제출 #143466

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
143466		icypiggy	Split the Attractions (IOI19_split)	C++14	100 / 100	186 ms	27768 KiB

split

#include <bits/stdc++.h>
using namespace std;
const int n_max = 2e5+1e3;
int visit[n_max];
int low[n_max];
int subtree[n_max];
vector<int> adj[n_max];
vector<int> child[n_max];
int parent[n_max];
int time_global = 0;
void dfs(int x) {
    subtree[x] = 1;
    visit[x] = time_global;
    time_global++;
    low[x] = visit[x];
    for(int i: adj[x]) {
        if(visit[i]==-1) {
            parent[i] = x;
            child[x].push_back(i);
            dfs(i);
            subtree[x] += subtree[i];
            low[x] = min(low[x], low[i]);
        } else if (i != parent[x]) {
            low[x] = min(low[x], visit[i]);
        }
    }
}
vector<int> find_split(int n, int a, int b, int c, vector<int> p, vector<int> q) {
    // assume a<=b<=c
    vector<pair<int,int>> vp = {make_pair(a,1),make_pair(b,2),make_pair(c,3)};
    sort(vp.begin(), vp.end());
    int small = vp[0].second;
    int mid = vp[1].second;
    int big = vp[2].second;
    a = vp[0].first;
    b = vp[1].first;
    c = vp[2].first;

    memset(visit, -1, sizeof(visit));
    for(int i=0; i<(int) p.size(); i++) {
        adj[p[i]].push_back(q[i]);
        adj[q[i]].push_back(p[i]);
    }
    dfs(0);
    vector<int> ans(n,0);
    for(int i=0; i<n; i++) {
        // assume c>=a
        // check: a+c or b+c?
        if(subtree[i]>=a && subtree[i]<=b+c) {
            if(subtree[i]>a+c) {
                swap(a,b);
                swap(small,mid);
            }
            queue<int> bfs;
            bfs.push(i);
            int ctr = 0;
            while(!bfs.empty()) {
                int next = bfs.front();
                ans[next] = (ctr<a ? small : big);
                ctr++;
                bfs.pop();
                for(int k: child[next]) {
                    bfs.push(k);
                }
            }
            bfs.push(0);
            ctr = 0;
            while(!bfs.empty()) {
                int next = bfs.front();
                ans[next] = (ctr<b ? mid : big);
                ctr++;
                bfs.pop();
                for(int k: child[next]) {
                    if(ans[k]==0) {
                        bfs.push(k);
                    }
                }
            }
            for(int k=0; k<n; k++) {
                assert(ans[k]!=0);
            }
            return ans;
        } else if(subtree[i]>b+c) {
            // if have heavy node with all light children
            bool all_light = true;
            for(int j: child[i]) {
                if(subtree[j]>=a) {
                    all_light = false;
                    // break;
                }
            }
            if(all_light) {
                if(i==0) {
                    assert(child[i].size()>=3);
                    return ans;
                }
                int minsize = 1;
                vector<int> tmp; // consists of all the children to be taken
                for(int j: child[i]) {
                    // add all those with bad lowtime
                    if(low[j]>= visit[i]) {
                        minsize += subtree[j];
                        tmp.push_back(j);
                    }
                }
                if(minsize > b+c) {
                    return ans;
                }
                for(int j: child[i]) {
                    // add the relevant subtrees
                    if(low[j]<visit[i] && minsize < b) {
                        minsize += subtree[j];
                        tmp.push_back(j);
                    }
                }
                assert(minsize >= b);
                assert(minsize <= b+c);
                // now for each subtree, color its children completely except that some are colored with 3
                int ctr = 1;
                ans[i] = mid;
                queue<int> bfs;
                for(int j: tmp) {
                    bfs.push(j);
                    while(!bfs.empty()) {
                        int next = bfs.front();
                        assert(ans[next]==0);
                        ans[next] = (ctr<b ? mid : big);
                        ctr++;
                        bfs.pop();
                        for(int k: child[next]) {
                            bfs.push(k);
                        }
                    }
                }
                //assert(ctr==b);
                ctr = 1;
                bfs.push(0); // we know that 0 is not the root
                ans[0] = small;
                while(!bfs.empty()) {
                    int next = bfs.front();
                    bfs.pop();
                    for(int k: adj[next]) {
                        if(ans[k]==0) {
                            bfs.push(k);
                            ans[k] = (ctr<a ? small : big);
                            ctr++;
                        }
                    }
                }
                //assert(ctr==a);
                for(int k=0; k<n; k++) {
                    //assert(ans[k]!=0);
                }
                return ans;
            }
        }
    }
    assert(false);
    return p;
}

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