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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#528533#1820. Contest ConstructionPetroTarnavskyi#WA 0ms3788kbC++202.5kb2024-08-23 15:56:162024-08-23 15:56:16

Judging History

你现在查看的是最新测评结果

  • [2024-08-23 15:56:16]
  • 评测
  • 测评结果:WA
  • 用时:0ms
  • 内存:3788kb
  • [2024-08-23 15:56:16]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define RFOR(i, a, b) for(int i = (a) - 1; i >= (b); i--)
#define SZ(a) int(a.size())
#define ALL(a) a.begin(), a.end()
#define PB push_back
#define MP make_pair
#define F first
#define S second

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef double db;

struct DSU
{
	int n;
	VI p, sz;
	
	void init(int _n)
	{
		n = _n;
		p.resize(n);
		iota(ALL(p), 0);
		sz.assign(n, 1);
	}
	
	int find(int v)
	{
		if (v == p[v])
			return v;
		return p[v] = find(p[v]);
	}
	bool unite(int u, int v)
	{
		u = find(u);
		v = find(v);
		if (u == v)
			return false;
		if (sz[u] > sz[v])
			swap(u, v);
		p[u] = v;
		sz[v] += sz[u];
		return true;
	}
};

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	int n, m;
	cin >> n >> m;
	int k;
	cin >> k;
	vector<VI> obst(n, VI(m));
	FOR (i, 0, k)
	{
		int x, y;
		cin >> x >> y;
		x--, y--;
		obst[x][y]++;
	}
	vector<VI> s(n + 1, VI(m + 1));
	FOR (i, 0, n)
	{
		FOR (j, 0, m)
		{
			s[i + 1][j + 1] = obst[i][j] + s[i + 1][j] + s[i][j + 1] - s[i + 1][j + 1];
		}
	}
	int l = 0, r = n + m;
	while (l + 1 < r)
	{
		int sz = (r + l) / 2;
		vector<VI> good(n, VI(m));
		FOR (i, 0, n)
		{
			FOR (j, 0, m)
			{
				if (i + sz < n && j + sz < m)
				{
					int sum = s[i + sz][j + sz] - s[i + sz][j] - s[i][j + sz] + s[i][j];
					if (sum == 0)
						good[i][j] = 1;
				}
			}
		}
		DSU d;
		d.init(n * m);
		int cnt = 0;
		int mx = 0;
		FOR (i, 0, n)
		{
			FOR (j, 0, m)
			{
				if (good[i][j])
				{
					cnt++;
					if (i && good[i - 1][j])
						d.unite(i * m + j, i * m - m + j);
					if (j && good[i][j - 1])
						d.unite(i * m + j, i * m - 1 + j);
					mx = max(mx, d.sz[d.find(i * m + j)]);
				}
			}
		}
		if (mx < cnt)
		{
			r = sz;
			continue;
		}
		vector<VI> pref(n + 1, VI(m + 1));
		FOR (i, 0, n)
		{
			FOR (j, 0, m)
			{
				pref[i + 1][j + 1] = good[i][j] + pref[i + 1][j] + pref[i][j + 1] - pref[i + 1][j + 1];
			}
		}
		bool ok = true;
		FOR (i, 0, n)
		{
			FOR (j, 0, m)
			{
				if (!obst[i][j])
				{
					int sum = pref[i + 1][j + 1] - pref[i + 1][max(j + 1 - sz, 0)] - pref[max(i + 1 - sz, 0)][j + 1] + pref[max(i + 1 - sz, 0)][max(j + 1 - sz, 0)];
					if (sum == 0)
						ok = false;
				}
			}
		}
		if (ok)
			l = sz;
		else
			r = sz;
	}
	if (l == 0)
		l = -1;
	cout << l << '\n';
	
	
	return 0;
}


Details

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Test #1:

score: 0
Wrong Answer
time: 0ms
memory: 3788kb

input:

5 4
2
1
4
3
5

output:

-1

result:

wrong answer expected '4', found '-1'