PKU 2446 Chessboard

题目

    

Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 20511		Accepted: 6427

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

大意

    

一个棋盘内，有些地方有洞，有些地方没有，没有洞的地方可以放东西，求是否可以用1*2的长方形填满所有格子（除了洞）。当然长方形不能相互覆盖。

分析

显然，每一块矩形相当于把相邻的两个点匹配起来。 根据前面题目的启发，应该可以相当实用黑白染色，点的编号类似。然后依然是黑点连白点。 如果相邻的点不是洞，那么就连一条边。显然这是一个二分图。 然后跑这个二分图最大匹配。假设有k对匹配对，洞的数目为c，说明有2*k个点匹配成功。 如果满足k*2+c=n*m，那么有解。 连边后如图，其中1-4,3-6,5-8,7-10,11-12即为一组大小为5的最大匹配。

  

代码 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
using namespace std;
int n,m,x,y,ans,link[4005],s[2050][2050],ss;
vector<int> f[4005];
bool cover[4005],a[2050][2050];
int fx[4][2]={{1,0},{0,1},{0,-1},{-1,0}};//4个方向
bool find(int i)
{
     for (int k=0;k<f[i].size();k++)
       if (!cover[f[i][k]])
       {
           int j=f[i][k];
          cover[j]=true;
             int q=link[j];
            link[j]=i;
           if (q==0||find(q)) return true;
           link[j]=q;
        }
    return false;
}
int main()
 {
     cin>>n>>m>>ss;
    for(int i=1;i<=n;i++)
     for(int j=1;j<=m;j++)
      s[i][j]=m*(i-1)+j;
    for (int i=1;i<=ss;i++)
     {
         cin>>x>>y;
       a[y][x]=1;  //attention！！！！！
    }
     for (int i=1;i<=n;i++)
     {
         for (int j=1;j<=m;j++)
         {
             if (!a[i][j])
             {
                 int ax,ay;
                   for(int kk=0;kk<4;kk++)//4个方向
                   {
                       ax=i+fx[kk][0],ay=j+fx[kk][1];
                       if(ax<1||ax>n||ay<1||ay>m) continue;//判断是否越界
                       if(a[ax][ay]) continue;//判断该可攻击的点是否被移除
                       f[s[i][j]].push_back(s[ax][ay]);//连边
                   }
             }
                
         }
     }
     int ans=0;
     for (int i=1;i<=n*m;i++)
             {
                 memset(cover,0,sizeof(cover));
                 ans+=find(i);
             }
     if (2*ans+ss==n*m) cout<<"YES";
     else cout<<"NO";     
 }