Codeforces 1059B Forgery

最新推荐文章于 2022-07-22 13:33:59 发布

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本文介绍了一种用于验证网格中填充元素是否满足特定条件的算法。该算法首先标记所有禁止放置元素的单元格，然后检查每个已填充的单元格是否至少有一个非禁止的邻居。时间复杂度为O(nm)，适用于快速验证大规模网格数据的有效性。

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附官方题解：

Each empty cell forbids to put a pen into every neighbor. Also, the border of the grid is forbidden. Let's mark all the forbidden cells. Now we have to check if for each filled cell there is at least one non-forbidden neighbor. Time complexity is O(nm).

我怎么就想不到呢！

#define  _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <algorithm>
#include <deque>
#include <iostream>
#include <string>
#include <math.h>

using namespace std;

int n, m;
char temp[1005][1005];
bool vis[1005][1005];

void x(int x, int y) {
	vis[x - 1][y - 1] = false;
	vis[x - 1][y] = false;
	vis[x - 1][y + 1] = false;
	vis[x][y - 1] = false;
	vis[x][y + 1] = false;
	vis[x + 1][y - 1] = false;
	vis[x + 1][y] = false;
	vis[x + 1][y + 1] = false;
}

bool check(int x,int y) {
	return vis[x - 1][y - 1] || vis[x - 1][y] || vis[x - 1][y + 1] || vis[x][y - 1] || vis[x][y + 1] || vis[x + 1][y - 1] || vis[x + 1][y] || vis[x + 1][y + 1];
}

int main() {
	cin >> n >> m;
	for (int i = 1;i <= n;++i) {
		for (int j = 1;j <= m;++j) {
			cin >> temp[i][j];
		}
	}
	for (int i = 1;i <= n;++i) {
		for (int j = 1;j <= m;++j) {
			vis[i][j] = true;
		}
	}
	for (int i = 1;i <= n;++i) {
		vis[i][1] = false;
		vis[i][m] = false;
	}
	for (int j = 1;j <= m;++j) {
		vis[1][j] = false;
		vis[n][j] = false;
	}
	for (int i = 1;i <= n;++i) {
		for (int j = 1;j <= m;++j) {
			if (temp[i][j] == '.') {
				x(i, j);
			}
		}
	}
	bool flag = true;
	for (int i = 1;i <= n;++i) {
		for (int j = 1;j <= m;++j) {
			if (temp[i][j] == '#') {
				if (check(i, j) == false) {
					flag = false;
				}
			}
		}
	}
	if (flag) {
		printf("YES\n");
	}
	else {
		printf("NO\n");
	}
	//system("pause");
	return 0;
}