poj Chessboard 2446 （最大匹配&转换）

Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15901		Accepted: 4926

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.
//题意：
给你一个m*n的矩形，现在在这个矩形上有k个洞，告诉你这k个洞的坐标位置，问是否可以用1*2的矩形纸片将除了洞的其他的位置都遮住，并且1*2的小矩形不能有相互遮盖。
//思路：

把这个棋盘看成一个类似国际象棋的棋盘，黑白相间的那种。任何2个相邻的格子肯定是不同颜色的（黑和白）。经过研究发现，如果一个格子的行号和列号加起来为奇数，那与它相邻的格子的行号和列号加起来一定是偶数，如果一个格子的行号和列号加起来为偶数，那与它相邻的格子的行号和列号加起来一定是奇数。

这样，我们就可以把这张棋盘分为奇数集合和偶数集合。加边的时候，我们只需要加它上面或左面就可以了。。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define IN __int64
#define ull unsigned long long
#define ll long long
#define N 1100
#define M 1000000007
using namespace std;
int head
,next[N*N],key[N*N],top;
int y
;
int vis
;
int map

;
int m,n,k;
void add(int u,int v)
{
key[top]=v;
next[top]=head[u];
head[u]=top++;
}
int dfs(int u)
{
int v;
for(int i=head[u];i!=-1;i=next[i])
{
v=key[i];
if(!vis[v])
{
vis[v]=1;
if(y[v]==-1||dfs(y[v]))
{
y[v]=u;
return 1;
}
}
}
return 0;
}
int main()
{
int xx,yy;
while(scanf("%d%d%d",&m,&n,&k)!=EOF)
{
if((m*n-k)&1)
{
printf("NO\n");
continue;
}
top=0;
memset(head,-1,sizeof(head));
memset(y,-1,sizeof(y));
memset(map,1,sizeof(map));
for(int i=0;i<k;i++)
{
scanf("%d%d",&yy,&xx);
map[xx-1][yy-1]=0;
}
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
{
if(map[i][j])
{
if(i-1>=0&&map[i-1][j])//上下方向连接
{
if((i+j)&1)//从下到上
add(i*n+j,(i-1)*n+j);
else//从上到下
add((i-1)*n+j,i*n+j);
}
if(j-1>=0&&map[i][j-1])//左右方向连接
{
if((i+j)&1)//从右向左
add(i*n+j,i*n+j-1);
else//从左向右
add(i*n+j-1,i*n+j);
}
}
}
}
int sum=0;
for(int i=0;i<n*m;i++)
{
memset(vis,0,sizeof(vis));
if(dfs(i))
sum++;
}
printf(sum==((m*n-k)/2)?"YES":"NO");
}
return 0;
}