Submission #1142730

#	Time	Username	Problem	Language	Result	Execution time	Memory
1142730		monkey133	Council (JOI23_council)	C++20	100 / 100	708 ms	99948 KiB

council

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	ll n, m;
	cin >> n >> m;
	vector<vector<ll> > a(n, vector<ll> (m));
	vector<ll> d1(n, 0), d2(n, 0), cnt(m, 0), ans(n, 0);
	vector<pi> dp1(1LL << m, mp(0, -1)), dp2(1LL << m, mp(0, -1));
	for (ll i=0; i<n; ++i)
	{
		for (ll j=0; j<m; ++j) cin >> a[i][j];
		for (ll j=0; j<m; ++j) if (a[i][j]==0) d1[i] |= (1LL << j);
	}
	for (ll i=0; i<m; ++i) for (ll j=0; j<n; ++j) cnt[i] += a[j][i];
	for (ll i=0; i<n; ++i) 
	{
		for (ll j=0; j<m; ++j) 
		{
			if (cnt[j] - n/2 - a[i][j] > 0) ans[i]++;
			if (cnt[j] - n/2 - a[i][j] == 0) d2[i] |= (1LL << j);
		}
	}
	for (ll i=0; i<n; ++i) 
	{
		pi now = mp(1, i);
		if (dp1[d1[i]].s == now.s && dp1[d1[i]].f < now.f) dp1[d1[i]].f = now.f;
		else if (dp1[d1[i]] < now) dp2[d1[i]] = dp1[d1[i]], dp1[d1[i]] = now;
		else if (dp1[d1[i]].s != now.s && dp2[d1[i]] < now) dp2[d1[i]] = now;
	}
	for (ll i=(1LL << m)-1; i>=0; --i)
	{
		for (ll j=0; j<m; ++j) if (i & (1LL << j))
		{
			ll mask = i ^ (1LL << j);
			pi now = dp1[i];
			if (dp1[mask].s == now.s && dp1[mask].f < now.f) dp1[mask].f = now.f;
			else if (dp1[mask] < now) dp2[mask] = dp1[mask], dp1[mask] = now;
			else if (dp1[mask].s != now.s && dp2[mask] < now) dp2[mask] = now;
			now = dp2[i];
			if (dp1[mask].s == now.s && dp1[mask].f < now.f) dp1[mask].f = now.f;
			else if (dp1[mask] < now) dp2[mask] = dp1[mask], dp1[mask] = now;
			else if (dp1[mask].s != now.s && dp2[mask] < now) dp2[mask] = now;
		}
	}
	for (ll i=0; i<(1LL << m); ++i) dp1[i].f*= __builtin_popcountll(i), dp2[i].f*= __builtin_popcountll(i);
	for (ll i=0; i<(1LL << m); ++i)
	{
		for (ll j=0; j<m; ++j) if (i & (1LL << j)) 
		{
			ll mask = i ^ (1LL << j);
			pi now = dp1[mask];
			if (dp1[i].s == now.s && dp1[i].f < now.f) dp1[i].f = now.f;
			else if (dp1[i] < now) dp2[i] = dp1[i], dp1[i] = now;
			else if (dp1[i].s != now.s && dp2[i] < now) dp2[i] = now;
			now = dp2[mask];
			if (dp1[i].s == now.s && dp1[i].f < now.f) dp1[i].f = now.f;
			else if (dp1[i] < now) dp2[i] = dp1[i], dp1[i] = now;
			else if (dp1[i].s != now.s && dp2[i] < now) dp2[i] = now;
		}
	}
	for (ll i=0; i<n; ++i)
	{
		if (dp1[d2[i]].s == i) ans[i] += dp2[d2[i]].f;
		else ans[i] += dp1[d2[i]].f;
		assert(dp1[d2[i]].s!=dp2[d2[i]].s);
		cout << ans[i] << endl;
	}
	
	
	
	return 0;
}

• Subtask 1: 8 / 8
• Subtask 2: 8 / 8
• Subtask 3: 6 / 6
• Subtask 4: 19 / 19
• Subtask 5: 15 / 15
• Subtask 6: 22 / 22
• Subtask 7: 22 / 22