POJ2446Chessboard

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 13 3

Sample Output

YES

Hint

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;}