제출 #815037

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
815037		physics07	Split the Attractions (IOI19_split)	C++17	18 / 100	110 ms	23448 KiB

split

#include "split.h"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<int> adj[100001], tree[100001];
bool visited[100001];
int sz[100001], n, c, cc[100001];
int dfs_tree(int x) {
    visited[x]=1;
    sz[x]=1;
    for(auto i: adj[x]) if(!visited[i]) {
        tree[x].push_back(i);
        tree[i].push_back(x);
        sz[x]+=dfs_tree(i);
    }
    return sz[x];
}
int centroid(int x) {
    visited[x]=1;
    for(auto i: tree[x]) if(!visited[i] && sz[i]*2>n) return centroid(i);
    return x;
}
int sol(int x) {
    visited[x]=1;
    cc[x]=1;
    for(auto i: adj[x]) if(!visited[i] && i!=c) cc[x]+=sol(i);
    return cc[x];
}
vector<int> ans;
int cnt;
void get_all(int x, int p, int avoid=-1) {
    visited[x]=1;
    ans[x]=p;
    cnt--;
    if(cnt<=0) return;
    for(auto i: adj[x]) if(!visited[i] && i!=avoid) {
        get_all(i, p, avoid);
        if(cnt<=0) return;
    }
}
vector<int> find_split(int N, int A, int B, int C, vector<int> p, vector<int> q) {
    n=N;
    vector<pair<int, int>> v;
    v.push_back({A, 1});
    v.push_back({B, 2});
    v.push_back({C, 3});
    sort(v.begin(), v.end());
    int a=v[0].first, b=v[1].first, c=v[2].first;
    int color[]={v[0].second, v[1].second, v[2].second};
    for(int i=0; i<n; i++) ans.push_back(0);
    int m=(int)p.size();
    for(int i=0; i<m; i++) {
        adj[p[i]].push_back(q[i]);
        adj[q[i]].push_back(p[i]);
    }
    dfs_tree(0);
    memset(visited, 0, sizeof(visited));
    c=centroid(0);
    memset(visited, 0, sizeof(visited));
    bool flag=0;
    int idx=0;
    for(int i=0; i<n; i++) if(!visited[i] && i!=c) {
        if(sol(i)>=a) {
            flag=1;
            idx=i;
            break;
        }
    }
    if(!flag) return ans;
    memset(visited, 0, sizeof(visited));
    cnt=a;
    get_all(idx, color[0], c);
    cnt=b;
    get_all(c, color[1]);
    for(int i=0; i<n; i++) if(!ans[i]) ans[i]=color[2];
    return ans;
}

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