poj 2446 Chessboard

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.

Key To Problem

题目大意：给定一个n∗m的方格，问是否可以用一些1∗2的方块在有一些格点无法覆盖的情况下将剩余格点全部覆盖。

题解：将这个方格想象成类似国际象棋棋盘的东西，则每个黑格子与之相邻的必定是一个白格子，这样，就可以构造出一个二分图，如果方格中两个点相邻且没有坏点，就可以将这两个格子连边，最后求方格的最大匹配就是最多可以放上的方块数量。

Code

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 35
using namespace std;
struct ss
{
int next,to;
};
ss Edge[N*N*2];
int n,m,k,tot;
bool used[N*N];
bool map

;
bool ll[N*N];
int head[N*N];
int f[N*N];

void clear()
{
tot=0;
memset(head,0,sizeof(head));
memset(map,false,sizeof(map));
memset(ll,0,sizeof(ll));
}

void add(int x,int y)
{
Edge[++tot].next=head[x];
Edge[tot].to=y;
head[x]=tot;
}

bool dfs(int u)
{
for(int i=head[u];i;i=Edge[i].next)
{
int to=Edge[i].to;
if(!used[to])
{
used[to]=true;
if(f[to]==-1||dfs(f[to]))
{
f[to]=u;
return true;
}
}
}
return false;
}

int hungary()
{
int cnt=0;
memset(f,-1,sizeof(f));
for(int i=1;i<=n*m;i++)
{
if(ll[i])continue;
memset(used,0,sizeof(used));
if(dfs(i))cnt++;
}
return cnt;
}

int main()
{
while(scanf("%d%d%d",&n,&m,&k)!=EOF)
{
clear();
for(int i=1;i<=k;i++)
{
int x,y;
scanf("%d%d",&y,&x);
map[x][y]=true;
}
int pp=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
ll[++pp]=map[i][j];
if(map[i][j])continue;
if(i-1>0&&!map[i-1][j])
{
int x=(i-2)*m+j;
if((i+j)&1)add(pp,x);
else add(x,pp);
}
if(j-1>0&&!map[i][j-1])
{
int x=(i-1)*m+j-1;
if((i+j)&1)add(pp,x);
else add(x,pp);
}
}
}
if(hungary()*2==n*m-k)puts("YES");
else puts("NO");
}
return 0;
}