poj-2446-Chessboard

poj-2446-Chessboard

 

Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19476		Accepted: 6147

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

 

17930785

2446

Accepted

4520K

63MS

G++

1399B

2017-12-08 10:55:23

 

二分图匹配问题。

题意：对于 Grid 中 非hole，能否用 1x2 的砖块铺满。

首先检验剩余砖块是否偶数。然后使用匈牙利算法进行匹配，解决。

 

 

// poj-2446 
#include <cstdio> 
#include <cstring>  

const int MAXN = 35; 
const int dx[4] = {0, 0, -1, 1}; 
const int dy[4] = {-1, 1, 0, 0}; 

int n, m, k, mp[MAXN][MAXN], vis[MAXN*MAXN], match[MAXN*MAXN]; 
int len[MAXN*MAXN], vt[MAXN*MAXN][MAXN*MAXN]; 

bool dfs(int x){
	for(int i=0; i<len[x]; ++i){
		int y = vt[x][i]; 
		if(vis[y] == 0){
			vis[y] = 1; 
			if(match[y] < 0 || dfs(match[y])){
				match[y] = x; 
				return true; 
			}
		}
	}
	return false; 
}

int main(){
	freopen("in.txt", "r", stdin); 

	int x, y, ans; 
	while(scanf("%d %d %d", &n, &m, &k) != EOF){
		memset(mp, 0, sizeof(mp)); 
		for(int i=0; i<k; ++i){
			scanf("%d %d", &x, &y);
			mp[y][x] = 1;  
		}

		if((m*n - k)%2 != 0){
			printf("NO\n");
			continue; 
		} 
		memset(len, 0, sizeof(len)); 
		for(int i=1; i<=n; ++i){
			for(int j=1; j<=m; ++j){
				if(mp[i][j] == 0){
					int c = (i - 1)*m + j; 
					for(int k=0; k<4; ++k){
						int nx = i + dx[k]; 
						int ny = j + dy[k]; 
						if(nx>=1 && nx<=n && ny>=1 && ny<=m && mp[nx][ny] == 0){
							vt[ c ][ len[c] ] =  m*(nx - 1) + ny; 
							++len[c]; 
						}
					}
				}
			}
		}

		memset(match, -1, sizeof(match)); 
		ans = 0; 
		for(int i=1; i<=n; ++i){
			for(int j=1; j<=m; ++j){ 
				int c = (i - 1)*m + j;
				memset(vis, 0, sizeof(vis)); 
				if(dfs(c)){
					++ans; 
				}  
			}
		}
		if(ans*2 == (n*m - k)){
			printf("YES\n");
		}else{
			printf("NO\n");
		}
	}
	return 0; 
}