Chessboard - POJ 2446 二分图匹配

最新推荐文章于 2018-12-26 23:00:11 发布

提比-我有特殊的AC技巧

最新推荐文章于 2018-12-26 23:00:11 发布

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分类专栏： POJ 图论

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本文探讨了一种利用1*2的矩形卡片覆盖带有洞孔的M*N大小棋盘的问题。文章提供了完整的AC代码实现，并介绍了如何通过深度优先搜索算法解决这一挑战。

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Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14001		Accepted: 4354

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

题意：是否存在一种方式，使得用1*2的长方形覆盖所有空白区域。

思路：先判断空白区域是否为奇数，然后就是模板了。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
int n,m,p,val[40][40],match[40][40][2],vis[40][40],VIS,dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
bool dfs(int x,int y)
{
    int i,j,k,x2,y2;
    for(i=0;i<4;i++)
    {
       x2=x+dx[i];y2=y+dy[i];
       if(val[x2][y2]==0 && vis[x2][y2]!=VIS)
       {
           vis[x2][y2]=VIS;
           if(match[x2][y2][0]==0 || dfs(match[x2][y2][0],match[x2][y2][1]))
           {
               match[x2][y2][0]=x;
               match[x2][y2][1]=y;
               return true;
           }
       }
    }
    return false;
}
int main()
{
    int i,j,k,x,y,ret;
    while(~scanf("%d%d%d",&n,&m,&p))
    {
        memset(val,0,sizeof(val));
        for(i=1;i<=p;i++)
        {
            scanf("%d%d",&x,&y);
            val[y][x]=1;
        }
        for(i=0;i<=n;i++)
           val[i][0]=val[i][m+1]=1;
        for(i=0;i<=m;i++)
           val[0][i]=val[n+1][i]=1;
        if((n*m-p)&1)
        {
            printf("NO\n");
            continue;
        }
        ret=0;
        memset(match,0,sizeof(match));
        for(i=1;i<=n;i++)
           for(j=1;j<=m;j++)
              if((i+j)%2==0 && val[i][j]==0)
              {
                  VIS++;
                  if(dfs(i,j))
                    ret++;

              }
        if(ret==(n*m-p)/2)
          printf("YES\n");
        else
          printf("NO\n");
    }
}