POJ 2446 Chessboard

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最新推荐文章于 2025-03-30 16:45:00 发布

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本文探讨了一个经典的棋盘覆盖问题，即使用1*2的矩形卡片覆盖带有随机空洞的M*N大小的棋盘，同时确保卡片不覆盖空洞且棋盘上的每个格子都被恰好一张卡片覆盖。文章提供了详细的解决方案及实现代码。

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Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11700		Accepted: 3653

Description

Alice and Bob often play games on chessboard. One day, Alice draws a board with size M * N. She wants Bob to use a lot of cards with size 1 * 2 to cover the board. However, she thinks it too easy to bob, so she makes some holes on the board (as shown in the figure below).

We call a grid, which doesn’t contain a hole, a normal grid. Bob has to follow the rules below:
1. Any normal grid should be covered with exactly one card.
2. One card should cover exactly 2 normal adjacent grids.

Some examples are given in the figures below:

A VALID solution.

An invalid solution, because the hole of red color is covered with a card.

An invalid solution, because there exists a grid, which is not covered.
Your task is to help Bob to decide whether or not the chessboard can be covered according to the rules above.

Input

There are 3 integers in the first line: m, n, k (0 < m, n <= 32, 0 <= K < m * n), the number of rows, column and holes. In the next k lines, there is a pair of integers (x, y) in each line, which represents a hole in the y-th row, the x-th column.

Output

If the board can be covered, output "YES". Otherwise, output "NO".

Sample Input

4 3 2
2 1
3 3

Sample Output

YES

Hint

A possible solution for the sample input.

Source

POJ Monthly,charlescpp

题意：

给出一个m*n的棋盘，其中有k个洞，要求用1*2的木块去盖住整个棋盘，规定木板不能重叠，而且洞口上面不能有木板。若能用木板将所有非洞处覆盖，则输出YES；反之则输出NO。

代码：

#include<cstdio>
#include<cstring>
#define N 150
#define MAX 2000

bool map[MAX][MAX];
bool vist[MAX];
int pre[MAX];
int mat[N][N];
int m,n,k;
int tm,tn;

bool find(int cur)
{
	for(int i=1;i<=tn;i++)
	{
		if(map[cur][i] && !vist[i])
		{
			vist[i]=true;
			if(pre[i]==-1 || find(pre[i]))
			{
				pre[i]=cur;
				return true;
			}
		}
	}
	return false;
}

int main()
{
	int x,y;
	while(scanf("%d%d%d",&m,&n,&k)!=EOF)
	{
		memset(mat,0,sizeof(mat));
		memset(map,false,sizeof(map));
		memset(pre,-1,sizeof(pre));

		if((m*n-k)%2!=0)
		{
			printf("NO\n");
			return 0;
		}

		for(int i=1;i<=k;i++)
		{
			scanf("%d%d",&x,&y);
			mat[y][x]=-1;
		}

		int j;
		tn=0;
		for(int i=1;i<=m;i++)
		{
			if(i%2==0) j=1;
			else j=2;
			for(;j<=n;j+=2)
				if(mat[i][j]!=-1)
					mat[i][j]=++tn;
			
		}

		tm=0;
		for(int i=1;i<=m;i++)
		{
			if(i%2==0) j=2;
			else j=1;
			for(;j<=n;j+=2)
				if(mat[i][j]!=-1)
				{
					mat[i][j]=++tm;
					if(j>1 && mat[i][j-1]!=-1)
						map[mat[i][j]][mat[i][j-1]]=true;
					if(j<n && mat[i][j+1]!=-1)
						map[mat[i][j]][mat[i][j+1]]=true;
					if(i>1 && mat[i-1][j]!=-1)
						map[mat[i][j]][mat[i-1][j]]=true;
					if(i<m && mat[i+1][j]!=-1)
						map[mat[i][j]][mat[i+1][j]]=true;
				}
		}

		int ans=0;
		for(int i=1;i<=tm;i++)
		{
			memset(vist,false,sizeof(vist));
			if(find(i)) ans++;
		}

		if((m*n-k)/2==ans) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}