Submission #294640

#	Time	Username	Problem	Language	Result	Execution time	Memory
294640		_7_7_	Split the Attractions (IOI19_split)	C++14	0 / 100	50 ms	6904 KiB

split

#include "split.h"
#include <bits/stdc++.h>                                           
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
//#define int long long
//#pragma GCC optimize("Ofast")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")
 
 
#define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout);
#define fastio ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define all(x) x.begin(), x.end()
#define sz(s) (int)s.size()
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define s second
#define f first
 
 
 
 
typedef pair < long long, long long > pll;    
typedef pair < int, int > pii;  
typedef unsigned long long ull;         
typedef vector < pii > vpii;                                   	
typedef vector < int > vi;
typedef long double ldb;  
typedef long long ll;  
typedef double db;
 
typedef tree < int, null_type, less < int >, rb_tree_tag, tree_order_statistics_node_update > ordered_set;
 
const int inf = 1e9, maxn = 2e5 + 48, mod = 998244353, N = 112;
const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, block = 300;
const pii base = mp(1171, 3307), Mod = mp(1e9 + 7, 1e9 + 9);
const db eps = 1e-12, pi = acos(-1);
const ll INF = 1e18;

using namespace std;

int n, m, root;
vi g[N], order;
bool was[N];

void dfs (int v) {
	was[v] = 1;
	order.pb(v);
	for (auto to : g[v])
		if (!was[to]) 
			dfs(to);		
}


vi find_split(int _n, int a, int b, int c, vi p, vi q) {
	n = _n;
	m = sz(p);
	for (int i = 0; i < m; ++i) {
		g[p[i]].pb(q[i]);
		g[q[i]].pb(p[i]);
	}		

	for (int i = 0; i < n; ++i)
		if (sz(g[i]) == 1)
			root = i;

	dfs(root);

	vi res;
	res.resize(n);
	for (int i = 0; i < a; ++i)
		res[order[i]] = 1;

	for (int i = a; i < a + b; ++i)
		res[order[i]] = 2;

	for (int i = a + b; i < n; ++i)
		res[order[i]] = 3;

	return res;			
}

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