제출 #657008

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
657008		Shahrad	T-Covering (eJOI19_covering)	C++17	15 / 100	112 ms	93768 KiB

covering

#include "bits/stdc++.h"
using namespace std;

typedef long long ll;
typedef long double ld;
#define endl '\n'
#define pb push_back
#define mk make_pair
#define sz size()
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define kill(x) return cout << x << endl, void();
#define int ll
typedef pair <int, int> pii;

#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target ("avx2")

const int N = 1e6 + 5;
const int MOD = 998244353, INF = 2e9, MOD2 = 1e9 + 7, sq = 350;

int dx[8] = {0, 0, 1, -1, -1, -1, 1, 1}, dy[8] = {-1, 1, 0, 0, -1, 1, 1, -1}, vis[N];
vector <int> a[N], mrk[N], adj[N];
int ans, mn, few, n, m;

bool val (int x, int y)
{
	return 0 <= x && x < n && 0 <= y && y < m;
}

void wef (int x, int y, int q)
{
	ans += a[x][y];
	mrk[x][y] = 0;
	a[x][y] = -INF;
	for (int i = 0; i < 4; i++)
	{
		if (dx[i] == dx[q])
			continue;
		if (!val (x + dx[i], y + dy[i]))
		{
			ans += -INF;
			continue;
		}
		ans += a[x + dx[i]][y + dy[i]];
		a[x + dx[i]][y + dy[i]] = -INF;
	}
}

void dfs (int v, int par)
{
	vis[v] = 1;
	int x = v / m, y = v % m;
	for (int i = 0; i < 4; i++)
	{
		if (val (x + dx[i], y + dy[i]))
			mn = min (mn, a[x + dx[i]][y + dy[i]]);
		else
			mn = -INF;
	}
	for (int u : adj[v])
	{
		if (u != par && !vis[u])
			dfs (u, v);
		else if (u != par)
			few++;
	}
}

void Solve ()
{
	int k, x, y;
	cin >> n >> m;
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
		{
			cin >> x;
			a[i].pb (x);
			mrk[i].pb (0);
		}
	cin >> k;
	for (int i = 0; i < k; i++)
	{
		cin >> x >> y;
		mrk[x][y] = 1;
	}
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
			for (int q = 0; q < 8; q++)
				if (val (i + dx[q], j + dy[q]) && mrk[i][j] && mrk[i + dx[q]][j + dy[q]])
				{
					wef (i, j, q);
					wef (i + dx[q], j + dy[q], (q + 2) % 4 + (q > 3 ? 4 : 0));
				}
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
		{
			if (!mrk[i][j])
				continue;
			ans += a[i][j];
			for (int q = 0; q < 4; q++)
			{
				if (val (i + dx[q], j + dy[q]))
					ans += a[i + dx[q]][j + dy[q]];
				else
					ans += -INF;
			}
			if (val (i + 2, j) && mrk[i + 2][j])
			{
				int v = i * m + j, u = (i + 2) * m + j;
				ans -= a[i + 1][j];
				adj[v].pb (u);
				adj[u].pb (v);
			}
			if (val (i, j + 2) && mrk[i][j + 2])
			{
				int v = i * m + j, u = i * m + j + 2;
				ans -= a[i][j + 1];
				adj[v].pb (u);
				adj[u].pb (v);
			}
		}
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
		{
			if (!mrk[i][j])
				continue;
			int v = i * m + j;
			if (vis[v])
				continue;
			mn = INF, few = 0;
			dfs (v, -1);
			if (few > 1)
				kill ("No");
			if (few == 0)
				ans -= mn;
		}
	if (ans < 0)
		kill ("No");
	cout << ans << endl;
}

int32_t main ()
{
	ios::sync_with_stdio (0), cin.tie (0), cout.tie (0);
	int tt = 1;
	//	cin >> tt;
	while (tt--)
		Solve ();
}

• Subtask 1: 5 / 5
• Subtask 2: 0 / 10
• Subtask 3: 0 / 10
• Subtask 4: 10 / 10
• Subtask 5: 0 / 15
• Subtask 6: 0 / 20
• Subtask 7: 0 / 30