POJ 2446-Chessboard(二分图_最大匹配)

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

题意：m*n的矩阵（m是列，n是行），有k个点有洞，问2*1的木板是否可以将剩下的铺满，如果可以，输出yes，否则输出no。

思路：将有洞的格子标记，然后将剩下的格子链接相邻的。差点超时。看到网上有人用奇偶分。

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>
using namespace std;
const int inf=0x3f3f3f3f;
int map[2010][2010];
int G[2010][2010];
int vis[2010];
int link[2010];
int n,m;
int cnt;
int jx[]={1,-1,0,0};
int jy[]={0,0,1,-1};
int dfs(int u)
{
    int i;
    for(i=1; i<=cnt; i++) {
        if(!vis[i]&&map[u][i]) {
            vis[i]=1;
            if(link[i]==-1||dfs(link[i])) {
                link[i]=u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int k,i,j,l;
    int a,b;
    while(~scanf("%d %d %d",&m,&n,&k)) {
        memset(map,0,sizeof(map));
        memset(link,-1,sizeof(link));
        memset(G,0,sizeof(G));
        cnt=0;
        for(i=1; i<=k; i++) {
            scanf("%d %d",&b,&a);
            G[a][b]=-1;
        }
        for(i=1;i<=m;i++) {
            for(j=1;j<=n;j++) {
                if(!G[i][j])
                    G[i][j]=++cnt;
            }
        }
        for(i=1; i<=m; i++) {
            for(j=1; j<=n; j++) {
                if(G[i][j]!=-1)
                    for(l=0;l<4;l++) {
                        int x=i+jx[l];
                        int y=j+jy[l];
                        if(x>=1&&x<=m&&y>=1&&y<=n) {
                            map[G[i][j]][G[x][y]]=1;
                        }
                    }
            }
        }
        int sum=0;
        for(i=1; i<=cnt; i++) {
            memset(vis,0,sizeof(vis));
            if(dfs(i))
                sum++;
        }
        if(sum==cnt)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}