poj 2446:Chessboard

2446:Chessboard

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总时间限制:

2000ms

内存限制:

65536kB

描述

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.

输入

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.

输出

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

样例输入

4 3 2
2 1
3 3

样例输出

YES

提示

A possible solution for the sample input.

题意：

       玩个游戏：给出一个m行n列的棋盘，里面有m*n个方格，其中有k个格子上有洞，我们称那些没洞的格子叫正常的格子（normal grid）,Bob要遵循两个规则去玩：（1）任何一个正常的格子都要被一张卡覆盖，（卡片是1*2规格的）（2）一张卡要正好覆盖两个相邻的正常格子

我们的任务是帮助Bob决定是否棋盘在上述两个规则下能被覆盖。

 

思路：

因为棋盘上都是两个格子放一张卡片，所以到最后肯定是两个点两个点连着的。由此想到了二分匹配，具体是这样的：

 给每个格子编号，从第一行到最后一行编号为1—12 ，然后每个点跟临近的正常点连接，这就建成了二分图

 代码：

#include<stdio.h>
#include<string.h>
#define N 2050
int dx[5]={1,0,0,-1};
int dy[5]={0,1,-1,0};
int map[N][N],vis[N][N],ID[N][N],mode[N],used[N];
int n,m,k,t=0;
int find(int x)//寻找增广路 
{
	int i;
	for(i=0;i<t;i++)
	 {
	 	if(map[x][i]&&!used[i])
	 	{
	 		used[i]=1;
	 		if(mode[i]==0||find(mode[i]))
	 		{
	 			mode[i]=x;
	 			return 1;
			 }
		 }
	 }
	 return 0;
}
int main()
{
	while(scanf("%d%d%d",&n,&m,&k)!=EOF)
	{
		memset(vis,0,sizeof(vis));
		memset(map,0,sizeof(map));
		memset(mode,0,sizeof(mode));
		int i,j,x,y;
		while(k--)
		{
			scanf("%d%d",&x,&y);
			if(x&&y)
			vis[y-1][x-1]=1;
		}
		 t=0;
		for(i=0;i<n;i++)
		 for(j=0;j<m;j++)
		  {
		  	if(!vis[i][j])
		    ID[i][j]=t++;
		  }
		  if(t%2!=0)
		  {
		  	printf("NO\n");
		  	continue;
		  }
		  for(i=0;i<n;i++)
		   {
		   	 for(j=0;j<m;j++)
		   	  {
		   	  	if(!vis[i][j])
		   	  	{
		   	  		for(k=0;k<4;k++)//四个方向 （可能双向，所以匹配数变成2倍） 
		   	  		 {
		   	  		 	int xi=i+dx[k];
		   	  		 	int yi=j+dy[k];
		   	  		 	if(xi>=0&&xi<n&&yi>=0&&yi<m&&!vis[xi][yi])
		   	  		 	map[ID[i][j]][ID[xi][yi]]=1;//建图 
					}
				 }
				 }
		   }
		   int s=0;
		   for(i=0;i<t;i++)
		   {
		   	memset(used,0,sizeof(used));
		   	if(find(i))
		   	{
		   	s++;//记录路的数量 
		    }
		   }
		   if(s==t)//是否 是匹配数的二倍 
		   printf("YES\n");
		   else
		   printf("NO\n");
	}
	return 0;
}