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).



POJ2446Chessboard_#include


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:




POJ2446Chessboard_i++_02


A VALID solution.




POJ2446Chessboard_i++_03


An invalid solution, because the hole of red color is covered with a card.




POJ2446Chessboard_匹配_04


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



POJ2446Chessboard_sed_05


A possible solution for the sample input.




这题太坑，坐标输进来是列，行的，开始没注意错了很久，其他没什么难的，只要相邻且不是洞就连边。

然后求最大匹配，如果最大匹配等于剩余格子数量，就是YES，否则是NO

#include<stdio.h>
#include<string.h>

const int maxn=1605;
struct node
{
    int to;
    int next;
}edge[4*maxn];
int head[maxn];
int mark[maxn];
bool used[maxn];
bool mat[55][55];
int cnt[55][55];
int tot;
int n,m;
int index;

void addedge(int from,int to)
{
    edge[tot].to=to;
    edge[tot].next=head[from];
    head[from]=tot++;
}

bool dfs(int x)
{
    for (int i=head[x];i!=-1;i=edge[i].next)
    {
        if (!used[edge[i].to])
        {
            used[edge[i].to]=1;
            if (mark[edge[i].to]==-1 || dfs(mark[edge[i].to]))
            {
                mark[edge[i].to]=x;
                return true;
            }
        }
    }
    return false;
}

int hungary()
{
    memset(mark,-1,sizeof(mark));
    int ans=0;
    for (int i=0;i<index;i++)
    {
        memset(used,0,sizeof(used));
        if (dfs(i))
            ans++;
    }
    return ans;
}

int main()
{
    int k;
    while (~scanf("%d%d%d",&n,&m,&k))
    {
        memset(head,-1,sizeof(head));
        memset(mat,false,sizeof(mat));
        tot=0;
        int x,y;
        for (int i=0;i<k;i++)
        {
            scanf("%d%d",&x,&y);
            x--;
            y--;
            mat[y][x]=true;
        }
        index=0;
        for (int i=0;i<n;i++)
            for (int j=0;j<m;j++)
                if (!mat[i][j])
                    cnt[i][j]=index++;
        for (int i=0;i<n;i++)
            for (int j=0;j<m;j++)
            {
                if (!mat[i][j])
                {
                    if (i>0 && !mat[i-1][j])
                        addedge(cnt[i][j],cnt[i-1][j]);
                    if (i<n-1 && !mat[i+1][j])
                        addedge(cnt[i][j],cnt[i+1][j]);
                    if (j>0 && !mat[i][j-1])
                        addedge(cnt[i][j],cnt[i][j-1]);
                    if (j<m-1 && !mat[i][j+1])
                        addedge(cnt[i][j],cnt[i][j+1]);
                }
            }
        int res=hungary();
        if (res==index)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}




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