x], stop[x]);
        }
        for(int i = 1; i < n; ++i) {
            smax[p][i] = max(smax[p][i - 1], a[p][i]);
        }
    }
    for(int i = 1; i < n; ++i) {
        for(int j = 1; j <= m; ++j) add(info[j][i]);
        int maxx = 0;
        for(int j = 1; j <= m; ++j) maxx = max(maxx, smax[j][i]);
        if(ok == 1 && maxx == i) {
            ++ans;
        }
    }

}

int solve(int N, int M, vector<int> F,vector<vector<int>> S) {
    n = N;
    m = M;
    for(int x = 1; x < n; ++x) {
        int j = F[x];
        adj[j + 1].pb(x + 1);
    }
    for(int x = 0; x < m; ++x) {
        int k = 0;
        for(int j : S[x]) info[x + 1][++k] = j + 1;
    }
    calc();
	return ans;
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 14
• Subtask 3: 0 / 5
• Subtask 4: 0 / 9
• Subtask 5: 0 / 5
• Subtask 6: 0 / 11
• Subtask 7: 0 / 9
• Subtask 8: 0 / 11
• Subtask 9: 0 / 9
• Subtask 10: 0 / 16