poj 2446 Chessboard(经典二分匹配)

题目：​​http://poj.org/problem?id=2446​​

Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15250		Accepted: 4751

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 22 1 3 3

Sample Output

YES

Hint

 

A possible solution for the sample input.

分析：一眼能看出这是二分匹配吗？所以我觉得这很经典（当然我也是听了别人的思路才懂的）。一张卡片是1*2的规模，所以一个格子就有上下左右四种可能的相邻的格子与他组合成卡片（没有hole），这就是匹配的雏形，于是用邻接表把所有格子的各种匹配情况都表示出来，最后用匈牙利最大匹配求出结果再与总的格子数，holes数做关系比较就能得到最终的判断。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=2000;
bool bmap[N][N],bmask[N],tag[35][35];
int nx,ny,cx[N],cy[N];
int find(int u){
    for(int j=1;j<=ny;j++){
        if(bmap[u][j]&&!bmask[j]){
            bmask[j]=1;
            if(cy[j]==-1||find(cy[j])){
                 cy[j]=u;
                 cx[u]=j;
                 return 1;
            }
        }
    }
    return 0;
}
int maxmatch(){
    int res=0;
    memset(cx,-1,sizeof(cx));
    memset(cy,-1,sizeof(cy));
    for(int i=1;i<=nx;i++){
        if(cx[i]==-1){
            for(int j=1;j<=ny;j++) bmask[j]=0;
            res+=find(i);
        }
    }
    return res;
}
int dir[4][2]={{0,-1},{0,1},{-1,0},{1,0}};
int main()
{
    //freopen("cin.txt","r",stdin);
    int m,n,k;
    while(cin>>m>>n>>k){
         int a,b;
         nx=ny=n*m;
         memset(tag,0,sizeof(tag));
         memset(bmap,0,sizeof(bmap));
         for(int i=0;i<k;i++){
            scanf("%d%d",&b,&a);
            tag[a][b]=1;
         }
         for(int i=1;i<=m;i++){
            for(int j=1;j<=n;j++){
                if(tag[i][j]==1) continue;
                int x,y;
                for(int k=0;k<4;k++){
                    x=i+dir[k][0];
                    y=j+dir[k][1];
                    if(x<=m&&x>=1&&y<=n&&y>=1&&tag[x][y]==0){
                         a=(i-1)*n+j;
                         b=(x-1)*n+y;
                         bmap[a][b]=1;
                    }
                }
            }
         }
         int res=maxmatch();
         //cout<<res<<endl;
         if(res==m*n-k) puts("YES");
         else puts("NO");
    }
    return 0;
}