Chessboard_poj2446_匹配



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

题目大意：

有那么一个棋盘和很多1*2大小的方块，棋盘上还有一些用坐标表示的洞，求能否不重叠且不覆盖破洞地用方块铺满整

思路：

因为对于每一个单独的点只能和其他四个临近点一起放上一个方块，所以对每个点四周可以一起放置的点连边，然后判断是否能铺满也就是是否完美匹配

优化：

假设有两个相邻点a(x,y)，b(s,t)，那么a->b和b->a其实是同一种，所以可以想到枚举偶数点然后对临近点连边，大概是少一半的时间。（然而我比较懒而且已经ac了不是吗:-P）

源代码/pas：

 

type  edge=record    x,y,next:longint;  end;const  dx:array[1..4]of integer=(-1,1,0,0);  dy:array[1..4]of integer=(0,0,-1,1);var  link,ls:array[0..1024]of longint;  f:array[0..33,0..33]of longint;  e:array[0..4096]of edge;  v:array[0..1024]of boolean;  maxE,n,m,num:longint;procedure add(x,y:longint);begin  inc(maxE);  e[maxE].x:=x;  e[maxE].y:=y;  e[maxE].next:=ls[x];  ls[x]:=maxE;end;function find(x:longint):boolean;var  i,k:longint;begin  find:=true;  i:=ls[x];  while i>0 do  with e[i] do  begin    if not v[y] then    begin      v[y]:=true;      k:=link[y];      link[y]:=x;      if (k=0)or find(k) then exit;      link[y]:=k;    end;    i:=next;  end;  find:=false;end;procedure main;var  i:longint;begin  for i:=1 to num do  begin    fillchar(v,sizeof(v),false);    if not find(i) then    begin      writeln('NO');      exit;    end;  end;  writeln('YES');end;procedure init;var  i,j,l,k,x,y:longint;begin  readln(m,n,k);  for i:=1 to k do  begin    readln(x,y);    f[x,y]:=-1;  end;  for i:=1 to n do  for j:=1 to m do  if f[i,j]=0 then  begin    inc(num);    f[i,j]:=num;  end;  for i:=1 to n do  for j:=1 to m do  if f[i,j]>0 then  for l:=1 to 4 do  if f[i+dx[l],j+dy[l]]>0 then  add(f[i,j],f[i+dx[l],j+dy[l]]);end;begin  init;  main;end.