Submission #1201988

#	Time	Username	Problem	Language	Result	Execution time	Memory
1201988		onbert	Split the Attractions (IOI19_split)	C++20	18 / 100	61 ms	27584 KiB

split

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

const int maxn = 1e5 + 5;
int n, m;
vector<int> adj[maxn];
vector<int> adj1[maxn];

bool vis[maxn];
int fa[maxn], sz[maxn];
void dfs(int u) {
    sz[u] = vis[u] = 1;
    for (int v:adj[u]) if (!vis[v]) {
        fa[v] = u;
        adj1[u].push_back(v); adj1[v].push_back(u);
        dfs(v);
        sz[u] += sz[v];
    }
}

vector<int> ord;
void dfs1(int u, int p) {
    ord.push_back(u);
    for (int v:adj1[u]) if (v != p) dfs1(v, u);
}

vector<int> ans;
int root = -1;
int dsu[maxn], nodes[maxn];
int rt(int u) {
    if (dsu[u] == u) return u;
    return dsu[u] = rt(dsu[u]);
}
void merge(int u, int v) {
    u = rt(u), v = rt(v);
    if (u != v) {
        nodes[v] += nodes[u];
        dsu[u] = v;
    }
}
void dfs2(int u, int p) {
    for (int v:adj1[u]) if (v != p) {
        merge(v, u);
        dfs2(v, u);
    }
}

void dfs3(int u, int RT) {
    ord.push_back(u);
    vis[u] = 1;
    for (int v:adj[u]) if (!vis[v] && rt(v) == RT) dfs3(v, RT);
}
void dfs4(int u) {
    ord.push_back(u);
    vis[u] = 1;
    for (int v:adj[u]) if (!vis[v] && !ans[v]) dfs4(v);
}

vector<int> find_split(int N, int a, int b, int c, vector<int> p, vector<int> q) {
    n = N, m = p.size();
    vector<pair<int,int>> vec = {{a, 1}, {b, 2}, {c, 3}};
    sort(vec.begin(), vec.end());
    for (int i=0;i<m;i++) {adj[p[i]].push_back(q[i]); adj[q[i]].push_back(p[i]);}
    fa[0] = -1;
    dfs(0);

    a = vec[0].first, b = vec[1].first;
    ans.resize(n);
    for (int i=0;i<n;i++) ans[i] = 0;

    for (int i=0;i<n;i++) {
        if (a <= sz[i] && sz[i] <= n-a) {
            if (sz[i] <= n-b) {
                for (int j=0;j<n;j++) ans[j] = vec[2].second;
                ord.clear();
                dfs1(i, fa[i]);
                for (int j=0;j<a;j++) ans[ord[j]] = vec[0].second;
                ord.clear();
                dfs1(fa[i], i);
                for (int j=0;j<b;j++) ans[ord[j]] = vec[1].second;
            } else {
                for (int j=0;j<n;j++) ans[j] = vec[2].second;
                ord.clear();
                dfs1(i, fa[i]);
                for (int j=0;j<b;j++) ans[ord[j]] = vec[1].second;
                ord.clear();
                dfs1(fa[i], i);
                for (int j=0;j<a;j++) ans[ord[j]] = vec[0].second;
            }
            return ans;
        }
    }

    for (int i=0;i<n;i++) if (n - sz[i] < a) {
        bool flag = true;
        for (int v:adj[i]) if (sz[v] >= a) flag = false;
        if (flag) root = i;
    }

    for (int i=0;i<n;i++) dsu[i] = i, nodes[i] = 1;
    for (int v:adj1[root]) dfs2(v, root);

    for (int i=0;i<m;i++) {
        int u = rt(p[i]), v = rt(q[i]);
        if (u == root || v == root || u == v) continue;
        merge(u, v);
        if (nodes[v] >= a) {
            ord.clear();
            for (int j=0;j<n;j++) vis[j] = 0;
            dfs3(v, v);
            for (int j=0;j<a;j++) ans[ord[j]] = vec[0].second;

            ord.clear();
            for (int j=0;j<n;j++) vis[j] = 0;
            dfs4(root);
            for (int j=0;j<b;j++) ans[ord[j]] = vec[1].second;

            for (int j=0;j<n;j++) if (!ans[j]) ans[j] = vec[2].second;
            return ans;
        }
    }
    return ans;
}

• Subtask 1: 7 / 7
• Subtask 2: 11 / 11
• Subtask 3: 0 / 22
• Subtask 4: 0 / 24
• Subtask 5: 0 / 36