Submission #124106

#	Time	Username	Problem	Language	Result	Execution time	Memory
124106		eriksuenderhauf	ICC (CEOI16_icc)	C++11	90 / 100	173 ms	632 KiB

icc

//#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#include "icc.h"
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
#define mem(a,v) memset((a), (v), sizeof (a))
#define enl printf("\n")
#define case(t) printf("Case #%d: ", (t))
#define ni(n) scanf("%d", &(n))
#define nl(n) scanf("%I64d", &(n))
#define nai(a, n) for (int i = 0; i < (n); i++) ni(a[i])
#define nal(a, n) for (int i = 0; i < (n); i++) nl(a[i])
#define pri(n) printf("%d\n", (n))
#define prl(n) printf("%I64d\n", (n))
#define pii pair<int, int>
#define pil pair<int, long long>
#define pll pair<long long, long long>
#define vii vector<pii>
#define vil vector<pil>
#define vll vector<pll>
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef cc_hash_table<int,int,hash<int>> ht;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> oset;
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int MAXN = 1e2 + 5;
const double eps = 1e-9;
int par[MAXN], sz[MAXN];
vi comp[MAXN];

int qry(int x) { return par[x] == x ? par[x] : qry(par[x]); }
void join(int u, int v) {
	u = qry(u), v = qry(v);
	if (u == v) return;
	if (sz[u] > sz[v]) swap(u, v);
	par[u] = v;
	comp[v].insert(comp[v].end(), comp[u].begin(), comp[u].end());
}

void print(vi cur) {
	for (int i: cur) cerr << i-1 << " "; cerr << "\n";
}

pii solve(int N) {
	pii ret;
	vi cand;
	for (int i = 1; i <= N; i++) {
		if (qry(i) != i)
			continue;
		cand.pb(i);
	}
	vi l, r;
	int msk = 0;
	for (int ind = 0; (1<<ind) < (int)cand.size(); ind++) {
		l.clear();
		r.clear();
		for (int j = 0; j < (int)cand.size(); j++) {
			if ((j >> ind) & 1)
				l.insert(l.end(), comp[cand[j]].begin(), comp[cand[j]].end());
			else
				r.insert(r.end(), comp[cand[j]].begin(), comp[cand[j]].end());
		}
		int la[l.size()], ra[r.size()];
		copy(l.begin(), l.end(), la);
		copy(r.begin(), r.end(), ra);
		int fl = query((int)l.size(),(int)r.size(), la, ra);
		msk |= fl * (1 << ind);
		//print(l); print(r);
		//cerr << fl << "\n";
		//if (fl == 1)
		//	break;
	}
	l.clear();
	r.clear();
	//cerr << msk << "\n";
	for (int i = 0; i < (int)cand.size(); i++) {
		//cerr << i << " " << (i^msk) << "\n";
		if (i & (msk & -msk))
			l.insert(l.end(), comp[cand[i]].begin(), comp[cand[i]].end());
		else
			r.insert(r.end(), comp[cand[i]].begin(), comp[cand[i]].end());
	}
	int lo = 0, hi = l.size()-1;
	if (hi == lo)
		hi--;
	while (lo <= hi) {
		int mi = (lo+hi) / 2;
		vi tmp;
		for (int i = 0; i <= mi; i++)
			tmp.pb(l[i]);
		int la[tmp.size()], ra[r.size()];
		copy(tmp.begin(), tmp.end(), la);
		copy(r.begin(), r.end(), ra);
		int fl = query((int)tmp.size(),(int)r.size(), la, ra);
		if (fl == 1)
			hi = mi - 1;
		else
			lo = mi + 1;
	}
	ret.fi = l[lo];
	lo = 0, hi = r.size()-1;
	if (hi == lo)
		hi--;
	while (lo <= hi) {
		int mi = (lo+hi) / 2;
		vi tmp;
		for (int i = 0; i <= mi; i++)
			tmp.pb(r[i]);
		int la[l.size()], ra[tmp.size()];
		copy(l.begin(), l.end(), la);
		copy(tmp.begin(), tmp.end(), ra);
		int fl = query((int)l.size(),(int)tmp.size(), la, ra);
		if (fl == 1)
			hi = mi - 1;
		else
			lo = mi + 1;
	}
	ret.se = r[lo];
	return ret;
}


void run(int N) {
	for (int i = 1; i <= N; i++)
		par[i] = i, sz[i] = 1, comp[i] = {i};
	for (int t = 1; t < N; t++) {
		int a, b;
		tie(a, b) = solve(N);
		setRoad(a, b);
		join(a, b);
	}
}

Compilation message (stderr)

icc.cpp: In function 'void print(std::vector<int>)':
icc.cpp:51:2: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
  for (int i: cur) cerr << i-1 << " "; cerr << "\n";
  ^~~
icc.cpp:51:39: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
  for (int i: cur) cerr << i-1 << " "; cerr << "\n";
                                       ^~~~

• Subtask 1: 7 / 7
• Subtask 2: 11 / 11
• Subtask 3: 22 / 22
• Subtask 4: 21 / 21
• Subtask 5: 29 / 29
• Subtask 6: 0 / 10